I have a problem using matplotlib in Tkinter.
Here is the plan:
1. Define a Button
and a Frame
(Frame_0) using Tkinter.
2. By clicking the Button
to plot the sin-function using matplotlib
, which is shown in Frame_0
.
But I don't know how to do it.
I am just a beginner. But I just think the easist way to do it might be, 'selecting' the widget by using get_tk_widget()
, is it possible? Or is there some other commands to do it?
Hope you understand my poor English and my beginners question. Here is the code. Thank you all.
# -*- coding: utf-8 -*-
from Tkinter import *
import matplotlib
matplotlib.use('TkAgg')
from numpy import arange, sin, pi
from matplotlib.backends.backend_tkagg import FigureCanvasTkAgg, NavigationToolbar2TkAgg
from matplotlib.backend_bases import key_press_handler
from matplotlib.figure import Figure
def command_0():
f = Figure(figsize=(5, 3), dpi=100)
a = f.add_subplot(111)
t = arange(0.0, 3.0, 0.01)
s = sin(2 * pi * t)
a.plot(t, s)
dataPlot = FigureCanvasTkAgg(f, master=root)
dataPlot.show()
dataPlot.get_tk_widget().pack(side=TOP, fill=BOTH, expand=1)
root.mainloop()
def Main():
global root
root = Tk()
root.title("Program")
root['background']='gray'
w = root.winfo_screenwidth()
h = root.winfo_screenheight()
root.geometry("%dx%d" %(w, h))
root.state("zoomed")
global Frame_0
Frame_0 = Frame(root)
Frame_0.place(height=600, width=800, x=10, y=10)
Frame_1 = Frame(root)
Frame_1.place(width=120, height=80, x=1000, y=20)
Button_1 = Button(Frame_1, text = "Click", width = 120, height=80, bg='green', command = command_0)
Button_1.place(x=1000, y=20)
Button_1.grid(row=0, column=0)
Button_1.pack()
root.mainloop()
return
Main()
If you wont to show it in frame_0
, i don't know why you are setting the master=root
. Simply change:
dataPlot = FigureCanvasTkAgg(f, master=root)
to:
dataPlot = FigureCanvasTkAgg(f, master=Frame_0)
Also you don't need that root.mainloop()
at the end of the function.
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.