Splitting Numpy array based on value

Question

Suppose I have this NumPy array:

a = np.array([0, 3, 5, 5, 0, 10, 14, 15, 56, 0, 12, 23, 45, 23, 12, 45, 
              0, 1, 0, 2, 3, 4, 0, 0 ,0])
<         


        

Answer 1

This is probably the worst way to do this, but you could also convert your array into a string and then split that string a couple times:

long_string = "_".join(a.astype(str))

while long_string.startswith("0_"):
    long_string = long_string.removeprefix("0_")
while long_string.endswith("_0"):
    long_string = long_string.removesuffix("_0")

result = [list(map(int, i.split("_"))) for i in long_string.split("_0_")]

# result: [[3, 5, 5], [10, 14, 15, 56], [12, 23, 45, 23, 12, 45], [1], [2, 3, 4]]

You would need Python 3.9 for .removeprefix() and .removesuffix().

Answer 2

No need for numpy, this lambda function works on a list, but we can convert your numpy array to and from a list on the way in and out:

cut = lambda x: [j for j in [cut(x[:x.index(0)])]+cut(x[x.index(0)+1:]) if j] if x.count(0) else x

numpy.array(cut(list(a)))

# array([[3, 5, 5], [10, 14, 15, 56], [12, 23, 45, 23, 12, 45], [1], [2, 3, 4]], dtype=object)

Answer 3

NumPy's split() and where() in a list compehension:

[x[x!=0] for x in np.split(a, np.where(a==0)[0]) if len(x[x!=0])]

[array([3, 5, 5]),
 array([10, 14, 15, 56]),
 array([12, 23, 45, 23, 12, 45]),
 array([1]),
 array([2, 3, 4])]

Answer 4

You can use groupby() function from itertools, and specify the key as the boolean condition of zero or nonzero. In such a way, all consecutive zeros and nonzeros will be grouped together. Use if filter to pick up groups of nonzeros and use list to convert the non zero groupers to lists.

from itertools import groupby
[list(g) for k, g in groupby(a, lambda x: x != 0) if k]

# [[3, 5], [10, 14, 15, 56], [12, 23, 45, 23, 12, 45], [1], [2, 3, 4]]

Answer 5

You can get the indices of zeros with np.where:

zeros = np.where(a == 0)[0]

And iterate over every pair to slice the array:

[a[i+1:j] for i, j in zip(zeros, zeros[1:]) if len(a[i+1:j])>0]

Out[46]: 
[array([3, 5]),
 array([10, 14, 15, 56]),
 array([12, 23, 45, 23, 12, 45]),
 array([1]),
 array([2, 3, 4])]

Answer 6

Here's a vectorized approach using np.where and np.split -

idx = np.where(a!=0)[0]
aout = np.split(a[idx],np.where(np.diff(idx)!=1)[0]+1)

Sample run -

In [23]: a
Out[23]: 
array([ 0,  3,  5,  5,  0, 10, 14, 15, 56,  0,  0,  0, 12, 23, 45, 23, 12,
       45,  0,  1,  0,  2,  3,  4,  0,  0,  0])

In [24]: idx = np.where(a!=0)[0]

In [25]: np.split(a[idx],np.where(np.diff(idx)!=1)[0]+1)
Out[25]: 
[array([3, 5, 5]),
 array([10, 14, 15, 56]),
 array([12, 23, 45, 23, 12, 45]),
 array([1]),
 array([2, 3, 4])]