If the size of a list is known, then pre-allocating the list can provide
performance improvements; especially for a large list. The example below
measures the performance of appending items to an empty list. Pre-allocating
a list of known size and assigning items to it gives better performance than
appending to the list. The list comprehension approach (see Example 3) gives
the best elapsed time.
Example 1 = 11.6552 ms
Example 2 = 8.9583 ms
Example 3 = 4.5061 ms
import time
# Size of the list
n = 100_000
# Example 1. Create an empty list and append items to it.
# ----------------------------------------------------------------------------
tic = time.perf_counter_ns()
list_one = []
for i in range(n):
list_one.append(i)
toc = time.perf_counter_ns()
print(f'Ex 1 elapsed time {(toc - tic) / 1_000_000:.4f} ms')
print(f'list_one = {list_one[:3]}...{list_one[-3:]}\n')
# Example 2. Pre-allocate list of None items then assign items at each index.
# ----------------------------------------------------------------------------
tic = time.perf_counter_ns()
list_two = [None] * n
for i in range(n):
list_two[i] = i
toc = time.perf_counter_ns()
print(f'Ex 2 elapsed time {(toc - tic) / 1_000_000:.4f} ms')
print(f'list_two = {list_two[:3]}...{list_two[-3:]}\n')
# Example 3. Pre-allocate list and assign items using list comprehension.
# ----------------------------------------------------------------------------
tic = time.perf_counter_ns()
list_three = [i for i in range(n)]
toc = time.perf_counter_ns()
print(f'Ex 3 elapsed time {(toc - tic) / 1_000_000:.4f} ms')
print(f'list_three = {list_three[:3]}...{list_three[-3:]}')
