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<h1 class="chapter" id="sec113">Chapter&#XA0;10&#XA0;&#XA0;Lists</h1>
<p>This chapter presents one of Python&#X2019;s most useful built-in types, lists.
You will also learn more about objects and what can happen when you have
more than one name for the same object.</p>
<h2 class="section" id="sec114">10.1&#XA0;&#XA0;A list is a sequence</h2>
<a id="sequence"></a></p><p>Like a string, a <span class="c010">list</span> is a sequence of values. In a string, the
values are characters; in a list, they can be any type. The values in
a list are called <span class="c010">elements</span> or sometimes <span class="c010">items</span>.
<a id="hevea_default696"></a>
<a id="hevea_default697"></a>
<a id="hevea_default698"></a>
<a id="hevea_default699"></a>
<a id="hevea_default700"></a></p><p>There are several ways to create a new list; the simplest is to
enclose the elements in square brackets (<code>[</code> and <code>]</code>):</p><pre class="verbatim">[10, 20, 30, 40]
['crunchy frog', 'ram bladder', 'lark vomit']
</pre><p>The first example is a list of four integers. The second is a list of
three strings. The elements of a list don&#X2019;t have to be the same type.
The following list contains a string, a float, an integer, and
(lo!) another list:</p><pre class="verbatim">['spam', 2.0, 5, [10, 20]]
</pre><p>A list within another list is <span class="c010">nested</span>.
<a id="hevea_default701"></a>
<a id="hevea_default702"></a></p><p>A list that contains no elements is
called an empty list; you can create one with empty
brackets, <code>[]</code>.
<a id="hevea_default703"></a>
<a id="hevea_default704"></a></p><p>As you might expect, you can assign list values to variables:</p><pre class="verbatim">&gt;&gt;&gt; cheeses = ['Cheddar', 'Edam', 'Gouda']
&gt;&gt;&gt; numbers = [42, 123]
&gt;&gt;&gt; empty = []
&gt;&gt;&gt; print(cheeses, numbers, empty)
['Cheddar', 'Edam', 'Gouda'] [42, 123] []
</pre><p><a id="hevea_default705"></a></p>
<h2 class="section" id="sec115">10.2&#XA0;&#XA0;Lists are mutable</h2>
<a id="mutable"></a>
<a id="hevea_default706"></a>
<a id="hevea_default707"></a>
<a id="hevea_default708"></a>
<a id="hevea_default709"></a>
<a id="hevea_default710"></a></p><p>The syntax for accessing the elements of a list is the same as for
accessing the characters of a string&#X2014;the bracket operator. The
expression inside the brackets specifies the index. Remember that the
indices start at 0:</p><pre class="verbatim">&gt;&gt;&gt; cheeses[0]
</pre><p>Unlike strings, lists are mutable. When the bracket operator appears
on the left side of an assignment, it identifies the element of the
list that will be assigned.
<a id="hevea_default711"></a></p><pre class="verbatim">&gt;&gt;&gt; numbers = [42, 123]
&gt;&gt;&gt; numbers[1] = 5
&gt;&gt;&gt; numbers
</pre><p>The one-eth element of <span class="c004">numbers</span>, which
used to be 123, is now 5.
<a id="hevea_default712"></a>
<a id="hevea_default713"></a></p><p>Figure&#XA0;<a href="thinkpython2011.html#fig.liststate">10.1</a> shows 
the state diagram for <span class="c004">cheeses</span>, <span class="c004">numbers</span> and <span class="c004">empty</span>:
<a id="hevea_default714"></a>
<a id="hevea_default715"></a></p><blockquote class="figure"><div class="center"><hr class="c019"></div>
<div class="center"><img src="thinkpython2011.png"></div>
<div class="caption"><table class="c001 cellpading0"><tr><td class="c018">Figure 10.1: State diagram.</td></tr>
</table></div>
<a id="fig.liststate"></a>
<div class="center"><hr class="c019"></div></blockquote><p>Lists are represented by boxes with the word &#X201C;list&#X201D; outside
and the elements of the list inside. <span class="c004">cheeses</span> refers to
a list with three elements indexed 0, 1 and 2.
<span class="c004">numbers</span> contains two elements; the diagram shows that the
value of the second element has been reassigned from 123 to 5.
<span class="c004">empty</span> refers to a list with no elements.
<a id="hevea_default716"></a>
<a id="hevea_default717"></a>
<a id="hevea_default718"></a></p><p>List indices work the same way as string indices:</p><ul class="itemize"><li class="li-itemize">Any integer expression can be used as an index.</li><li class="li-itemize">If you try to read or write an element that does not exist, you
get an <span class="c004">IndexError</span>.
