在matplotlib和Python中是否有办法返回在饼图中单击的值/标签。例如,如果用户单击饼图的条子A,则返回值A.如果用户单击B饼图的条子B,则返回值B.
答案 0 :(得分:3)
from matplotlib import pyplot as plt
# make a square figure and axes
plt.figure(figsize=(6,6))
ax = plt.axes([0.1, 0.1, 0.8, 0.8])
labels = 'Frogs', 'Hogs', 'Dogs', 'Logs'
fracs = [15,30,45, 10]
explode=(0, 0.05, 0, 0)
p = plt.pie(fracs, explode=explode, labels=labels, autopct='%1.1f%%', shadow=True)
plt.title('Raining Hogs and Dogs', bbox={'facecolor':'0.8', 'pad':5})
w = p[0][0]
plt.show()
class PieEventHandler:
def __init__(self,p):
self.p = p
self.fig = p[0].figure
self.ax = p[0].axes
self.fig.canvas.mpl_connect('button_press_event', self.onpress)
def onpress(self, event):
if event.inaxes!=self.ax:
return
for w in self.p:
(hit,_) = w.contains(event)
if hit:
print w.get_label()
handler = PieEventHandler(p[0])
的引用:
答案 1 :(得分:1)
:)
import matplotlib.pyplot as plt
def main():
# Make an example pie plot
fig = plt.figure()
ax = fig.add_subplot(111)
labels = ['Apple', 'Mango', 'Orange']
wedges, plt_labels = ax.pie([20, 40, 60], labels=labels)
ax.axis('equal')
make_picker(fig, wedges)
plt.show()
def make_picker(fig, wedges):
def onclick(event):
wedge = event.artist
label = wedge.get_label()
print label
# Make wedges selectable
for wedge in wedges:
wedge.set_picker(True)
fig.canvas.mpl_connect('pick_event', onclick)
if __name__ == '__main__':
main()