我正在尝试使用Matplotlib中的imshow
或matshow
创建10x10网格。下面的函数将numpy数组作为输入,并绘制网格。但是,我希望数组中的值也显示在网格定义的单元格内。到目前为止,我找不到合适的方法来做到这一点。我可以使用plt.text
将内容放在网格上,但这需要每个单元格的坐标,这非常不方便。有没有更好的方法来做我想要完成的事情?
谢谢!
注意:下面的代码还没有从数组中获取值,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:
答案 0 :(得分:21)
你能做点什么:
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:
以下是具有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()
答案 1 :(得分:2)
对于您的图表,您应该尝试使用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)
答案 2 :(得分:2)
对@wflynny的代码进行了一些阐述,使其成为一个函数,无论大小是多少都会占用任何矩阵并绘制其值。
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)