python把四个图画在一个窗口_Python matplotlib如何在一个窗口画多张图，使用forward to next view按钮切换...

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# -*- coding: utf-8 -*-

from matplotlib.figure import Figure

from matplotlib.backends.backend_tkagg import FigureCanvasTkAgg

from Tkinter import *

class test(object):

def __init__(self,parent):

self.parent = parent

self.n = 0

Button(self.parent,text = 'change',command = self.change).pack()

'''表格个数'''

num = 2

fig = []

self.canvas = []

for i in range(num):

fig.append(Figure())

self.canvas.append(FigureCanvasTkAgg(fig[i],master = self.parent))

self.canvas[0]._tkcanvas.pack()

'''以下可以创2113建不同的图5261'''

axe = fig[0].add_subplot(111)

axe.set_title(u'第一个图',{'fontname':'STSong'})

axe2 = fig[1].add_subplot(211)

axe2_2 = fig[1].add_subplot(212)

axe2.set_title(u'第二个图',{'fontname':'STSong'})

def change(self):

self.n +=1

if self.n == 2:

self.n = 0

self.canvas[self.n -1]._tkcanvas.pack_forget()

self.canvas[self.n]._tkcanvas.pack()

if __name__ == "__main__":

root = Tk()

test(root)

root.mainloop()

您看着样4102能不能满足您的要求1653？

单击change后：