<a id="hevea_default719"></a>
<a id="hevea_default720"></a></li><li class="li-itemize">If an index has a negative value, it counts backward from the
end of the list.</li></ul><p>
<a id="hevea_default721"></a></p><p><a id="hevea_default722"></a>
<a id="hevea_default723"></a>
<a id="hevea_default724"></a>
<a id="hevea_default725"></a></p><p>The <span class="c004">in</span> operator also works on lists.</p><pre class="verbatim">&gt;&gt;&gt; cheeses = ['Cheddar', 'Edam', 'Gouda']
&gt;&gt;&gt; 'Edam' in cheeses
&gt;&gt;&gt; 'Brie' in cheeses
<h2 class="section" id="sec116">10.3&#XA0;&#XA0;Traversing a list</h2>
<a id="hevea_default726"></a>
<a id="hevea_default727"></a>
<a id="hevea_default728"></a>
<a id="hevea_default729"></a>
<a id="hevea_default730"></a></p><p>The most common way to traverse the elements of a list is
with a <span class="c004">for</span> loop. The syntax is the same as for strings:</p><pre class="verbatim">for cheese in cheeses:
    print(cheese)
</pre><p>This works well if you only need to read the elements of the
list. But if you want to write or update the elements, you
need the indices. A common way to do that is to combine
the built-in functions <span class="c004">range</span> and <span class="c004">len</span>:
<a id="hevea_default731"></a>
<a id="hevea_default732"></a></p><pre class="verbatim">for i in range(len(numbers)):
    numbers[i] = numbers[i] * 2
</pre><p>This loop traverses the list and updates each element. <span class="c004">len</span>
returns the number of elements in the list. <span class="c004">range</span> returns
a list of indices from 0 to <span class="c009">n</span>&#X2212;1, where <span class="c009">n</span> is the length of
the list. Each time through the loop <span class="c004">i</span> gets the index
of the next element. The assignment statement in the body uses
<span class="c004">i</span> to read the old value of the element and to assign the
<a id="hevea_default733"></a>
<a id="hevea_default734"></a></p><p>A <span class="c004">for</span> loop over an empty list never runs the body:</p><pre class="verbatim">for x in []:
    print('This never happens.')
</pre><p>Although a list can contain another list, the nested
list still counts as a single element. The length of this list is
<a id="hevea_default735"></a>
<a id="hevea_default736"></a></p><pre class="verbatim">['spam', 1, ['Brie', 'Roquefort', 'Pol le Veq'], [1, 2, 3]]
<h2 class="section" id="sec117">10.4&#XA0;&#XA0;List operations</h2>
<a id="hevea_default737"></a></p><p>The <span class="c004">+</span> operator concatenates lists:
<a id="hevea_default738"></a>
<a id="hevea_default739"></a></p><pre class="verbatim">&gt;&gt;&gt; a = [1, 2, 3]
&gt;&gt;&gt; b = [4, 5, 6]
&gt;&gt;&gt; c = a + b
&gt;&gt;&gt; c
[1, 2, 3, 4, 5, 6]
</pre><p>The <span class="c004">*</span> operator repeats a list a given number of times:
<a id="hevea_default740"></a>
<a id="hevea_default741"></a></p><pre class="verbatim">&gt;&gt;&gt; [0] * 4
[0, 0, 0, 0]
&gt;&gt;&gt; [1, 2, 3] * 3
[1, 2, 3, 1, 2, 3, 1, 2, 3]
</pre><p>The first example repeats <span class="c004">[0]</span> four times. The second example
repeats the list <span class="c004">[1, 2, 3]</span> three times.</p>
<h2 class="section" id="sec118">10.5&#XA0;&#XA0;List slices</h2>
<a id="hevea_default742"></a>
<a id="hevea_default743"></a>
<a id="hevea_default744"></a>
<a id="hevea_default745"></a>
<a id="hevea_default746"></a></p><p>The slice operator also works on lists:</p><pre class="verbatim">&gt;&gt;&gt; t = ['a', 'b', 'c', 'd', 'e', 'f']
&gt;&gt;&gt; t[1:3]
&gt;&gt;&gt; t[:4]
['a', 'b', 'c', 'd']
&gt;&gt;&gt; t[3:]
['d', 'e', 'f']
</pre><p>If you omit the first index, the slice starts at the beginning.
If you omit the second, the slice goes to the end. So if you
omit both, the slice is a copy of the whole list.
<a id="hevea_default747"></a>
<a id="hevea_default748"></a>
<a id="hevea_default749"></a></p><pre class="verbatim">&gt;&gt;&gt; t[:]
['a', 'b', 'c', 'd', 'e', 'f']
</pre><p>Since lists are mutable, it is often useful to make a copy
before performing operations that modify lists.
<a id="hevea_default750"></a></p><p>A slice operator on the left side of an assignment
can update multiple elements:
<a id="hevea_default751"></a>
<a id="hevea_default752"></a></p><pre class="verbatim">&gt;&gt;&gt; t = ['a', 'b', 'c', 'd', 'e', 'f']
&gt;&gt;&gt; t[1:3] = ['x', 'y']
&gt;&gt;&gt; t
['a', 'x', 'y', 'd', 'e', 'f']
<h2 class="section" id="sec119">10.6&#XA0;&#XA0;List methods</h2>
<a id="hevea_default753"></a>
<a id="hevea_default754"></a></p><p>Python provides methods that operate on lists. For example,
<span class="c004">append</span> adds a new element to the end of a list:
<a id="hevea_default755"></a>
<a id="hevea_default756"></a></p><pre class="verbatim">&gt;&gt;&gt; t = ['a', 'b', 'c']
&gt;&gt;&gt; t.append('d')
&gt;&gt;&gt; t
['a', 'b', 'c', 'd']
</pre><p><span class="c004">extend</span> takes a list as an argument and appends all of
the elements:
<a id="hevea_default757"></a>
<a id="hevea_default758"></a></p><pre class="verbatim">&gt;&gt;&gt; t1 = ['a', 'b', 'c']
&gt;&gt;&gt; t2 = ['d', 'e']
&gt;&gt;&gt; t1.extend(t2)
&gt;&gt;&gt; t1
['a', 'b', 'c', 'd', 'e']
</pre><p>This example leaves <span class="c004">t2</span> unmodified.</p><p><span class="c004">sort</span> arranges the elements of the list from low to high:
<a id="hevea_default759"></a>
<a id="hevea_default760"></a></p><pre class="verbatim">&gt;&gt;&gt; t = ['d', 'c', 'e', 'b', 'a']
&gt;&gt;&gt; t.sort()
&gt;&gt;&gt; t
['a', 'b', 'c', 'd', 'e']
</pre><p>Most list methods are void; they modify the list and return <span class="c004">None</span>.
