How to efficiently get the row nr for each column where condition is met in a numpy matrix?

Question

Following this earlier question about a numpy matrix transformation.

np_array = np.matrix(
   [[0,0,0,0,1,0,0,0,0,0,0],
    [0,0,0,1,0,1,0,0,0,0,0],
    [0,0,0,0,0,0,0,0,0,0,0],
    [0,0,1,0,0,0,1,0,0,0,0],
    [0,0,0,0,0,0,0,0,0,0,0],
    [0,1,0,0,0,0,0,1,0,1,1],
    [0,0,0,0,0,0,0,0,0,0,0],
    [1,0,0,0,0,0,0,0,1,0,0]]
)

What would be the most efficient way of getting the row nr for each column where there is 1 without looping? If a column has no row with number 1 in it should return np.nan for that column. The result if this numpy matrix would be a

np.array([7,5,3,1,0,1,3,5,7,5,5])

Answer 1

i,j=np.where(np_array==1)

其中，i是标识该行的索引列表，j是表示该列的索引列表