Removing an item from a list of lists based on each of the lists first element

Question

Given:

a = [[1,2],[3,4],[5,6],[7,8]]
b = 3

I would like to remove an item of a that has b as it's first item. So in this case we would remove [3,4] to give:

a = [[1,2],[5,6],[7,8]]

My current code is:

if b in [i[0] for i in a]:
    pos = [i[0] for i in a].index(b)
       del a[pos]

This works but it is slow. What would a better way to do this be?

EDIT: I've not tested performance before so I may be doing this wrong but I get this:

def fun1():
    lst = [[x, 2*x] for x in range(1000000)]
    lst = [x for x in lst if x[0] != 500]
    return lst

def fun2():
    lst = [[x, 2*x] for x in range(1000000)]
    for i in reversed(range(len(lst))):
        if lst[i][0] == 500:
            del lst[i]
    return lst

cProfile.runctx('fun1()', None, locals())
        6 function calls in 0.460 seconds

cProfile.runctx('fun2()', None, locals())
        6 function calls in 0.502 seconds

Answer 1

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.

Answer 2

You can use a list comprehension:

>>> a = [[1,2],[3,4],[5,6],[7,8]]
>>> b = 3
>>> a = [x for x in a if x[0] != b]
>>> a
[[1, 2], [5, 6], [7, 8]]

Answer 3

for i in a[:-1]:
    if i[0]==b:
        a.remove(i)

What about this?

The output is

[[1, 2], [5, 6], [7, 8]]

Answer 4

If your list are small then you are also try filter ,

a = [[1,2],[3,4],[5,6],[7,8]]
b = 3

print(list(filter(lambda x:x[0]!=b,a)))

output:

[[1, 2], [5, 6], [7, 8]]