Reverse delete a
, modifying it in-place:
for i in reversed(range(len(a))):
if a[i][0] == 3:
del a[i]
An in-place modification means that this is more efficient, since it does not create a new list (as a list comprehension would).
Since OP requests a performant solution, here’s a timeit
comparison between the two top voted answers here.
Setup –
a = np.random.choice(4, (100000, 2)).tolist()
print(a[:5])
[[2, 1], [2, 2], [3, 2], [3, 3], [3, 1]]
List comprehension –
%timeit [x for x in a if x[0] != b]
11.1 ms ± 685 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
Reverse delete –
%%timeit
for i in reversed(range(len(a))):
if a[i][0] == 3:
del a[i]
10.1 ms ± 146 µs per loop (mean ± std. dev. of 7 runs, 1 loop each)
They’re really close, but reverse delete has a 1UP on performance because it doesn’t have to generate a new list in memory, as the list comprehension would.