如何从二维numpy数组的每一行中获取元素的首次出现？

Question

I have a 2d numpy array. I am looking for the indices of the first occurance of a specific element in each row. The output will be an (nx 2) array, where n is the number of rows and every entry contains the x and y coordinate of the first occurance of that specific element.

Thanks a lot.

Answer 1

If I understand the question correctly, you want something like this:

import numpy as N
#
nrows=5
ncols=10 
#
a=N.random.random((nrows,ncols))
b=-99*N.ones((nrows,2))
#
for j in range(nrows):
    for i in range(ncols):
        if(a[j,i]<0.5):
            b[j,0]=i
            b[j,1]=j
            continue

Answer 2

I'm not sure exactly what you mean by the "s and y coordinate" so I'm assuming you mean the row and column position.

import numpy as np
np.array([(s, list(row).index(your_element)) for s,row in enumerate(your_array)])

Note it will raise ValueError if your_element wasn't contained in some row.

The following version will give you an output which may contain fewer rows than the input, but won't raise ValueError for the case where your_element was missing from a row.

np.array([(s, list(row).index(your_element)) for s,row in enumerate(your_array) if your_element in row])

Answer 3

>>> # generate some fake data:
>>> A = NP.random.randint(5, 10, 100).reshape(10, 10)
>>> A
  array([[5, 7, 8, 8, 5, 6, 6, 9, 6, 9],
        [9, 8, 8, 9, 5, 6, 6, 9, 8, 9],
        [8, 5, 6, 7, 8, 9, 5, 8, 6, 7],
        [5, 8, 8, 6, 9, 6, 8, 5, 8, 9],
        [6, 9, 8, 8, 5, 7, 6, 5, 7, 6],
        [7, 8, 6, 7, 6, 6, 7, 8, 6, 8],
        [8, 6, 8, 9, 8, 8, 9, 6, 8, 7],
        [8, 7, 8, 5, 9, 5, 7, 8, 6, 9],
        [9, 6, 5, 9, 9, 8, 8, 9, 8, 8],
        [6, 8, 5, 8, 6, 5, 8, 6, 8, 5]])

>>> # sort this 2D array along one axis (i chose row-wise)
>>> A = NP.sort(A, axis=1)
>>> A
  array([[5, 5, 6, 6, 6, 7, 8, 8, 9, 9],
         [5, 6, 6, 8, 8, 8, 9, 9, 9, 9],
         [5, 5, 6, 6, 7, 7, 8, 8, 8, 9],
         [5, 5, 6, 6, 8, 8, 8, 8, 9, 9],
         [5, 5, 6, 6, 6, 7, 7, 8, 8, 9],
         [6, 6, 6, 6, 7, 7, 7, 8, 8, 8],
         [6, 6, 7, 8, 8, 8, 8, 8, 9, 9],
         [5, 5, 6, 7, 7, 8, 8, 8, 9, 9],
         [5, 6, 8, 8, 8, 8, 9, 9, 9, 9],
         [5, 5, 5, 6, 6, 6, 8, 8, 8, 8]])

>>> # now diff the sorted array along the same axis
>>> A1 = NP.diff(A ,axis=1)

>>> # A1 contains non-zero values for "first occurrences" and
>>> # zero values for repeat values
>>> A1
  array([[0, 1, 0, 0, 1, 1, 0, 1, 0],
         [1, 0, 2, 0, 0, 1, 0, 0, 0],
         [0, 1, 0, 1, 0, 1, 0, 0, 1],
         [0, 1, 0, 2, 0, 0, 0, 1, 0],
         [0, 1, 0, 0, 1, 0, 1, 0, 1],
         [0, 0, 0, 1, 0, 0, 1, 0, 0],
         [0, 1, 1, 0, 0, 0, 0, 1, 0],
         [0, 1, 1, 0, 1, 0, 0, 1, 0],
         [1, 2, 0, 0, 0, 1, 0, 0, 0],
         [0, 0, 1, 0, 0, 2, 0, 0, 0]])

you can reformulate the result, A1, as necessary, eg, as a boolean array having the same shape as A1 in wich each cell is either T/F depending on whether the value in the original matrix represents the first occurrence of that value:

>>> ndx = A1==0
>>> ndx
  array([[ True, False,  True,  True, False, False,  True, False,  True],
         [False,  True, False,  True,  True, False,  True,  True,  True],
         [ True, False,  True, False,  True, False,  True,  True, False],
         [ True, False,  True, False,  True,  True,  True, False,  True],
         [ True, False,  True,  True, False,  True, False,  True, False],
         [ True,  True,  True, False,  True,  True, False,  True,  True],
         [ True, False, False,  True,  True,  True,  True, False,  True],
         [ True, False, False,  True, False,  True,  True, False,  True],
         [False, False,  True,  True,  True, False,  True,  True,  True],
         [ True,  True, False,  True,  True, False,  True,  True,  True]], dtype=bool)