for i in range(a.count(' ')):
    a.remove(' ')

我相信要简单得多。

更好的解决方案与列表理解

x = [ i for i in x if i!=2 ]

功能的方法:

Python 3. x

>>> x = [1,2,3,2,2,2,3,4]
>>> list(filter((2).__ne__, x))
[1, 3, 3, 4]

or

>>> x = [1,2,3,2,2,2,3,4]
>>> list(filter(lambda a: a != 2, x))
[1, 3, 3, 4]

or

>>> [i for i in x if i != 2]

Python 2. x

>>> x = [1,2,3,2,2,2,3,4]
>>> filter(lambda a: a != 2, x)
[1, 3, 3, 4]

Let

>>> x = [1, 2, 3, 4, 2, 2, 3]

最简单有效的解决方案是

>>> x[:] = [v for v in x if v != 2]
>>> x
[1, 3, 4, 3]

另一种使用较少内存但速度较慢的方法是

>>> for i in range(len(x) - 1, -1, -1):
        if x[i] == 2:
            x.pop(i)  # takes time ~ len(x) - i
>>> x
[1, 3, 4, 3]

长度为1000和100000且有10%匹配条目的列表的计时结果:0.16 vs 0.25 ms, 23 vs 123 ms。

Numpy方法和对包含1.000.000个元素的列表/数组的计时:

计时:

In [10]: a.shape
Out[10]: (1000000,)

In [13]: len(lst)
Out[13]: 1000000

In [18]: %timeit a[a != 2]
100 loops, best of 3: 2.94 ms per loop

In [19]: %timeit [x for x in lst if x != 2]
10 loops, best of 3: 79.7 ms per loop

结论:numpy(在我的笔记本上)比列表理解方法快27倍

PS如果你想将常规的Python列表lst转换为numpy数组:

arr = np.array(lst)

设置:

import numpy as np
a = np.random.randint(0, 1000, 10**6)

In [10]: a.shape
Out[10]: (1000000,)

In [12]: lst = a.tolist()

In [13]: len(lst)
Out[13]: 1000000

检查:

In [14]: a[a != 2].shape
Out[14]: (998949,)

In [15]: len([x for x in lst if x != 2])
Out[15]: 998949

以更抽象的方式重复第一篇文章的解决方案:

>>> x = [1, 2, 3, 4, 2, 2, 3]
>>> while 2 in x: x.remove(2)
>>> x
[1, 3, 4, 3]