If you accidentally write <span class="c004">t = t.sort()</span>, you will be disappointed
with the result.
<a id="hevea_default761"></a>
<a id="hevea_default762"></a>
<a id="hevea_default763"></a>
<a id="hevea_default764"></a></p>
<h2 class="section" id="sec120">10.7&#XA0;&#XA0;Map, filter and reduce</h2>
<a id="filter"></a></p><p>To add up all the numbers in a list, you can use a loop like this:</p><pre class="verbatim">def add_all(t):
    total = 0
    for x in t:
        total += x
    return total
</pre><p><span class="c004">total</span> is initialized to 0. Each time through the loop,
<span class="c004">x</span> gets one element from the list. The <span class="c004">+=</span> operator
provides a short way to update a variable. This 
<span class="c010">augmented assignment statement</span>,
<a id="hevea_default765"></a>
<a id="hevea_default766"></a>
<a id="hevea_default767"></a>
<a id="hevea_default768"></a></p><pre class="verbatim">    total += x
</pre><p>is equivalent to</p><pre class="verbatim">    total = total + x
</pre><p>As the loop runs, <span class="c004">total</span> accumulates the sum of the
elements; a variable used this way is sometimes called an
<span class="c010">accumulator</span>.
<a id="hevea_default769"></a></p><p>Adding up the elements of a list is such a common operation
that Python provides it as a built-in function, <span class="c004">sum</span>:</p><pre class="verbatim">&gt;&gt;&gt; t = [1, 2, 3]
&gt;&gt;&gt; sum(t)
</pre><p>An operation like this that combines a sequence of elements into
a single value is sometimes called <span class="c010">reduce</span>.
<a id="hevea_default770"></a>
<a id="hevea_default771"></a>
<a id="hevea_default772"></a></p><p>Sometimes you want to traverse one list while building
another. For example, the following function takes a list of strings
and returns a new list that contains capitalized strings:</p><pre class="verbatim">def capitalize_all(t):
    res = []
    for s in t:
        res.append(s.capitalize())
    return res
</pre><p><span class="c004">res</span> is initialized with an empty list; each time through
the loop, we append the next element. So <span class="c004">res</span> is another
kind of accumulator.
<a id="hevea_default773"></a></p><p>An operation like <code>capitalize_all</code> is sometimes called a <span class="c010">map</span> because it &#X201C;maps&#X201D; a function (in this case the method <span class="c004">capitalize</span>) onto each of the elements in a sequence.
<a id="hevea_default774"></a>
<a id="hevea_default775"></a>
<a id="hevea_default776"></a>
<a id="hevea_default777"></a></p><p>Another common operation is to select some of the elements from
a list and return a sublist. For example, the following
function takes a list of strings and returns a list that contains
only the uppercase strings:</p><pre class="verbatim">def only_upper(t):
    res = []
    for s in t:
        if s.isupper():
            res.append(s)
    return res
</pre><p><span class="c004">isupper</span> is a string method that returns <span class="c004">True</span> if
the string contains only upper case letters.</p><p>An operation like <code>only_upper</code> is called a <span class="c010">filter</span> because
it selects some of the elements and filters out the others.</p><p>Most common list operations can be expressed as a combination
of map, filter and reduce.</p>
<h2 class="section" id="sec121">10.8&#XA0;&#XA0;Deleting elements</h2>
<a id="hevea_default778"></a>
<a id="hevea_default779"></a></p><p>There are several ways to delete elements from a list. If you
know the index of the element you want, you can use
<span class="c004">pop</span>:
<a id="hevea_default780"></a>
<a id="hevea_default781"></a></p><pre class="verbatim">&gt;&gt;&gt; t = ['a', 'b', 'c']
&gt;&gt;&gt; x = t.pop(1)
&gt;&gt;&gt; t
&gt;&gt;&gt; x
</pre><p><span class="c004">pop</span> modifies the list and returns the element that was removed.
If you don&#X2019;t provide an index, it deletes and returns the
last element.</p><p>If you don&#X2019;t need the removed value, you can use the <span class="c004">del</span>
<a id="hevea_default782"></a>
<a id="hevea_default783"></a></p><pre class="verbatim">&gt;&gt;&gt; t = ['a', 'b', 'c']
&gt;&gt;&gt; del t[1]
&gt;&gt;&gt; t
</pre><p>If you know the element you want to remove (but not the index), you
can use <span class="c004">remove</span>:
<a id="hevea_default784"></a>
<a id="hevea_default785"></a></p><pre class="verbatim">&gt;&gt;&gt; t = ['a', 'b', 'c']
&gt;&gt;&gt; t.remove('b')
&gt;&gt;&gt; t
</pre><p>The return value from <span class="c004">remove</span> is <span class="c004">None</span>.
