Numpy多维数组切片

Question

Suppose I have defined a 3x3x3 numpy array with

x = numpy.arange(27).reshape((3, 3, 3))

Now, I can get an array containing the (0,1) element of each 3x3 subarray with x[:, 0, 1] , which returns array([ 1, 10, 19]) . What if I have a tuple (m,n) and want to retrieve the (m,n) element of each subarray(0,1) stored in a tuple?

For example, suppose that I have t = (0, 1) . I tried x[:, t] , but it doesn't have the right behaviour - it returns rows 0 and 1 of each subarray. The simplest solution I have found is

x.transpose()[tuple(reversed(t))].transpose()

but I am sure there must be a better way. Of course, in this case, I could do x[:, t[0], t[1]] , but that can't be generalised to the case where I don't know how many dimensions x and t have.

Answer 1

你可以先创建索引元组：

index = (numpy.s_[:],)+t 
x[index]

Answer 2

HYRY solution is correct, but I have always found numpy's r_ , c_ and s_ index tricks to be a bit strange looking. So here is the equivalent thing using a slice object:

x[(slice(None),) + t]

That single argument to slice is the stop position (ie None meaning all in the same way that x[:] is equivalent to x[None:None] )