Transform Matrix

I got the following small numpy matrix, the values of the matrix can only be 0 or 1. The size of the actual matrix i am using is actually much bigger but for demonstration purposes

Solution 1:

Here's one approach -

def per_col(a):
    idx = a.argmax(0)
    out = np.zeros_like(a)
    r = np.arange(a.shape[1])
    out[idx, r] = a[idx, r]
    return out

Sample runs

Case #1 :

In [41]: a
Out[41]: 
array([[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, 1, 0, 1, 0, 0, 0, 0, 0],
       [0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 0],
       [0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0],
       [0, 1, 0, 0, 0, 0, 1, 1, 0, 1, 1],
       [0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0],
       [1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0]])

In [42]: per_col(a)
Out[42]: 
array([[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]])

Case #2 (Insert an all zeros column):

In [78]: a[:,1] = 0

In [79]: a
Out[79]: 
array([[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, 1, 0, 1, 0, 0, 0, 0, 0],
       [0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 0],
       [0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1],
       [0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0],
       [1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0]])

In [80]: per_col(a)
Out[80]: 
array([[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, 0, 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]])

If you are crazy about one-liners or a fan of broadcasting, here's another -

((a.argmax(0) == np.arange(a.shape[0])[:,None]).astype(int))*a.any(0)

Sample run -

In [89]: a
Out[89]: 
array([[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, 1, 0, 1, 0, 0, 0, 0, 0],
       [0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 0],
       [0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1],
       [0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0],
       [1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0]])

In [90]: ((a.argmax(0) == np.arange(a.shape[0])[:,None]).astype(int))*a.any(0)
Out[90]: 
array([[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, 0, 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]])

Runtime test -

In [98]: a = np.random.randint(0,2,(100,10000))

# @DSM's soln
In [99]: %timeit ((a == 1) & (a.cumsum(axis=0) == 1)).astype(int)
100 loops, best of 3: 5.19 ms per loop

# Proposed in this post : soln1
In [100]: %timeit per_col(a)
100 loops, best of 3: 3.4 ms per loop

# Proposed in this post : soln2
In [101]: %timeit ((a.argmax(0) == np.arange(a.shape[0])[:,None]).astype(int))*a.any(0)
100 loops, best of 3: 7.73 ms per loop

Solution 2:

You can use cumsum to count the number of 1s you see, and then select the first:

In [42]: arr.cumsum(axis=0)
Out[42]: 
matrix([[0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0],
        [0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0],
        [0, 0, 0, 2, 1, 2, 0, 0, 0, 0, 0],
        [0, 0, 1, 2, 1, 3, 1, 0, 0, 0, 0],
        [0, 0, 2, 2, 1, 3, 2, 0, 0, 0, 0],
        [0, 1, 2, 2, 1, 3, 3, 1, 0, 1, 1],
        [0, 2, 2, 2, 1, 3, 3, 2, 0, 2, 1],
        [1, 2, 2, 2, 1, 3, 3, 3, 1, 3, 1]])

and thus

In [43]: ((arr == 1) & (arr.cumsum(axis=0) == 1)).astype(int)
Out[43]: 
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]])

Solution 3:

another approach is:

for i in range(a.shape[1]):
    a[np.where(a[:,i]==1)[0][1:],i] = 0

output:

[[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]]

Solution 4:

You can use nonzero and unique function:

c, r = np.nonzero(np_array.T)
_, ind = np.unique(c, return_index=True)
np_array[:] = 0
np_array[r[ind], c[ind]] = 1

Given the example, the result:

[[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]]