Question

假设您有一个2D numpy数组，您已经切片以提取其核心，就像您从较大的框架中删除内部框架一样。

更大的框架：

In[0]: import numpy
In[1]: a=numpy.array([[0,1,2,3,4],[5,6,7,8,9],[10,11,12,13,14],[15,16,17,18,19]])
In[2]: a
Out[2]: 
array([[ 0,  1,  2,  3,  4],
       [ 5,  6,  7,  8,  9],
       [10, 11, 12, 13, 14],
       [15, 16, 17, 18, 19]])

内框：

In[3]: b=a[1:-1,1:-1]
Out[3]: 
array([[ 6,  7,  8],
       [11, 12, 13]])

我的问题：如果我想检索原始数组b中a中每个值的位置，是否有比这更好的方法？

c=numpy.ravel(a) #This will flatten my values in a, so to have a sequential order
d=numpy.ravel(b) #Each element in b will tell me what its corresponding position in a was

Answer 1

 
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Python：将2D数组映射到更大的2D数组的一般规则

1 个答案: