Trying to remove all the leading zeroes from a list of array using next() and enumerate within a list comprehension. Came across the below code which works. Can anyone explain clearly what the code does.
example : result = [0,0,1,2,0,0,3] returns result = [1,2,0,0,3]
Edited* - the code just removes the leading zeroes
result = result[next((i for i, x in enumerate(result) if x != 0), len(result)):]
print(result)
Trying to remove all the leading zeroes from a list of array using next() and enumerate within a list comprehension.
Are you obligated to use next(), enumerate() and a list comprehension? An alternate approach:
from itertools import dropwhile
from operator import not_ as is_zero
result = dropwhile(is_zero, [0, 0, 1, 2, 0, 0, 3])
print(*result)
OUTPUT
% python3 test.py
1 2 0 0 3
%
We can potentially explain the original code:
result = [0, 0, 1, 2, 0, 0, 3]
result[next((i for i, x in enumerate(result) if x != 0), len(result)):]
By breaking it down into pieces and executing them:
enumerate(result) # list of indexes and values [(i0, x0), (i1, x1), ...]
[(0, 0), (1, 0), (2, 1), (3, 2), (4, 0), (5, 0), (6, 3)]
[i for i, x in enumerate(result)] # just the indexes
[i for i, x in [(0, 0), (1, 0), ..., (5, 0), (6, 3)]] # what effectively happens
[0, 1, 2, 3, 4, 5, 6]
[i for i, x in enumerate(result) if x != 0] # just the indexes of non-zero values
[2, 3, 6]
# not needed with this example input, used to make an all
# zero list like [0, 0, ..., 0] return the empty list []
len(result)
7
# pull off the first element of list of indexes of non-zero values
next((i for i, x in enumerate(result) if x != 0), len(result))
next(iter([2, 3, 6]), 7) # what effectively happens
2
result[next((i for i, x in enumerate(result) if x != 0), len(result)):] # slice
result[2:] # what effectively happens
[1, 2, 0, 0, 3]