<a id="hevea_default786"></a>
<a id="hevea_default787"></a></p><p>To remove more than one element, you can use <span class="c004">del</span> with
a slice index:</p><pre class="verbatim">&gt;&gt;&gt; t = ['a', 'b', 'c', 'd', 'e', 'f']
&gt;&gt;&gt; del t[1:5]
&gt;&gt;&gt; t
</pre><p>As usual, the slice selects all the elements up to but not
including the second index.</p>
<h2 class="section" id="sec122">10.9&#XA0;&#XA0;Lists and strings</h2>
<a id="hevea_default788"></a>
<a id="hevea_default789"></a>
<a id="hevea_default790"></a></p><p>A string is a sequence of characters and a list is a sequence
of values, but a list of characters is not the same as a
string. To convert from a string to a list of characters,
you can use <span class="c004">list</span>:
<a id="hevea_default791"></a>
<a id="hevea_default792"></a></p><pre class="verbatim">&gt;&gt;&gt; s = 'spam'
&gt;&gt;&gt; t = list(s)
&gt;&gt;&gt; t
['s', 'p', 'a', 'm']
</pre><p>Because <span class="c004">list</span> is the name of a built-in function, you should
avoid using it as a variable name. I also avoid <span class="c004">l</span> because
it looks too much like <span class="c004">1</span>. So that&#X2019;s why I use <span class="c004">t</span>.</p><p>The <span class="c004">list</span> function breaks a string into individual letters. If
you want to break a string into words, you can use the <span class="c004">split</span>
<a id="hevea_default793"></a>
<a id="hevea_default794"></a></p><pre class="verbatim">&gt;&gt;&gt; s = 'pining for the fjords'
&gt;&gt;&gt; t = s.split()
&gt;&gt;&gt; t
['pining', 'for', 'the', 'fjords']
</pre><p>An optional argument called a <span class="c010">delimiter</span> specifies which
characters to use as word boundaries.
The following example
uses a hyphen as a delimiter:
<a id="hevea_default795"></a>
<a id="hevea_default796"></a>
<a id="hevea_default797"></a></p><pre class="verbatim">&gt;&gt;&gt; s = 'spam-spam-spam'
&gt;&gt;&gt; delimiter = '-'
&gt;&gt;&gt; t = s.split(delimiter)
&gt;&gt;&gt; t
['spam', 'spam', 'spam']
</pre><p><span class="c004">join</span> is the inverse of <span class="c004">split</span>. It
takes a list of strings and
concatenates the elements. <span class="c004">join</span> is a string method,
so you have to invoke it on the delimiter and pass the
list as a parameter:
<a id="hevea_default798"></a>
<a id="hevea_default799"></a>
<a id="hevea_default800"></a></p><pre class="verbatim">&gt;&gt;&gt; t = ['pining', 'for', 'the', 'fjords']
&gt;&gt;&gt; delimiter = ' '
&gt;&gt;&gt; s = delimiter.join(t)
&gt;&gt;&gt; s
'pining for the fjords'
</pre><p>In this case the delimiter is a space character, so
<span class="c004">join</span> puts a space between words. To concatenate
strings without spaces, you can use the empty string,
<code>''</code>, as a delimiter. 
<a id="hevea_default801"></a>
<a id="hevea_default802"></a></p>
<h2 class="section" id="sec123">10.10&#XA0;&#XA0;Objects and values</h2>
<a id="equivalence"></a>
<a id="hevea_default803"></a>
<a id="hevea_default804"></a></p><p>If we run these assignment statements:</p><pre class="verbatim">a = 'banana'
b = 'banana'
</pre><p>We know that <span class="c004">a</span> and <span class="c004">b</span> both refer to a
string, but we don&#X2019;t
know whether they refer to the <em>same</em> string.
There are two possible states, shown in Figure&#XA0;<a href="thinkpython2011.html#fig.list1">10.2</a>.
<a id="hevea_default805"></a></p><blockquote class="figure"><div class="center"><hr class="c019"></div>
<div class="center"><img src="thinkpython2012.png"></div>
<div class="caption"><table class="c001 cellpading0"><tr><td class="c018">Figure 10.2: State diagram.</td></tr>
</table></div>
<a id="fig.list1"></a>
<div class="center"><hr class="c019"></div></blockquote><p>In one case, <span class="c004">a</span> and <span class="c004">b</span> refer to two different objects that
have the same value. In the second case, they refer to the same
<a id="hevea_default806"></a>
<a id="hevea_default807"></a></p><p>To check whether two variables refer to the same object, you can
use the <span class="c004">is</span> operator.</p><pre class="verbatim">&gt;&gt;&gt; a = 'banana'
&gt;&gt;&gt; b = 'banana'
&gt;&gt;&gt; a is b
</pre><p>In this example, Python only created one string object, and both <span class="c004">a</span> and <span class="c004">b</span> refer to it. But when you create two lists, you get
two objects:</p><pre class="verbatim">&gt;&gt;&gt; a = [1, 2, 3]
&gt;&gt;&gt; b = [1, 2, 3]
&gt;&gt;&gt; a is b
</pre><p>So the state diagram looks like Figure&#XA0;<a href="thinkpython2011.html#fig.list2">10.3</a>.
