Show only certain items in legend

Question

I currently am plotting a stacked bar graph of a large amount of taxonomic data, and only wish to show significant species in the legend (out of ~500 I wish to show ~25). Is there a simple way to do this? Below is the code I have:

labels=['0','20','40','60','80','100','120']
ax1=subj1df.plot(kind='barh', stacked=True,legend=True,cmap='Paired', grid=False)
legend(ncol=2,loc=2, bbox_to_anchor=(1.05, 1), borderaxespad=0.)
label1=['Baseline','8h','24h','48h','96h','120h']
ax1.set_yticklabels(label1, fontdict=None, minor=False)
plt.title('Subject 1 Phyla',fontweight='bold')
plt.savefig('Subject1Phyla.eps', format='eps', dpi=1000)
ax1.set_xticklabels(labels)

Edit: tried adding this to show only one legend entry, however only returns an empty legend:

h, l = ax1.get_legend_handles_labels()
legend(l[4],h[4],ncol=2,loc=2, bbox_to_anchor=(1.05, 1), borderaxespad=0.)

Thordia

Answer 1

I often insert an empty label for legends I don't want to show. I made an extremely simple example which I hope will help you. You will need to tweak this to your own data but the elements you need should be there.

import matplotlib.pyplot as plt 
import numpy as np

myY=np.random.randint(20, size=10)
myX=np.arange(0,len(myY))

selected=[5,10,15]

fig = plt.figure()
for X,Y in zip(myX,myY):
    if Y in selected:
        mylabel="label = %s"%(Y); mycolor='blue'
    else:
        mylabel=None; mycolor='red'
    plt.scatter(X,Y,50, color=mycolor, label=mylabel)
plt.legend()
plt.show()

This creates the following plot:

Assuming

legend also takes a list of artists and a list of labels to precisely control what goes into your legend – Bandy

This could work, but is there a way for selected to be a list of strings of the legend entries I want? my plot is a stacked bar plot of many entries. – Thordia

Sure. For each of your entries (identified by X,Y, and name) check whether your current entry is in a list of selected labels that you want to show the legend for. Its difficult to show you how to do this unless you show more info as to how your data looks. – Assuming

Try the indicated solution with label='nolegend' – Musa

Philipp means label='_nolegend_' (note the underscores), see the answer from jlansey. – Superincumbent

Answer 2

This works:

plt.plot(x, y, label='_nolegend_')

source

Algology

Hmm. Tough luck to the poor bastard who wants his label to say _nolegend_. – Pinta

Is there any reason the developers chose this over label=None ? – Comminute

Everything that starts with _ is skipped from the legend. – Zennas

How can I label my plot as _nolegend_ ? :P – Galahad

as shown, by adding the parameter: label='_nolegend_' – Algology

I'm plotting data from multiple columns of a dataframe on the same chart. Your solution doesn't work if I implement it as df.plot(..., label='nolegend'). It works if I rename the dataframe columns as 'nolegend' before running df.plot(...) but I get the following warning, repeated for each column I'm plotting: UserWarning: The handle <matplotlib.lines.Line2D object at 0x000001B26EB57370> has a label of 'nolegend' which cannot be automatically added to the legend. ax.legend(handles, labels, loc="best", title=title) Is there a way to avoid that warning? – Weatherworn

you are missing an underscore – Algology

Thank you for your response. I meant _nolegend but the editor here automatically removed the underscores and converted the text in between to italics. Anyway, as I said, your solution with the underscore works but gives me the warnings. Have you come across them? Not sure why I get them. – Weatherworn

Link doesn't work anymore :( – Ashlynashman

Unfortunately, I noticed today that it now says MatplotlibDeprecationWarning: An artist whose label starts with an underscore was passed to legend(); such artists will no longer be ignored in the future. for this solution. The use of None seems to be the best solution. – Breaststroke

Answer 3

For whatever reason both answers didn't work for mine situation. What worked, and actually was indicated above:

legend also takes a list of artists and a list of labels to precisely control what goes into your legend – tacaswell Jul 11 '14 at 4:46

import pandas as pd
import matplotlib.pyplot as plt
import pylab

pd.Series(range(10)).plot(color = 'grey')
x = list(range(10))
y = [i + 1 for i in x]  
scat1 = plt.scatter(x, y)

pylab.legend([scat1],['moved points'], loc = 'upper left')

plt.show()

The result of the code:

Mauro

Clean solution that scales properly, thank you for sharing – Launceston

Answer 4

You can also use an empty string variable:

    plt.plot(xData, yData, ..., label=str())

By passing an empty str() object, it doesn't write down anything.

Matplotlib API reference

Cynthea

or save yourself 3 characters and write label="" – Wigeon

Answer 5

I know this post is coming up on a decade old, but I was looking for a way to do something similar and this post was on top of the google results. I was working with data tracking heap usage of a process and the processes that it spawned. Most of the processes used up a negligible amount of memory, and displaying the entire legend would have been unwieldy. I ended up with the following solution:

label_names = heap_df.columns.values.tolist()
max_per_proc = heap_df.max()
max_heap_overall = max(max_per_proc)

fig, ax = plt.subplots()
plt.stackplot(heap_df.index, heap_df.values.T, labels=label_names)
lines, labels = ax.get_legend_handles_labels()
lbl_idx = []
for idx in range(len(labels)):
    if max_per_proc[labels[idx]] / max_heap_overall > 0.01:
        lbl_idx.append(idx)
ax.legend([lines[i] for i in lbl_idx], [labels[i] for i in lbl_idx])
        
plt.show()

Ferous

Answer 6

In the case where you are plotting multiple lines simultaneously with a Pandas dataframe and the Pandas plot method, it's not as easy to control which lines make it onto the legend.

Here is a solution to this problem:

df_line_data.plot(legend=False)

labels = [(col if col in cols_to_plot else '_Hidden label') for col in df_line_data.columns]

ax.legend(labels=labels)

Breaststroke