Matplotlib imshow/matshow在地图上显示值

Question

我想在Matplotlib中使用imshow或matshow创建一个10x10网格。下面的函数将numpy数组作为输入，并绘制网格图。但是，我想从阵列中的值也显示在网格定义的单元格内。到目前为止，我找不到一个正确的方法来做到这一点。我可以使用plt.text将东西放在网格上，但这需要每个单元的坐标，完全不方便。有没有更好的方式去做我想要完成的事情？

谢谢！

注意：下面的代码没有从数组中取值，我只是在玩plt.text。

import numpy as np 
import matplotlib.pyplot as plt 
from matplotlib import colors 

board = np.zeros((10, 10)) 

def visBoard(board): 
    cmap = colors.ListedColormap(['white', 'red']) 
    bounds=[0,0.5,1] 
    norm = colors.BoundaryNorm(bounds, cmap.N) 
    plt.figure(figsize=(4,4)) 
    plt.matshow(board, cmap=cmap, norm=norm, interpolation='none', vmin=0, vmax=1) 
    plt.xticks(np.arange(0.5,10.5), []) 
    plt.yticks(np.arange(0.5,10.5), []) 
    plt.text(-0.1, 0.2, 'x') 
    plt.text(0.9, 0.2, 'o') 
    plt.text(1.9, 0.2, 'x') 
    plt.grid() 

    visBoard(board) 

输出：

Answer 1

你可以这样做：

import numpy as np 
import matplotlib.pyplot as plt 

fig, ax = plt.subplots() 

min_val, max_val = 0, 10 
ind_array = np.arange(min_val + 0.5, max_val + 0.5, 1.0) 
x, y = np.meshgrid(ind_array, ind_array) 

for i, (x_val, y_val) in enumerate(zip(x.flatten(), y.flatten())): 
    c = 'x' if i%2 else 'o' 
    ax.text(x_val, y_val, c, va='center', ha='center') 
#alternatively, you could do something like 
#for x_val, y_val in zip(x.flatten(), y.flatten()): 
# c = 'x' if (x_val + y_val)%2 else 'o' 

ax.set_xlim(min_val, max_val) 
ax.set_ylim(min_val, max_val) 
ax.set_xticks(np.arange(max_val)) 
ax.set_yticks(np.arange(max_val)) 
ax.grid() 

---

编辑：

以下是带有imshow背景的更新示例。

import numpy as np 
import matplotlib.pyplot as plt 

fig, ax = plt.subplots() 

min_val, max_val, diff = 0., 10., 1. 

#imshow portion 
N_points = (max_val - min_val)/diff 
imshow_data = np.random.rand(N_points, N_points) 
ax.imshow(imshow_data, interpolation='nearest') 

#text portion 
ind_array = np.arange(min_val, max_val, diff) 
x, y = np.meshgrid(ind_array, ind_array) 

for x_val, y_val in zip(x.flatten(), y.flatten()): 
    c = 'x' if (x_val + y_val)%2 else 'o' 
    ax.text(x_val, y_val, c, va='center', ha='center') 

#set tick marks for grid 
ax.set_xticks(np.arange(min_val-diff/2, max_val-diff/2)) 
ax.set_yticks(np.arange(min_val-diff/2, max_val-diff/2)) 
ax.set_xticklabels([]) 
ax.set_yticklabels([]) 
ax.set_xlim(min_val-diff/2, max_val-diff/2) 
ax.set_ylim(min_val-diff/2, max_val-diff/2) 
ax.grid() 
plt.show() 

Answer 2

为了您的图形，你应该应该尝试与pyplot.table：

import matplotlib.pyplot as plt 
import numpy as np 

board = np.zeros((10, 10)) 
board[0,0] = 1 
board[0,1] = -1 
board[0,2] = 1 
def visBoard(board): 
    data = np.empty(board.shape,dtype=np.str) 
    data[:,:] = ' ' 
    data[board==1.0] = 'X' 
    data[board==-1.0] = 'O' 
    plt.axis('off') 
    size = np.ones(board.shape[0])/board.shape[0] 
    plt.table(cellText=data,loc='center',colWidths=size,cellLoc='center',bbox=[0,0,1,1]) 
    plt.show() 

visBoard(board) 

Answer 3

上@wflynny的代码一些阐述使它成为可以接收任意矩阵不管是什么尺寸的功能并绘制其价值。

import numpy as np 
import matplotlib.pyplot as plt 

cols = np.random.randint(low=1,high=30) 
rows = np.random.randint(low=1,high=30) 
X = np.random.rand(rows,cols) 

def plotMat(X): 
    fig, ax = plt.subplots() 
    #imshow portion 
    ax.imshow(X, interpolation='nearest') 
    #text portion 
    diff = 1. 
    min_val = 0. 
    rows = X.shape[0] 
    cols = X.shape[1] 
    col_array = np.arange(min_val, cols, diff) 
    row_array = np.arange(min_val, rows, diff) 
    x, y = np.meshgrid(col_array, row_array) 
    for col_val, row_val in zip(x.flatten(), y.flatten()): 
     c = '+' if X[row_val.astype(int),col_val.astype(int)] < 0.5 else '-' 
     ax.text(col_val, row_val, c, va='center', ha='center') 
    #set tick marks for grid 
    ax.set_xticks(np.arange(min_val-diff/2, cols-diff/2)) 
    ax.set_yticks(np.arange(min_val-diff/2, rows-diff/2)) 
    ax.set_xticklabels([]) 
    ax.set_yticklabels([]) 
    ax.set_xlim(min_val-diff/2, cols-diff/2) 
    ax.set_ylim(min_val-diff/2, rows-diff/2) 
    ax.grid() 
    plt.show() 

plotMat(X)