2d数组的每列的第一个和最后一个数字1的索引

Question

I have a 2d array ( Q ) consisting of only zeros and ones (binary matrix). For each column of Q I want to find the index of the first and last row which the value 1 occurs. Each column contains at least one 1 .

Here's an example:

[[1, 1, 1, 0, 0, 0, 0],
 [0, 1, 1, 1, 0, 0, 0],
 [1, 0, 0, 0, 1, 0, 1],
 [0, 0, 0, 1, 0, 1, 1],
 [1, 0, 1, 0, 0, 0, 0],
 [0, 0, 1, 0, 0, 0, 1],
 [0, 0, 0, 1, 0, 1, 0]]

boundsList = {0: (0, 4), 1: (0, 1), 2: (0, 5), 3: (1, 6), 4: (2, 2), 5: (3, 6), 6: (2, 5)}

I implemented an algorithm, it works, but for large arrays it is not efficient:

boundsList = {}
for i in range (0, len(Q)):
    column = Q[:,i]
    indexesList = []
    for idx, pos in enumerate (column):
        if pos == 1:
            indexesList.append(idx)
    boundsList[i] = (indexesList[0], indexesList[-1])

Can anybody could suggest another simple solution to this problem?

Answer 1

Let's start with your array:

>>> Q
array([[1, 1, 1, 0, 0, 0, 0],
       [0, 1, 1, 1, 0, 0, 0],
       [1, 0, 0, 0, 1, 0, 1],
       [0, 0, 0, 1, 0, 1, 1],
       [1, 0, 1, 0, 0, 0, 0],
       [0, 0, 1, 0, 0, 0, 1],
       [0, 0, 0, 1, 0, 1, 0]])

To obtain the indices for each column of the first row which contains 1:

>>> np.argmax(Q, axis=0) # Index of first appearance of 1
array([0, 0, 0, 1, 2, 3, 2])

To obtain the indices for each column of the last row which contains 1:

>>> Q.shape[0] - np.argmax(Q[::-1, :], axis=0) - 1 # Index of last appearance of 1
array([4, 1, 5, 6, 2, 6, 5])

To combine them into a dictionary of your liking:

>>> dict(enumerate(zip( np.argmax(Q, axis=0), Q.shape[0] - np.argmax(Q[::-1, :], axis=0) - 1)))
{0: (0, 4), 1: (0, 1), 2: (0, 5), 3: (1, 6), 4: (2, 2), 5: (3, 6), 6: (2, 5)}

Answer 2

Probably the fastest way would be to use the argmax method (it works because it finds the position of the first maximum) from both sides and then put that in a dictionary. The argmax method has far less overhead (constant factor) than using np.argmax so, especially for small arrays, the method will be much faster.

Because dict , enumerate and zip are faster on lists than arrays I also convert the intermediate arrays to lists (the tolist method is the fastest way to achieve that):

>>> dict(enumerate(zip(Q.argmax(axis=0).tolist(), 
...                    (Q.shape[0]-1-Q[::-1].argmax(axis=0)).tolist())))
{0: (0, 4), 1: (0, 1), 2: (0, 5), 3: (1, 6), 4: (2, 2), 5: (3, 6), 6: (2, 5)}

The Q[::-1] is the reversed array and to get the "not-reversed" indices I have to subtract them from Q.shape[0]-1 .