根据每个列表的第一个元素从列表中删除项目 [英] Removing an item from a list of lists based on each of the lists first element

问题描述

给出:

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

我要删除以b作为第一项的a项.因此，在这种情况下，我们将删除[3,4]给予:

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]]

我当前的代码是:

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

这有效，但是速度很慢.更好的方法是什么?

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

我之前没有测试过性能，所以我可能做错了，但是我得到了:

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

推荐答案

反向删除a，对其进行就地修改:

Reverse delete a, modifying it in-place:

for i in reversed(range(len(a))):
    if a[i][0] == 3:
        del a[i]

就地修改意味着它效率更高，因为它不会创建新列表(如列表理解那样).

An in-place modification means that this is more efficient, since it does not create a new list (as a list comprehension would).

由于OP请求一种高效的解决方案，因此下面是两个投票最高的答案之间的timeit比较.

Since OP requests a performant solution, here's a timeit comparison between the two top voted answers here.

设置-

a = np.random.choice(4, (100000, 2)).tolist()

print(a[:5])
[[2, 1], [2, 2], [3, 2], [3, 3], [3, 1]]

列表理解-

%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)

反向删除-

%%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)

它们确实很接近，但是反向删除的性能提高了1UP，因为它不必像列表理解那样在内存中生成新列表.

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.

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