是否有 Numpy 函数将数组的索引映射到稀疏向量？

Question

I have a data structure like this:

• my source arrays are a sorted arrays like [2,3,4,5,7,8,9,10,11]
• I know a priori the max number of this array collection, in this case it's 17

What I need to do is to build a sparse matrix with 17 rows (the max number mentioned above) and n cols where n is the number of arrays, and each column vector should contain a mapping of the index+1 of the source vector in position [value of the source vector's element], and 0 when it's not present. In the mentioned example the output vector should be [0,1,2,3,4,0,5,6,7,8,9,10,11,0,0,0,0] . Is there an efficient way to do that in numpy without having to loop through cols and rows which would have a dramatic computational cost?

Answer 1

from scipy import sparse
import numpy as np

in_list = [2,3,4,5,7,8,9,10,11]
in_list_len = len(in_list)
max_num = 17
a = sparse.csr_matrix((max_num, in_list_len), dtype=np.int)

for ind, val in enumerate(in_list):
    a[val, ind] = ind + 1

and

Out[23]: array([[0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0, 0, 0, 0], [0, 2, 0, 0, 0, 0, 0, 0, 0], [0, 0, 3, 0, 0, 0, 0, 0, 0], [0, 0, 0, 4, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 5, 0, 0, 0, 0], [0, 0, 0, 0, 0, 6, 0, 0, 0], [0, 0, 0, 0, 0, 0, 7, 0, 0], [0, 0, 0, 0, 0, 0, 0, 8, 0], [0, 0, 0, 0, 0, 0, 0, 0, 9], [0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0]])

Complexity is O(len(in_list))

Your desired output makes no sense, because you asked for a matrix but specified a list.
I am pretty sure this is what you wanted.

The closest would be

a.data Out[18]: array([1, 2, 3, 4, 5, 6, 7, 8, 9])

Answer 2

This functionality appears to exist in pandas :

matteos_sorted_arrays_with_nans = mateos_sorted_arrays
matteos_sorted_arrays[2:-2] = np.nan
sdf = pd.Series(pd.SparseArray(matteos_sorted_arrays_with_nans))

Without further particulars, I haven't the foggiest what to recommend as a next step, though.

Answer 3

Using x to assign consecutive numbers to elements of an array:

In [16]: x
Out[16]: array([ 2,  3,  4,  5,  7,  8,  9, 10, 11])
In [17]: arr = np.zeros(17,int)
In [18]: arr
Out[18]: array([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0])
In [19]: arr[x-1] = np.arange(1,len(x)+1)
In [20]: arr
Out[20]: array([0, 1, 2, 3, 4, 0, 5, 6, 7, 8, 9, 0, 0, 0, 0, 0, 0])