<a id="hevea_default808"></a>
<a id="hevea_default809"></a></p><blockquote class="figure"><div class="center"><hr class="c019"></div>
<div class="center"><img src="thinkpython2013.png"></div>
<div class="caption"><table class="c001 cellpading0"><tr><td class="c018">Figure 10.3: State diagram.</td></tr>
</table></div>
<a id="fig.list2"></a>
<div class="center"><hr class="c019"></div></blockquote><p>In this case we would say that the two lists are <span class="c010">equivalent</span>,
because they have the same elements, but not <span class="c010">identical</span>, because
they are not the same object. If two objects are identical, they are
also equivalent, but if they are equivalent, they are not necessarily
<a id="hevea_default810"></a>
<a id="hevea_default811"></a></p><p>Until now, we have been using &#X201C;object&#X201D; and &#X201C;value&#X201D;
interchangeably, but it is more precise to say that an object has a
value. If you evaluate <span class="c004">[1, 2, 3]</span>, you get a list
object whose value is a sequence of integers. If another
list has the same elements, we say it has the same value, but
it is not the same object.
<a id="hevea_default812"></a>
<a id="hevea_default813"></a></p>
<h2 class="section" id="sec124">10.11&#XA0;&#XA0;Aliasing</h2>
<a id="hevea_default814"></a>
<a id="hevea_default815"></a></p><p>If <span class="c004">a</span> refers to an object and you assign <span class="c004">b = a</span>,
then both variables refer to the same object:</p><pre class="verbatim">&gt;&gt;&gt; a = [1, 2, 3]
&gt;&gt;&gt; b = a
&gt;&gt;&gt; b is a
</pre><p>The state diagram looks like Figure&#XA0;<a href="thinkpython2011.html#fig.list3">10.4</a>.
<a id="hevea_default816"></a>
<a id="hevea_default817"></a></p><blockquote class="figure"><div class="center"><hr class="c019"></div>
<div class="center"><img src="thinkpython2014.png"></div>
<div class="caption"><table class="c001 cellpading0"><tr><td class="c018">Figure 10.4: State diagram.</td></tr>
</table></div>
<a id="fig.list3"></a>
<div class="center"><hr class="c019"></div></blockquote><p>The association of a variable with an object is called a <span class="c010">reference</span>. In this example, there are two references to the same
<a id="hevea_default818"></a></p><p>An object with more than one reference has more
than one name, so we say that the object is <span class="c010">aliased</span>.
<a id="hevea_default819"></a></p><p>If the aliased object is mutable, changes made with one alias affect
the other:</p><pre class="verbatim">&gt;&gt;&gt; b[0] = 42
&gt;&gt;&gt; a
</pre><p>Although this behavior can be useful, it is error-prone. In general,
it is safer to avoid aliasing when you are working with mutable
<a id="hevea_default820"></a></p><p>For immutable objects like strings, aliasing is not as much of a
problem. In this example:</p><pre class="verbatim">a = 'banana'
b = 'banana'
</pre><p>It almost never makes a difference whether <span class="c004">a</span> and <span class="c004">b</span> refer
to the same string or not.</p>
<h2 class="section" id="sec125">10.12&#XA0;&#XA0;List arguments</h2>
<a id="list.arguments"></a>
<a id="hevea_default821"></a>
<a id="hevea_default822"></a>
<a id="hevea_default823"></a>
<a id="hevea_default824"></a>
<a id="hevea_default825"></a></p><p>When you pass a list to a function, the function gets a reference to
the list. If the function modifies the list, the caller sees
the change. For example, <code>delete_head</code> removes the first element
from a list:</p><pre class="verbatim">def delete_head(t):
    del t[0]
</pre><p>Here&#X2019;s how it is used:</p><pre class="verbatim">&gt;&gt;&gt; letters = ['a', 'b', 'c']
&gt;&gt;&gt; delete_head(letters)
&gt;&gt;&gt; letters
</pre><p>The parameter <span class="c004">t</span> and the variable <span class="c004">letters</span> are
aliases for the same object. The stack diagram looks like
Figure&#XA0;<a href="thinkpython2011.html#fig.stack5">10.5</a>.
<a id="hevea_default826"></a>
<a id="hevea_default827"></a></p><blockquote class="figure"><div class="center"><hr class="c019"></div>
<div class="center"><img src="thinkpython2015.png"></div>
<div class="caption"><table class="c001 cellpading0"><tr><td class="c018">Figure 10.5: Stack diagram.</td></tr>
</table></div>
<a id="fig.stack5"></a>
<div class="center"><hr class="c019"></div></blockquote><p>Since the list is shared by two frames, I drew
it between them.</p><p>It is important to distinguish between operations that
modify lists and operations that create new lists. For
example, the <span class="c004">append</span> method modifies a list, but the
<span class="c004">+</span> operator creates a new list.
