for循环中matplotlib中的多个图例

Question

The following program executes fine but only one legend is displayed. How can I have all the four legends displayed? kindly see the image attached.

import matplotlib.pyplot as plt
dct = {'list_1' : [1,2,4,3,1],'list_2' : [2,4,5,1,2],'list_3' : [1,1,3,4,6],'list_4' : [1,1,2,2,1]}

xs = [0,1,2,3,4]


for i in [1,2,3,4]:
    plt.plot(xs,dct['list_%s' %i])
    plt.legend(['%s data' %i])

plt.show()

Answer 1

import matplotlib.pyplot as plt
dct = {'list_1' : [1,2,4,3,1],'list_2' : [2,4,5,1,2],'list_3' : [1,1,3,4,6],'list_4' : [1,1,2,2,1]}

xs = [0,1,2,3,4]


for i in [1,2,3,4]:
    plt.plot(xs,dct['list_%s' %i], label='%s data' % i)

plt.legend()

plt.show()

You are running up against the way that legend works, each time it is called it destroys the current legend and replaces it with the new one. If you only give legend a list of strings it iterates through the artists (the objects that represent the data to be drawn) in the axes until it runs out of labels (hence why your first curve is labeled as the 4th). If you include the kwarg label in the plot command, when you call legend with out any arguments, it will iterate through the artists* and generate legend entries for the artists with labels.

[*] there are some exceptions on which artists it will pick up

Answer 2

AFAIK, you need to call legend once, with all the arguments.

import matplotlib.pyplot as plt
dct = {'list_1' : [1,2,4,3,1],'list_2' : [2,4,5,1,2],
       'list_3' : [1,1,3,4,6],'list_4' : [1,1,2,2,1]}

xs = [0,1,2,3,4]

lines = []    
for i in range(1,5):
    lines += plt.plot(xs,dct['list_%s' %i], label="{} data".format(i))

Note that I have included the label here as one of the arguments to the plot function, so that later we can call get_label().

labels = [l.get_label() for l in lines]
plt.legend(lines, labels)
plt.show()

This will also work if you have separate axes (such as twinx), and all of the legend information will come through on one legend. By the way, I seem to recall that the % notation is old and one should prefer str.format(), but I'm afraid I can't remember why.