Efficient way to create a list of Point. #2117
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Please noteI want to create a list of p = [Point(x, y) for x, y in zip(xq, yq)]
df = pd.DataFrame({'x':xq, 'y':yq, 'coords': p}) df = pd.DataFrame({'x':xq, 'y':yq})
df['coords'] = list(zip(df['x'],df['y']))
df['coords'] = df['coords'].apply(Point) |
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Replies: 2 comments 5 replies
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Assuming you are using Shapely 2.x, you can use the import numpy as np
import shapely
rng = np.random.default_rng(123)
n = 10_000
xq = rng.random(n)
yq = rng.random(n)
pts = shapely.points(xq, yq)and when you get to do the point-in-polygon check, also use vectorized functions like poly = shapely.from_wkt("POLYGON((0 0, 0 0.1, 0.1 0.1, 0 0))")
pt_in_poly = shapely.contains(poly, pts)
# or poly.contains(pts)
pt_in_poly.sum() # 60 |
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Thank you so much, @mwtoews! Just checked, the run time drops significantly (750s to 16s). |
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I am trying to create a "equivalent" https://www.mathworks.com/help/matlab/ref/inpolygon.html def inpolygon(xq, yq, xv, yv):
polygon = Polygon(list(zip(xv, yv)))
if not polygon.is_valid:
polygon = polygon.buffer(0)
pts = shapely.points(xq, yq)
return polygon.touches(pts) | polygon.contains(pts)And this is the result with line_profile. Is there some way to improve it? Thank you so much! |
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You could try:
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Thank you so much, @theroggy. It really helps. For the testcase, originally it (the above
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If it is still taking 11 seconds, this can most likely, depending on your data and/or if your code can be structured a bit differently be further optimised by using a spatial index... |
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Got it. Right, with all your changes, it is close to matlab now. Thank you so much! |
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Assuming you are using Shapely 2.x, you can use the
pointsfunction, which is vectorized to return numpy array types.and when you get to do the point-in-polygon check, also use vectorized functions like
contains: