a = [1, 2, 2, 3, 1]
to_remove = 1
a = [i for i in a if i != to_remove]
print(a)

也许不是最蟒蛇的，但对我来说仍然是最简单的，哈哈

关于速度!

import time
s_time = time.time()

print 'start'
a = range(100000000)
del a[:]
print 'finished in %0.2f' % (time.time() - s_time)
# start
# finished in 3.25

s_time = time.time()
print 'start'
a = range(100000000)
a = []
print 'finished in %0.2f' % (time.time() - s_time)
# start
# finished in 2.11

删除所有重复的出现，并在列表中保留一个:

test = [1, 1, 2, 3]

newlist = list(set(test))

print newlist

[1, 2, 3]

下面是我在Project Euler中使用的函数:

def removeOccurrences(e):
  return list(set(e))

以可读性为代价，我认为这个版本稍微快一点，因为它不强迫while重新检查列表，因此做完全相同的工作删除必须做的事情:

x = [1, 2, 3, 4, 2, 2, 3]
def remove_values_from_list(the_list, val):
    for i in range(the_list.count(val)):
        the_list.remove(val)

remove_values_from_list(x, 2)

print(x)

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

我相信要简单得多。

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