<a id="hevea_default828"></a>
<a id="hevea_default829"></a>
<a id="hevea_default830"></a>
<a id="hevea_default831"></a></p><p>Here&#X2019;s an example using <span class="c004">append</span>:
</p><pre class="verbatim">&gt;&gt;&gt; t1 = [1, 2]
&gt;&gt;&gt; t2 = t1.append(3)
&gt;&gt;&gt; t1
&gt;&gt;&gt; t2
</pre><p>The return value from <span class="c004">append</span> is <span class="c004">None</span>.</p><p>Here&#X2019;s an example using the <span class="c004">+</span> operator:
</p><pre class="verbatim">&gt;&gt;&gt; t3 = t1 + [4]
&gt;&gt;&gt; t1
&gt;&gt;&gt; t3
[1, 2, 3, 4]
</pre><p>The result of the operator is a new list, and the original list is
unchanged.</p><p>This difference is important when you write functions that
are supposed to modify lists. For example, this function
<em>does not</em> delete the head of a list:
</p><pre class="verbatim">def bad_delete_head(t):
    t = t[1:]              # WRONG!
</pre><p>The slice operator creates a new list and the assignment
makes <span class="c004">t</span> refer to it, but that doesn&#X2019;t affect the caller.
<a id="hevea_default832"></a>
<a id="hevea_default833"></a>
</p><pre class="verbatim">&gt;&gt;&gt; t4 = [1, 2, 3]
&gt;&gt;&gt; bad_delete_head(t4)
&gt;&gt;&gt; t4
</pre><p>At the beginning of <code>bad_delete_head</code>, <span class="c004">t</span> and <span class="c004">t4</span>
refer to the same list. At the end, <span class="c004">t</span> refers to a new list,
but <span class="c004">t4</span> still refers to the original, unmodified list.</p><p>An alternative is to write a function that creates and
returns a new list. For
example, <span class="c004">tail</span> returns all but the first
element of a list:</p><pre class="verbatim">def tail(t):
    return t[1:]
</pre><p>This function leaves the original list unmodified.
Here&#X2019;s how it is used:</p><pre class="verbatim">&gt;&gt;&gt; letters = ['a', 'b', 'c']
&gt;&gt;&gt; rest = tail(letters)
&gt;&gt;&gt; rest
<h2 class="section" id="sec126">10.13&#XA0;&#XA0;Debugging</h2>
<a id="hevea_default834"></a></p><p>Careless use of lists (and other mutable objects)
can lead to long hours of debugging. Here are some common
pitfalls and ways to avoid them:</p><ol class="enumerate" type=1><li class="li-enumerate">Most list methods modify the argument and
return <span class="c004">None</span>. This is the opposite of the string methods,
which return a new string and leave the original alone.<p>If you are used to writing string code like this:</p><pre class="verbatim">word = word.strip()
</pre><p>It is tempting to write list code like this:</p><pre class="verbatim">t = t.sort()           # WRONG!
</pre><p><a id="hevea_default835"></a>
<a id="hevea_default836"></a></p><p>Because <span class="c004">sort</span> returns <span class="c004">None</span>, the
next operation you perform with <span class="c004">t</span> is likely to fail.</p><p>Before using list methods and operators, you should read the
documentation carefully and then test them in interactive mode.</p></li><li class="li-enumerate">Pick an idiom and stick with it.<p>Part of the problem with lists is that there are too many
ways to do things. For example, to remove an element from
a list, you can use <span class="c004">pop</span>, <span class="c004">remove</span>, <span class="c004">del</span>,
or even a slice assignment.</p><p>To add an element, you can use the <span class="c004">append</span> method or
the <span class="c004">+</span> operator. Assuming that <span class="c004">t</span> is a list and
<span class="c004">x</span> is a list element, these are correct: </p><pre class="verbatim">t.append(x)
t = t + [x]
</pre><p>And these are wrong:</p><pre class="verbatim">t.append([x])          # WRONG!
t = t.append(x)        # WRONG!
t + [x]                # WRONG!
t = t + x              # WRONG!
</pre><p>Try out each of these examples in interactive mode to make sure
you understand what they do. Notice that only the last
one causes a runtime error; the other three are legal, but they
do the wrong thing.</p></li><li class="li-enumerate">Make copies to avoid aliasing.
<a id="hevea_default837"></a>
<a id="hevea_default838"></a><p>If you want to use a method like <span class="c004">sort</span> that modifies
the argument, but you need to keep the original list as
well, you can make a copy.</p><pre class="verbatim">&gt;&gt;&gt; t = [3, 1, 2]
&gt;&gt;&gt; t2 = t[:]
&gt;&gt;&gt; t2.sort()
&gt;&gt;&gt; t
&gt;&gt;&gt; t2
</pre><p>In this example you could also use the built-in function <span class="c004">sorted</span>,
which returns a new, sorted list and leaves the original alone.
<a id="hevea_default839"></a>
<a id="hevea_default840"></a></p><pre class="verbatim">&gt;&gt;&gt; t2 = sorted(t)
&gt;&gt;&gt; t
&gt;&gt;&gt; t2
</pre></li></ol>
<h2 class="section" id="sec127">10.14&#XA0;&#XA0;Glossary</h2>
<dl class="description"><dt class="dt-description"><span class="c010">list:</span></dt><dd class="dd-description"> A sequence of values.
<a id="hevea_default841"></a></dd><dt class="dt-description"><span class="c010">element:</span></dt><dd class="dd-description"> One of the values in a list (or other sequence),
also called items.
<a id="hevea_default842"></a></dd><dt class="dt-description"><span class="c010">nested list:</span></dt><dd class="dd-description"> A list that is an element of another list.
