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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.)
Trenton McKinney
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Funsaized
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5 Answers5

291

This works:

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

source

jlansey
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    Hmm. Tough luck to the poor bastard who wants his label to say `_nolegend_`. – Jason S Jan 06 '19 at 16:26
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    Is there any reason the developers chose this over `label=None` ? – rovyko Apr 16 '19 at 22:44
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    Everything that starts with `_` is skipped from the legend. – StSav012 May 15 '19 at 17:21
  • How can I label my plot as `_nolegend_` ? :P – Black Jack 21 Jun 19 '20 at 10:46
  • as shown, by adding the parameter: `label='_nolegend_'` – jlansey Jun 24 '20 at 13:04
  • 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 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? – ric Nov 28 '22 at 17:37
  • you are missing an underscore – jlansey Nov 29 '22 at 18:42
  • 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. – ric Nov 30 '22 at 12:15
39

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: enter image description here

Trond Kristiansen
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    `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 04:46
  • 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. – Funsaized Jul 11 '14 at 14:10
  • 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. – Trond Kristiansen Jul 11 '14 at 15:50
  • Try the indicated solution with label='_nolegend_' – Philipp Schwarz Oct 07 '16 at 23:13
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    Philipp means `label='_nolegend_'` (note the underscores), see the answer from [jlansey](https://stackoverflow.com/a/35710894/523100). – Czechnology Aug 23 '18 at 19:23
13

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: The result of the code:

Andrey Selivanov
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6

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

Enrique Millán Valbuena
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1

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()
Paul
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