为什么我不能在tkinter画布上绘制numpy矩阵？

Question

我正在尝试用numpy、matplotlib和tkinter创建生活游戏。我计划首先询问用户游戏板上有多少行和列，以及一个单元格活产卵的概率。输入这些信息后，用户将按“生成”按钮开始游戏。但是，只要按下按钮，就会弹出错误：“AttributeError：‘numpy.ndarray’object has no attribute‘set\u canvas’”。我不知道为什么FigureCanvasTkAgg不能绘制numpy矩阵。有没有办法在tkinter窗口中绘制numpy矩阵？这是我的代码：

from tkinter import *
import numpy as np
import matplotlib.pyplot as plt
from board import Board
from PIL import Image, ImageTk
from matplotlib.backends.backend_tkagg import (
    FigureCanvasTkAgg, NavigationToolbar2Tk)

def num():
    n1 = int(t1.get())
    n2 = int(t2.get())
    n3 = int(t3.get()) / 100.00
    initBoard = np.zeros((n1, n2))
    for row in range(0,n1):
            for column in range(0,n2):
                initBoard[row][column] = np.random.choice(np.arange(0, 2), p = [1 - n3, n3])
    #game_board = Board(n1, n2, initBoard)
    canvas = FigureCanvasTkAgg(initBoard, master=root)  # A tk.DrawingArea.
    canvas.draw()
    canvas.get_tk_widget().grid(row = 4, column = 0)
    root.quit()

root = Tk()
root.title('Game of Life')
root.geometry('800x600')
Label(root, text="How many rows?: ").grid(row = 0)
Label(root, text="How many columns?: ").grid(row = 1)
Label(root, text="Probability of spawn (between 0 and 100): ").grid(row = 2)

t1 = Entry(root)
t2 = Entry(root)
t3 = Entry(root)

t1.grid(row = 0, column = 1)
t2.grid(row = 1, column = 1)
t3.grid(row = 2, column = 1)


Button(root, text = 'Generate', command = num).grid(row = 3, column = 1, sticky = W, pady = 4)

fig= plt.figure()
canvas = FigureCanvasTkAgg(fig, master=root)  # A tk.DrawingArea.
canvas.draw()
canvas.get_tk_widget().grid(row = 4, column = 0)

mainloop()

Answer 1

代码中至少有两个问题：

>

fig = plt.figure()  # matplotlib figure
ax = fig.subplot_add(111)  # subplot inside figure
canvas = FigureCanvasTkAgg(fig, master=root)  # integrate the figure in a tkinter widget
canvas.get_tk_widget().grid(row = 4, column = 0)  # display figure in tkinter app

现在，如果您想在图中绘制一些东西，只需使用ax。绘图（…） （或任何其他打印功能）。在您的情况下，由于要显示2D数组，因此可能需要使用ax。imshow（初始化板）。

您在num（）中退出了root，这也会破坏包含您的图形的窗口。我更愿意创建一个Toplevel来询问用户有关参数的信息，然后关闭这个Toplevel，而不是num（）中的root窗口。

完整代码：

from tkinter import *  # not a good idea in my opinion, can lead to accidental 
                       # naming conflicts, "import tkinter as tk" is a better alternative
import numpy as np
import matplotlib.pyplot as plt
#~from board import Board
from PIL import Image, ImageTk
from matplotlib.backends.backend_tkagg import (
    FigureCanvasTkAgg, NavigationToolbar2Tk)

def num():
    n1 = int(t1.get())
    n2 = int(t2.get())
    n3 = int(t3.get()) / 100.00
    top.destroy()  # close dialog
    initBoard = np.zeros((n1, n2))
    for row in range(0, n1):
        for column in range(0, n2):
            initBoard[row][column] = np.random.choice(np.arange(0, 2), p=[1 - n3, n3])
    #game_board = Board(n1, n2, initBoard)
    ax.imshow(initBoard)  # draw board
    canvas.draw_idle()  # update matplotlib figure

root = Tk()
root.title('Game of Life')
root.geometry('800x600')
# create matplotlib figure
fig = plt.figure()
ax = fig.add_subplot(111)  # create axis
ax.axis('off')
canvas = FigureCanvasTkAgg(fig, master=root)  # A tk.DrawingArea.
canvas.get_tk_widget().grid(row=4, column=0)

# toplevel to choose parameters
top = Toplevel(root)
Label(top, text="How many rows?: ").grid(row=0)
Label(top, text="How many columns?: ").grid(row=1)
Label(top, text="Probability of spawn (between 0 and 100): ").grid(row=2)

t1 = Entry(top)
t2 = Entry(top)
t3 = Entry(top)

t1.grid(row=0, column=1)
t2.grid(row=1, column=1)
t3.grid(row=2, column=1)

Button(top, text='Generate', command=num).grid(row=3, column=1, sticky=W, pady=4)

mainloop()