No handles with labels found to put in legend

Question

I'm trying to create a parallelogram in PyPlot. I'm not up to drawing the parallelogram--first I'm putting in the vector arrows--using the following code:

fig = plt.figure()
ax = fig.add_subplot(111)
ax.spines['left'].set_position('zero')
ax.spines['right'].set_color('none')
ax.spines['bottom'].set_position('zero')
ax.spines['top'].set_color('none')
plt.axis([-5,5,-5,5])
ax.xaxis.set_ticks_position('bottom')
ax.yaxis.set_ticks_position('left')
plt.grid()
plt.arrow(0,0, 3,1, head_width=0.2, color='r', length_includes_head=True, label='u')
plt.arrow(0,0, 1,3, head_width=0.2, color='r', length_includes_head=True, label='v')
plt.arrow(0,0, 4,4, head_width=0.2, color='r', length_includes_head=True, label='u+v')
plt.legend()

This returns the following error:

No handles with labels found to put in legend.

I'm not sure why, because, based on the documentation for plt.arrow(), label is an acceptable kwarg, and plt.legend() should ostensibly be reading that. The rest of the figure draws fine; it's just missing the legend.

For some reason, I cannot reproduce this error. After including the import statement, I can run this code with any errors and it displays the legend. — akalanka, Mar 29 '23 at 22:22

score 51 · Accepted Answer · edited Feb 08 '21 at 00:47

It might be late but for anyone with the same issue the solution is using the method legend() for the corresponding ax not as for plt

fig = plt.figure()
ax = fig.add_subplot(111)
ax.spines['left'].set_position('zero')
ax.spines['right'].set_color('none')
ax.spines['bottom'].set_position('zero')
ax.spines['top'].set_color('none')
plt.axis([-5,5,-5,5])
ax.xaxis.set_ticks_position('bottom')
ax.yaxis.set_ticks_position('left')
plt.grid()
plt.arrow(0,0, 3,1, head_width=0.2, color='r', length_includes_head=True, label='u')
plt.arrow(0,0, 1,3, head_width=0.2, color='r', length_includes_head=True, label='v')
plt.arrow(0,0, 4,4, head_width=0.2, color='r', length_includes_head=True, label='u+v')
ax.legend()

Its very easy to forget that you have subplots. Thanks for this. Really helped me out. — preetam, Dec 15 '20 at 21:59

score 25 · Answer 2 · answered Dec 15 '19 at 18:24

25

You can explicitly define the elements in the legend.

For full control of which artists have a legend entry, it is possible to pass an iterable of legend artists followed by an iterable of legend labels respectively. Reference

Example:

arr1 = plt.arrow(0,0, 3,1, head_width=0.2, color='r', length_includes_head=True)
arr2 = plt.arrow(0,0, 1,3, head_width=0.2, color='g', length_includes_head=True)
arr3 = plt.arrow(0,0, 4,4, head_width=0.2, color='b', length_includes_head=True)

plt.xlim(0,5)
plt.ylim(0,5)

plt.legend([arr1, arr2, arr3], ['u','v','u+v'])

answered Dec 15 '19 at 18:24

abc

11,579
2
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51

13

That being said, shouldn't the legend be working with the labels provided in the `plt.arrow()`s? – Yehuda Dec 16 '19 at 00:52
@Yehuda Yes, plt.arrow() works like plt.plot(), and that is being asked for. It is strange why this solution is upvoted so much since it does not explain why plt.legend() does not understand the setting of the label in plt.arrow(), although it should. If it is the same way as in plt.plot(), see https://stackoverflow.com/questions/66862848/no-handles-with-labels-found-to-put-in-legend-error-after-plotting-two-plots-on or https://stackoverflow.com/questions/64555759/no-handles-with-labels-found-to-put-in-legend-linechart. – questionto42 Jul 11 '21 at 19:57
1

@questionto42 1. Thanks for the links and the answers 2. This was the only answer available for almost a year, hence the upvotes ;) – Yehuda Jul 12 '21 at 20:22

score 8 · Answer 3 · edited Jul 15 '21 at 21:49

The error is thrown because you haven't specified the label text

Either do something like this

plt.hist([x01, x02,x03], color=["lightcoral","lightskyblue","slategrey"], stacked=True, 
             label=['Supressed','Active','Resolved'])
plt.legend()

Or

Do not use plt.legend() if you haven't specified the label text as in the following WRONG example:

plt.hist([x01])
plt.legend()

The above will throw the same error, so either remove legend function or provide what it needs -> label. Side note: Here x01 is just a list of number for which I am creating a histogram, in the first example they are three list of numbers to create stacked bar chart

The bottom line is this error is thrown because of not specifying legend text and calling/initializing a legend

Each arrow has a specified label though. – Yehuda Dec 08 '20 at 16:02 — Yehuda, Dec 08 '20 at 16:02

score 5 · Answer 4 · answered Sep 30 '21 at 10:24

5

I had this error when using labels which started with an underscore

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

Removing the front underscore from the labels solved the issue

answered Sep 30 '21 at 10:24

Oliver Zendel

2,695
34
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score 1 · Answer 5 · answered May 21 '22 at 13:17

1

Assuming you have 2 plots ax and ax2, we can:

get the labels from each y-axis via ax.get_label()
.legend allows an array to be ingested

fig.legend([ax.get_ylabel(), ax2.get_ylabel()], loc='upper right')

answered May 21 '22 at 13:17

dimButTries

661
7
15

score -1 · Answer 6 · answered Sep 09 '21 at 12:40

I face similar problem like No handles with labels found to put in legend.

First My code look like

figure, axis = pyplot.subplots(nrows=1,ncols=2, figsize=(15, 6), tight_layout=True)

axis[0].legend(title='Country', title_fontsize = 12) #this line
axis[0].pie(x=piechart_result['value_eur'],labels=piechart_result['short_name'])

axis[1].pie(x=piechart_result['value_eur'],labels=piechart_result['short_name')

pyplot.show()

Then I changed to

figure, axis = pyplot.subplots(nrows=1,ncols=2, figsize=(15, 6), tight_layout=True)

axis[0].pie(x=piechart_result['value_eur'],labels=piechart_result['short_name'])
axis[0].legend(title='Country', title_fontsize = 12) # this line 

axis[1].pie(x=piechart_result['value_eur'],labels=piechart_result['short_name')

pyplot.show()

this work for me in colab notebook

score -1 · Answer 7 · edited Jun 14 '22 at 23:07

I had this same issue and solved it with an understanding that .legend() function has to be after all the instructions that deal with the label attribute. This includes both plt and ax.

So moving ax.legend(*) as the last command.

I hope this helps you too.

Change

ax.plot(-trip_df.stop_lat, -trip_df.stop_lon, label = trip_id)
plt.legend()

to

ax.plot(-trip_df.stop_lat, -trip_df.stop_lon, label = trip_id)
ax.legend()
plt.legend()

No handles with labels found to put in legend

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