How to determine python array element periodically-CodePudding

I have a python array of a certain number of elements and I want to keep 5 elements, then discard 15, keep 5, discard 15, etc.

Is there a fast way to do this?

CodePudding user response：

You can use a good old list comprehension and enumerate. Take the modulo. Values 1-5 are modulo 0-4:

[v for i,v in enumerate(your_list) if i%(15 5)<5]

Example:

your_list = list(range(100))
[v for i,v in enumerate(your_list) if i%20<5]

output:

[0, 1, 2, 3, 4, 20, 21, 22, 23, 24, 40, 41, 42, 43, 44, 60, 61, 62, 63, 64, 80, 81, 82, 83, 84]

CodePudding user response：

You can use simple slicing and appropriate start/stop/step calculations:

arr = list(range(100))

from itertools import chain

take, skip = 5, 15
list(chain.from_iterable(arr[i:i take] for i in range(0, len(arr), take skip)))
# [0, 1, 2, 3, 4, 
#  20, 21, 22, 23, 24,
#  40, 41, 42, 43, 44,
#  60, 61, 62, 63, 64, 
#  80, 81, 82, 83, 84]

CodePudding user response：

If you use `numpy`.

Given that l is:

l = np.arange(100)

Use:

>>> np.split(l, range(5, len(l), 5))[::4]
[array([0, 1, 2, 3, 4]), array([20, 21, 22, 23, 24]), array([40, 41, 42, 43, 44]), array([60, 61, 62, 63, 64]), array([80, 81, 82, 83, 84])]
>>>

For an 2d array instead of a list of 1d arrays, try:

>>> np.array(np.split(l, range(5, len(l), 5))[::4])
array([[ 0,  1,  2,  3,  4],
       [20, 21, 22, 23, 24],
       [40, 41, 42, 43, 44],
       [60, 61, 62, 63, 64],
       [80, 81, 82, 83, 84]])
>>>

CodePudding user response：

The following solution, in addition to not using any import, is also quite fast:

def skip_elements(l, n1, n2):
    i1 = range(len(l))[::n1 n2]
    i2 = [x   n1 for x in i1]
    return [l[x:y] for x, y in zip(i1, i2)]

test = list(range(1011))
n_elements_to_keep = 5
n_elements_to_skip = 15

print(skip_elements(test, n_elements_to_keep, n_elements_to_skip))

If you need the result as one list (rather that a list of lists), you can change the return statement in def skip_elements as follows:

return sum([l[x:y] for x, y in zip(i1, i2)],[])

Anyway, if you need a list of lists, this is the fastest solution from what I can see. Vice versa, if one list is the desired output, the itertools one is the best. Despite being quite elegant (I must admit - I voted it ), the list comprehension one is the slowest among the ones shown.

If you use numpy.

If you use `numpy`.