<a id="hevea_default843"></a></dd><dt class="dt-description"><span class="c010">accumulator:</span></dt><dd class="dd-description"> A variable used in a loop to add up or
accumulate a result.
<a id="hevea_default844"></a></dd><dt class="dt-description"><span class="c010">augmented assignment:</span></dt><dd class="dd-description"> A statement that updates the value
of a variable using an operator like <code>+=</code>.
<a id="hevea_default845"></a>
<a id="hevea_default846"></a>
<a id="hevea_default847"></a></dd><dt class="dt-description"><span class="c010">reduce:</span></dt><dd class="dd-description"> A processing pattern that traverses a sequence 
and accumulates the elements into a single result.
<a id="hevea_default848"></a>
<a id="hevea_default849"></a></dd><dt class="dt-description"><span class="c010">map:</span></dt><dd class="dd-description"> A processing pattern that traverses a sequence and
performs an operation on each element.
<a id="hevea_default850"></a>
<a id="hevea_default851"></a></dd><dt class="dt-description"><span class="c010">filter:</span></dt><dd class="dd-description"> A processing pattern that traverses a list and
selects the elements that satisfy some criterion.
<a id="hevea_default852"></a>
<a id="hevea_default853"></a></dd><dt class="dt-description"><span class="c010">object:</span></dt><dd class="dd-description"> Something a variable can refer to. An object
has a type and a value.
<a id="hevea_default854"></a></dd><dt class="dt-description"><span class="c010">equivalent:</span></dt><dd class="dd-description"> Having the same value.
<a id="hevea_default855"></a></dd><dt class="dt-description"><span class="c010">identical:</span></dt><dd class="dd-description"> Being the same object (which implies equivalence).
<a id="hevea_default856"></a></dd><dt class="dt-description"><span class="c010">reference:</span></dt><dd class="dd-description"> The association between a variable and its value.
<a id="hevea_default857"></a></dd><dt class="dt-description"><span class="c010">aliasing:</span></dt><dd class="dd-description"> A circumstance where two or more variables refer to the same
<a id="hevea_default858"></a></dd><dt class="dt-description"><span class="c010">delimiter:</span></dt><dd class="dd-description"> A character or string used to indicate where a
string should be split.
<a id="hevea_default859"></a></dd></dl>
<h2 class="section" id="sec128">10.15&#XA0;&#XA0;Exercises</h2>
<p>You can download solutions to these exercises from
<a href="http://thinkpython2.com/code/list_exercises.py"><span class="c004">http://thinkpython2.com/code/list_exercises.py</span></a>.</p><div class="theorem"><span class="c010">Exercise&#XA0;1</span>&#XA0;&#XA0;<p><em>Write a function called <code>nested_sum</code> that takes a list of lists
of integers and adds up the elements from all of the nested lists.
For example:</em></p><pre class="verbatim"><em>&gt;&gt;&gt; t = [[1, 2], [3], [4, 5, 6]]
&gt;&gt;&gt; nested_sum(t)
</em></pre></div><div class="theorem"><span class="c010">Exercise&#XA0;2</span>&#XA0;&#XA0;
<a id="cumulative"></a>
<a id="hevea_default860"></a><p><em>Write a function called <span class="c004">cumsum</span> that takes a list of numbers and
returns the cumulative sum; that is, a new list where the </em><span class="c009">i</span><em>th
element is the sum of the first </em><span class="c009">i</span>+1<em> elements from the original list.
For example:</em></p><pre class="verbatim"><em>&gt;&gt;&gt; t = [1, 2, 3]
&gt;&gt;&gt; cumsum(t)
</em></pre></div><div class="theorem"><span class="c010">Exercise&#XA0;3</span>&#XA0;&#XA0;<p><em>Write a function called <code>middle</code> that takes a list and
returns a new list that contains all but the first and last
elements. For example:</em></p><pre class="verbatim"><em>&gt;&gt;&gt; t = [1, 2, 3, 4]
&gt;&gt;&gt; middle(t)
</em></pre></div><div class="theorem"><span class="c010">Exercise&#XA0;4</span>&#XA0;&#XA0;<p><em>Write a function called <code>chop</code> that takes a list, modifies it
by removing the first and last elements, and returns <span class="c004">None</span>.
For example:</em></p><pre class="verbatim"><em>&gt;&gt;&gt; t = [1, 2, 3, 4]
&gt;&gt;&gt; chop(t)
&gt;&gt;&gt; t
</em></pre></div><div class="theorem"><span class="c010">Exercise&#XA0;5</span>&#XA0;&#XA0;<em>
Write a function called <code>is_sorted</code> that takes a list as a
parameter and returns <span class="c004">True</span> if the list is sorted in ascending
order and <span class="c004">False</span> otherwise. For example:</em><pre class="verbatim"><em>&gt;&gt;&gt; is_sorted([1, 2, 2])
&gt;&gt;&gt; is_sorted(['b', 'a'])
</em></pre></div><div class="theorem"><span class="c010">Exercise&#XA0;6</span>&#XA0;&#XA0;
<a id="anagram"></a>
<a id="hevea_default861"></a><p><em>Two words are anagrams if you can rearrange the letters from one
to spell the other. Write a function called <code>is_anagram</code>
that takes two strings and returns <span class="c004">True</span> if they are anagrams.
