Description
According to Theming and templates in Python, a figure can be created using a predefined template or updated to match it. However, the way updates happen is confusing if not inconsistent. Consider the examples below for plotly[express] == 6.2.0.
Plotly Express
import plotly.express as px
df = px.data.iris()
# uses the default `plotly` template
fig = px.scatter(df, x="sepal_width", y="sepal_length", color="petal_length", title="Figure 1")
fig.show()
# update to use another template, note that the colour scale is unchanged
fig.update_layout(template="seaborn", title="Figure 2")
fig.show()
# this is not equivalent to Figure 2
fig = px.scatter(df, x="sepal_width", y="sepal_length", color="petal_length", template="seaborn", title="Figure 3")
fig.show()So Figure 3 is not the same as Figure 2, at least because the colour scales don't match. For custom themes, I can imagine that there may be other components of the figure object that would differ. As far as I know, there is no way to update traces using a template, e.g., fig.update_traces(template="seaborn"), to adjust non-layout settings.
Plotly Graph Object
To make things more confusing, there is a Specifying themes in graph object figures example in the documentation that does the same, but it actually works as expected. Adjusting the example we get:
import plotly.express as px
import plotly.graph_objects as go
df = px.data.iris()
# uses the default `plotly` template
fig = go.Figure(
data=go.Scatter(
x=df["sepal_width"],
y=df["sepal_length"],
mode="markers",
marker={
"color": df["petal_length"],
"showscale": True,
},
)
)
fig.update_layout(title="Figure 4")
fig.show()
# update to use another template, note that the colour scale has changed as expected
fig.update_layout(template="seaborn", title="Figure 5")
fig.show()However, specifying an explicit colorscale will prevent a template update from setting its own colour scheme:
# same as Figure 4 but with an explicit `colorscale`
fig = go.Figure(
data=go.Scatter(
x=df["sepal_width"],
y=df["sepal_length"],
mode="markers",
marker={
"color": df["petal_length"],
"colorscale": "Viridis", # or any other colour scale
"showscale": True,
},
)
)
fig.update_layout(title="Figure 6")
fig.show()
# no longer matches Figure 5
fig.update_layout(template="seaborn", title="Figure 7")
fig.show()Suggestions
- I would expect
fig.update_layout(template="seaborn")to have the same effect, regardless of whetherpxorgointerface is used. From this perspective, to result for Figure 2 is a bug (the colour scale should have changed forseaborntemplate). - When an existing figure is updated, the settings from a template should take precedence over any existing settings. So Figure 5 and Figure 7 should produce the same result. This would entail that an update in Figure 7 overwrites whatever the
colorscalesetting Figure 6 has with that of a template provided inupdate_layout.