I am using Seaborn to make a boxplot using data from a pandas dataframe.
colorpalette = sns.hls_palette(8,h=.9)
g = sns.boxplot(x="estimator", y="mean_score", data=dFrame, palette=colorpalette)
g.set(ylabel='Mean Accuracy', xlabel='')
plt.show()
This results me in the previous figure. As you can see the ticklabels are too long to be in one line. So, I plan to use textwrap on the xticklabels to span them over multiple rows. In order to get the labels, I tried using
g.xaxis.get_ticklabels()
Returns me the following
<a list of 9 Text major ticklabel objects>
If I try it in a loop like this
for item in g.xaxis.get_ticklabels():
print(item)
I get the following output
Text(0,0,'ExtraTreesClassifier')
Text(1,0,'RandomForestClassifier')
Text(2,0,'GradientBoostingClassifier')
Text(3,0,'LogisticRegression')
Text(4,0,'DecisionTreeClassifier')
Text(5,0,'kNearestNeighbors')
Text(6,0,'LinearSVC')
Text(7,0,'Perceptron')
Is there a way to do it more efficiently using default functions/methods in seaborn.
Having a matplotlib axes instance ax (as it is e.g. returned by seaborn plots),
ax = sns.boxplot(...)
allows to obtain the ticklabels as
ax.get_xticklabels()
The easiest way to get the texts out of the list would be
texts = [t.get_text() for t in ax.get_xticklabels()]
Wrapping the text could be done as well on the fly
texts = [textwrap.fill(t.get_text(), 10) for t in ax.get_xticklabels()]
and even setting the text back as ticklabels can be done in the same line
ax.set_xticklabels([textwrap.fill(t.get_text(), 10) for t in ax.get_xticklabels()])
The accepted answer didn't work for me (there was message like: 'FacetGrid' object has no attribute 'get_xticklabels'.
but this worked:
g.fig.autofmt_xdate()
Related
I have a matplotlib boxplot with very long strings as xticks. Is there a way to automatically split them into multiple lines to make the plot more clean? I am using the seaborn function barplot to create the graph.
This is the plot:
And the code i use to create it:
plt.figure(figsize=(15,13))
sns.barplot(x="Component",y="TTTR",color = "C0",data=new,dodge=False,edgecolor="black",zorder=3)
plt.xticks(rotation=90)
plt.grid(axis='y',zorder=0)
plt.title("10 most impactful components",size=30,y=1.04,**pfont)
plt.ylabel("Impact (Sum TTR in h)")
plt.xlabel('Component')
plt.tight_layout()
seaborn.barplot returns a matplotlib.Axes object so you could use Axes.set_xticklabels to update the labels afterwards, e.g.
import textwrap
max_width = 20
ax = sns.barplot(...)
ax.set_xticklabels(textwrap.fill(x.get_text(), max_width) for x in ax.get_xticklabels())
Found a workaround: Replacing the label column in the pandas dataframe with the labels from this answer
new.Component = [re.sub("(.{20})", "\\1\n", label, 0, re.DOTALL) for label in new.Component]
sns.barplot(x="Component",y="TTTR",color ="C0",data=new,dodge=False,edgecolor="black",zorder=3)
I've seen Creating multi column legend in python seaborn plot but I think my question is a bit different. In short, I've got a dataframe that I'm plotting in seaborn's lmplot and getting a FacetGrid. Trouble is, there are tons of values for hue so I get a super long, single column legend. Code example below:
ers = sns.lmplot(
data=emorb,
x="Pb",
y="Nd",
row="Ridge Sys",
hue="Seg Name",
scatter=True,
fit_reg=False,
scatter_kws={"alpha":0.7, "edgecolor": "w"},
palette=sns.color_palette("bright", 20),
legend=True
)
ers.set(ylim=(0.5122,0.5134))
I can access the legend object that is created by calling ers._legend and this returns an object with type Legend (basically, a matplotlib object). However, I can't then call to this legend object to change the number of columns, e.g., with:
l = ers._legend
l(ncols=9)
Any suggestions, or am I missing something perhaps more obvious, such as a way to redraw the legend and specify any parameters?
Thanks.
Whoops, figured it out:
The FacetGrid object has an attribute fig, i.e.
g = sns.lmplot()
parent_mpl_figure = g.fig
And so if I set legend=False in sns.lmplot(), I can then specify parent_mpl_figure.legend(labels=[], ncol=9, bbox_to_anchor=(1,1)).
Written cleanly:
g = sns.lmplot(legend = False)
parent_mpl_figure = g.fig
parent_mpl_figure.legend(labels = [], ncol = 9, bbox_to_anchor = (1,1))
Hope this is instructive for someone else / now to figure out how to have each Facet span the full color palette so that different hue groups within each Facet group are easier to distinguish...
I am trying to create a FacetGrid in Seaborn
My code is currently:
g = sns.FacetGrid(df_reduced, col="ActualExternal", margin_titles=True)
bins = np.linspace(0, 100, 20)
g.map(plt.hist, "ActualDepth", color="steelblue", bins=bins, width=4.5)
This gives my the Figure
Now, instead of "ActualExternal =0.0" and "ActualExternal =1.0" I would like the titles "Internal" and "External"
And, instead of "ActualDepth" I would like the xlabel to say "Percentage Depth"
Finally, I would like to add a ylabel of "Number of Defects".
I've tried Googling and have tried a few things but so far no success. Please can you help me?