</em></p></div><div class="theorem"><span class="c010">Exercise&#XA0;7</span>&#XA0;&#XA0;
<a id="duplicate"></a>
<a id="hevea_default862"></a>
<a id="hevea_default863"></a><p><em>Write a function called <code>has_duplicates</code> that takes
a list and returns <span class="c004">True</span> if there is any element that
appears more than once. It should not modify the original
list.</em></p></div><div class="theorem"><span class="c010">Exercise&#XA0;8</span>&#XA0;&#XA0;<p><em>This exercise pertains to the so-called Birthday Paradox, which you
can read about at </em><a href="http://en.wikipedia.org/wiki/Birthday_paradox"><em><span class="c004">http://en.wikipedia.org/wiki/Birthday_paradox</span></em></a><em>.
</em><a id="hevea_default864"></a></p><p><em>If there are 23 students in your class, what are the chances
that two of you have the same birthday? You can estimate this
probability by generating random samples of 23 birthdays
and checking for matches. Hint: you can generate random birthdays
with the <span class="c004">randint</span> function in the <span class="c004">random</span> module.
</em><a id="hevea_default865"></a>
<a id="hevea_default866"></a>
<a id="hevea_default867"></a>
<a id="hevea_default868"></a></p><p><em>You can download my
solution from </em><a href="http://thinkpython2.com/code/birthday.py"><em><span class="c004">http://thinkpython2.com/code/birthday.py</span></em></a><em>.</em></p></div><div class="theorem"><span class="c010">Exercise&#XA0;9</span>&#XA0;&#XA0;
<a id="hevea_default869"></a>
<a id="hevea_default870"></a>
<a id="hevea_default871"></a>
<a id="hevea_default872"></a><p><em>Write a function that reads the file <span class="c004">words.txt</span> and builds
a list with one element per word. Write two versions of
this function, one using the <span class="c004">append</span> method and the
other using the idiom <span class="c004">t = t + [x]</span>. Which one takes
longer to run? Why?</em></p><p><em>Solution: </em><a href="http://thinkpython2.com/code/wordlist.py"><span class="c004"><em>http://thinkpython2.com/code/wordlist.py</em></span></a><em>.
</em><a id="hevea_default873"></a>
<a id="hevea_default874"></a></p></div><div class="theorem"><span class="c010">Exercise&#XA0;10</span>&#XA0;&#XA0;
<a id="wordlist1"></a>
<a id="bisection"></a>
<a id="hevea_default875"></a>
<a id="hevea_default876"></a>
<a id="hevea_default877"></a>
<a id="hevea_default878"></a>
<a id="hevea_default879"></a>
<a id="hevea_default880"></a><p><em>To check whether a word is in the word list, you could use
the <span class="c004">in</span> operator, but it would be slow because it searches
through the words in order.</em></p><p><em>Because the words are in alphabetical order, we can speed things up
with a bisection search (also known as binary search), which is
similar to what you do when you look a word up in the dictionary. You
start in the middle and check to see whether the word you are looking
for comes before the word in the middle of the list. If so, you
search the first half of the list the same way. Otherwise you search
the second half.</em></p><p><em>Either way, you cut the remaining search space in half. If the
word list has 113,809 words, it will take about 17 steps to
find the word or conclude that it&#X2019;s not there.</em></p><p><em>Write a function called <code>in_bisect</code> that takes a sorted list
and a target value and returns the index of the value
in the list if it&#X2019;s there, or <span class="c004">None</span> if it&#X2019;s not.
</em><a id="hevea_default881"></a>
<a id="hevea_default882"></a></p><p><em>Or you could read the documentation of the <span class="c004">bisect</span> module
and use that! Solution: </em><a href="http://thinkpython2.com/code/inlist.py"><span class="c004"><em>http://thinkpython2.com/code/inlist.py</em></span></a><em>.</em></p></div><div class="theorem"><span class="c010">Exercise&#XA0;11</span>&#XA0;&#XA0;
<a id="hevea_default883"></a><p><em>Two words are a &#X201C;reverse pair&#X201D; if each is the reverse of the
other. Write a program that finds all the reverse pairs in the
word list. Solution: </em><a href="http://thinkpython2.com/code/reverse_pair.py"><em><span class="c004">http://thinkpython2.com/code/reverse_pair.py</span></em></a><em>.</em></p></div><div class="theorem"><span class="c010">Exercise&#XA0;12</span>&#XA0;&#XA0;
<a id="hevea_default884"></a><p><em>Two words &#X201C;interlock&#X201D; if taking alternating letters from each forms
a new word. For example, &#X201C;shoe&#X201D; and &#X201C;cold&#X201D;
interlock to form &#X201C;schooled&#X201D;.
Solution: </em><a href="http://thinkpython2.com/code/interlock.py"><em><span class="c004">http://thinkpython2.com/code/interlock.py</span></em></a><em>.
Credit: This exercise is inspired by an example at </em><a href="http://puzzlers.org"><em><span class="c004">http://puzzlers.org</span></em></a><em>.</em></p><ol class="enumerate" type=1><li class="li-enumerate"><em>Write a program that finds all pairs of words that interlock.
Hint: don&#X2019;t enumerate all pairs!</em></li><li class="li-enumerate"><em>Can you find any words that are three-way interlocked; that is,
every third letter forms a word, starting from the first, second or
third?</em></li></ol></div>
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