Accessing the j-th dimension of an n-dimensional matrix in python

Question

Given:

• an integer input j,d such that 0 < j < d+1

• integer vectors -1 < a < b of dimension d.

• d-dimensional matrix (i.e. tensor) T as a numpy array

I would like to read certain information that would variably depend on the integer j.

For example,

  c[1]>u[,1]

I would like to access

  T[(a[0]):(b[0]),...,(a[j]-1):(b[j]+1),...,(a[n-1]):(b[n-1])]

I am wondering if there is a generic way of doing this, especially in the case where d and j can be variable.

A similar question can be found here: Access n-th dimension in python.

Answer 1

Constructing an indexing tuple from slices:

In [88]: a = [1,0,4]; b = [4,1,None]
In [89]: idx = [slice(i,j) for i,j in zip(a,b)]
In [90]: idx
Out[90]: [slice(1, 4, None), slice(0, 1, None), slice(4, None, None)]
In [91]: arr = np.arange(5*3*7).reshape(5,3,7)
In [92]: arr[tuple(idx)]
Out[92]: 
array([[[25, 26, 27]],

       [[46, 47, 48]],

       [[67, 68, 69]]])
In [93]: _.shape
Out[93]: (3, 1, 3)