我写了一个例程,将你的矩阵切成块。这个例子很容易理解。我以简单的形式编写了它以显示结果(仅用于检查目的)。如果您对此感兴趣,可以在输出中包含块数或任何内容。
import matplotlib.pyplot as plt
import numpy as np
def cut_array2d(array, shape):
arr_shape = np.shape(array)
xcut = np.linspace(0,arr_shape[0],shape[0]+1).astype(np.int)
ycut = np.linspace(0,arr_shape[1],shape[1]+1).astype(np.int)
blocks = []; xextent = []; yextent = []
for i in range(shape[0]):
for j in range(shape[1]):
blocks.append(array[xcut[i]:xcut[i+1],ycut[j]:ycut[j+1]])
xextent.append([xcut[i],xcut[i+1]])
yextent.append([ycut[j],ycut[j+1]])
return xextent,yextent,blocks
nx = 900; ny = 650
X, Y = np.meshgrid(np.linspace(-5,5,nx), np.linspace(-5,5,ny))
arr = X**2+Y**2
x,y,blocks = cut_array2d(arr,(10,10))
n = 0
for x,y,block in zip(x,y,blocks):
n += 1
plt.imshow(block,extent=[y[0],y[1],x[0],x[1]],
interpolation='nearest',origin='lower',
vmin = arr.min(), vmax=arr.max(),
cmap=plt.cm.Blues_r)
plt.text(0.5*(y[0]+y[1]),0.5*(x[0]+x[1]),str(n),
horizontalalignment='center',
verticalalignment='center')
plt.xlim([0,900])
plt.ylim([0,650])
plt.savefig("blocks.png",dpi=72)
plt.show()
输出是:
问候
注意:我认为您可以使用 np.meshgrid 优化此例程,而不是使用 xextent 和 yextent 进行大量附加。