The problem
Create a operate, that accepts an arbitrary variety of arrays and returns a single array generated by alternately appending components from the passed-in arguments. If certainly one of them is shorter than the others, the outcome needs to be padded with empty components.
Examples:
interleave([1, 2, 3], ["c", "d", "e"]) == [1, "c", 2, "d", 3, "e"]
interleave([1, 2, 3], [4, 5]) == [1, 4, 2, 5, 3, None]
interleave([1, 2, 3], [4, 5, 6], [7, 8, 9]) == [1, 4, 7, 2, 5, 8, 3, 6, 9]
interleave([]) == []
The answer in Python code
Possibility 1:
from itertools import chain, zip_longest
def interleave(*args):
return record(chain.from_iterable(zip_longest(*args)))
Possibility 2:
def interleave(*args):
max_len = max(map(len,args))
interleaved = []
for i in vary(max_len):
for arr in args:
if i < len(arr):
interleaved.append(arr[i])
else:
interleaved.append(None)
return interleaved
Possibility 3:
interleave=lambda *a:sum([list(i) for i in __import__('itertools').zip_longest(*a)],[])
Take a look at circumstances to validate our answer
check.assert_equals(interleave([1, 2, 3], ["c", "d", "e"]), [1, "c", 2, "d", 3, "e"])
check.assert_equals(interleave([1, 2, 3], [4, 5]), [1, 4, 2, 5, 3, None])
check.assert_equals(interleave([1, 2, 3], [4, 5, 6], [7, 8, 9]), [1, 4, 7, 2, 5, 8, 3, 6, 9])
check.assert_equals(interleave([]), [])
