How to obtain only the first True value from each row of a numpy array?
Hire the world's top talent on demand or became one of them at Toptal: https://topt.al/25cXVn
and get $2,000 discount on your first invoice
--------------------------------------------------
Music by Eric Matyas
https://www.soundimage.org
Track title: Cool Puzzler LoFi
--
Chapters
00:00 How To Obtain Only The First True Value From Each Row Of A Numpy Array?
01:00 Accepted Answer Score 16
02:12 Answer 2 Score 4
02:57 Thank you
--
Full question
https://stackoverflow.com/questions/5415...
--
Content licensed under CC BY-SA
https://meta.stackexchange.com/help/lice...
--
Tags
#python #arrays #numpy
#avk47
ACCEPTED ANSWER
Score 16
You can use cumsum, and find the first bool by comparing the result with 1.
all_bools.cumsum(axis=1).cumsum(axis=1) == 1 
array([[False,  True, False],
       [ True, False, False],
       [False, False,  True],
       [False, False, False]])
This also accounts for the issue @a_guest pointed out. The second cumsum call is needed to avoid matching all False values between the first and second True value.
If performance is important, use argmax and set values:
y = np.zeros_like(all_bools, dtype=bool)
idx = np.arange(len(x)), x.argmax(axis=1)
y[idx] = x[idx]
y
array([[False,  True, False],
       [ True, False, False],
       [False, False,  True],
       [False, False, False]])
Perfplot Performance Timings
I'll take this opportunity to show off perfplot, with some timings, since it is good to see how our solutions vary with different sized inputs.
import numpy as np
import perfplot
def cs1(x):
    return  x.cumsum(axis=1).cumsum(axis=1) == 1 
def cs2(x):
    y = np.zeros_like(x, dtype=bool)
    idx = np.arange(len(x)), x.argmax(axis=1)
    y[idx] = x[idx]
    return y
def a_guest(x):
    b = np.zeros_like(x, dtype=bool)
    i = np.argmax(x, axis=1)
    b[np.arange(i.size), i] = np.logical_or.reduce(x, axis=1)
    return b
perfplot.show(
    setup=lambda n: np.random.randint(0, 2, size=(n, n)).astype(bool),
    kernels=[cs1, cs2, a_guest],
    labels=['cs1', 'cs2', 'a_guest'],
    n_range=[2**k for k in range(1, 8)],
    xlabel='N'
)
The trend carries forward to larger N. cumsum is very expensive, while there is a constant time difference between my second solution, and @a_guest's.
ANSWER 2
Score 4
You can use the following approach using np.argmax and a product with np.logical_or.reduce for dealing with rows that are all False:
b = np.zeros_like(a, dtype=bool)
i = np.argmax(a, axis=1)
b[np.arange(i.size), i] = np.logical_or.reduce(a, axis=1)
Timing results
Different versions in increasing performance, i.e. fastest approach comes last:
In [1]: import numpy as np
In [2]: def f(a):
   ...:     return a.cumsum(axis=1).cumsum(axis=1) == 1
   ...: 
   ...: 
In [3]: def g(a):
   ...:     b = np.zeros_like(a, dtype=bool)
   ...:     i = np.argmax(a, axis=1)
   ...:     b[np.arange(i.size), i] = np.logical_or.reduce(a, axis=1)
   ...:     return b
   ...: 
   ...: 
In [4]: x = np.random.randint(0, 2, size=(1000, 1000)).astype(bool)
In [5]: %timeit f(x)
10.4 ms ± 155 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
In [6]: %timeit g(x)
120 µs ± 184 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
In [7]: def h(a):
   ...:     y = np.zeros_like(x)
   ...:     idx = np.arange(len(x)), x.argmax(axis=1)
   ...:     y[idx] += x[idx]
   ...:     return y
   ...: 
   ...: 
In [8]: %timeit h(x)
92.1 µs ± 3.51 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
In [9]: def h2(a):
    ...:     y = np.zeros_like(x)
    ...:     idx = np.arange(len(x)), x.argmax(axis=1)
    ...:     y[idx] = x[idx]
    ...:     return y
    ...: 
    ...: 
In [10]: %timeit h2(x)
78.5 µs ± 353 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
