I have a list of size < N and I want to pad it up to the size N with a value.
Certainly, I can use something like the following, but I feel that there should be something I missed:
>>> N = 5
>>> a = [1]
>>> map(lambda x, y: y if x is None else x, a, ['']*N)
[1, '', '', '', '']
If you want to pad with None instead of '', map() does the job:
>>> map(None,[1,2,3],xrange(7))
[(1, 0), (2, 1), (3, 2), (None, 3), (None, 4), (None, 5), (None, 6)]
>>> zip(*map(None,[1,2,3],xrange(7)))[0]
(1, 2, 3, None, None, None, None)
extra_length = desired_length - len(l)
l.extend(value for _ in range(extra_length))
This avoids any extra allocation, unlike any solution that depends on creating and appending the list [value] * extra_length. The "extend" method first calls __length_hint__ on the iterator, and extends the allocation for l by that much before filling it in from the iterator.
you can use * iterable unpacking operator:
N = 5
a = [1]
pad_value = ''
pad_size = N - len(a)
final_list = [*a, *[pad_value] * pad_size]
print(final_list)
output:
[1, '', '', '', '']
Using iterators and taking advantage of the default argument for next:
i = iter(a)
a = [next(i, '') for _ in range(N)]
Either way we want to produce N items. Hence the for _ in range(N). Then the elements should be as much as we can from a and the rest ''. Using an iterator over a we grab all possible elements, and when we get StopIteration, the default will be returned which is ''.
Adding to the existing list with np.repeat:
import numpy as np
a + list(np.repeat([''], (N - len(a))))
Adding the padding before the list of elements
a[:0] += [''] * (N - len(a))
Adding the padding after the list of elements
a += [''] * (N - len(a))
A pythonic way to pad your list with empty elements is using list comprehension.
my_list = [1,2]
desired_len = 3
# Ensure that the length of my list is 3 elements
[my_list.extend(['']) for _ in range(desired_len - len(my_list))]
[my_list.pop() for _ in range(len(my_list)-desired_len )]