Thanks
Although you can iterate through the axes and set the titles individually using matplotlib commands, it is cleaner to use seaborn's built-in tools to control the title. For example:
# Add a column of appropriate labels
df_reduced['measure'] = df_reduced['ActualExternal'].replace({0: 'Internal',
1: 'External'}
g = sns.FacetGrid(df_reduced, col="measure", margin_titles=True)
g.map(plt.hist, "ActualDepth", color="steelblue", bins=bins, width=4.5)
# Adjust title and axis labels directly
g.set_titles("{col_name}") # use this argument literally
g.set_axis_labels(x_var="Percentage Depth", y_var="Number of Defects")
This has the benefit of not needing modification regardless of whether you have 1D or 2D facets.
You can access the axes of a FacetGrid (g = sns.FacetGrid(...)) via g.axes. With that you are free to use any matplotlib method you like to tweak the plot.
Change titles:
axes = g.axes.flatten()
axes[0].set_title("Internal")
axes[1].set_title("External")
Change labels:
axes = g.axes.flatten()
axes[0].set_ylabel("Number of Defects")
for ax in axes:
ax.set_xlabel("Percentage Depth")
Note that I prefer those above the FacetGrid's internal g.set_axis_labels and set_titles methods, because it makes it more obvious which axes is to be labelled.
The easiest way to set multiple titles would be:
titles = ['Internal','External']
for ax,title in zip(g.axes.flatten(),titles):
ax.set_title(title )
I am trying to create a FacetGrid in Seaborn
My code is currently:
g = sns.FacetGrid(df_reduced, col="ActualExternal", margin_titles=True)
bins = np.linspace(0, 100, 20)
g.map(plt.hist, "ActualDepth", color="steelblue", bins=bins, width=4.5)
This gives my the Figure
Now, instead of "ActualExternal =0.0" and "ActualExternal =1.0" I would like the titles "Internal" and "External"
And, instead of "ActualDepth" I would like the xlabel to say "Percentage Depth"
Finally, I would like to add a ylabel of "Number of Defects".
I've tried Googling and have tried a few things but so far no success. Please can you help me?
Thanks
Although you can iterate through the axes and set the titles individually using matplotlib commands, it is cleaner to use seaborn's built-in tools to control the title. For example:
# Add a column of appropriate labels
df_reduced['measure'] = df_reduced['ActualExternal'].replace({0: 'Internal',
1: 'External'}
g = sns.FacetGrid(df_reduced, col="measure", margin_titles=True)
g.map(plt.hist, "ActualDepth", color="steelblue", bins=bins, width=4.5)
# Adjust title and axis labels directly
g.set_titles("{col_name}") # use this argument literally
g.set_axis_labels(x_var="Percentage Depth", y_var="Number of Defects")
This has the benefit of not needing modification regardless of whether you have 1D or 2D facets.
You can access the axes of a FacetGrid (g = sns.FacetGrid(...)) via g.axes. With that you are free to use any matplotlib method you like to tweak the plot.
Change titles:
axes = g.axes.flatten()
axes[0].set_title("Internal")
axes[1].set_title("External")
Change labels:
axes = g.axes.flatten()
axes[0].set_ylabel("Number of Defects")
for ax in axes:
ax.set_xlabel("Percentage Depth")
Note that I prefer those above the FacetGrid's internal g.set_axis_labels and set_titles methods, because it makes it more obvious which axes is to be labelled.
The easiest way to set multiple titles would be:
titles = ['Internal','External']
for ax,title in zip(g.axes.flatten(),titles):
ax.set_title(title )
I have produced a very simple pie chart in Python using Matplotlib and I am wanting to edit the alignment of my labels. I have used \n within my labels to split the line as the labels are too long for one line. But as you can see from the picture called 'pie chart image', it's a mix of weird alignments at the moment. I would really like to have it center alignment.
For other chart/graph types in Matplotlib there is an argument called align where you can set it to center, however, plt.pie(...) does not seem to have this attribute.
Here is my code:
import matplotlib.pyplot as plt
k = [7,15]
labels = 'Strongly and Mostly \n Agree', 'Strongly/Mostly Disagree \n and In the Middle'
plt.pie(k, labels= labels)
plt.show()
Any ideas?
You can pass a dictionary of text properties to plt.pie via the textprops argument. For example:
plt.pie(k, labels=labels, textprops={'weight': 'bold'})
However, if you try to specify the horizontalalignment property, you'll get an error saying that you provided that parameter twice. Obviously you didn't, but matplotlib passed both it's hard-coded value and your value to some internal function.
But that's probably a good thing. The way I see it, there's not so much a mix of alignments, but a consistent alignment of the text against the pie.
Back to your question
pie returns both the patches and the labels for each wedge. So you can loop through the labels after your initial call to pie to modify their alignment. That looks like this:
k = [7, 15]
labels = 'Strongly and Mostly\nAgree', 'Strongly/Mostly Disagree\nand In the Middle'
fig, ax = plt.subplots()
ax.set_aspect('equal')
wedges, labels = ax.pie(k, labels=labels, textprops={'weight': 'bold'})
for label in labels:
label.set_horizontalalignment('center')
As you can see, the labels now overlap with the wedges, diminishing legibility.
The labels also have a set_position method (i.e., label.set_position((x, y))), but recomputing the positions for N labels in a pie chart sounds like a Bad Time to me.