hello =  ['h', 'e', 'l', 'l', 'o', ' ', 'w', 'o', 'r', 'l', 'd']
#chech every item for a match
for item in range(len(hello)-1):
     if hello[item] == ' ': 
#if there is a match, rebuild the list with the list before the item + the list after the item
         hello = hello[:item] + hello [item + 1:]
print hello

[' h ',‘e’,‘l’,‘l’,‘o’,‘w’,‘o’,‘r’,‘l’,' d ')

你可以使用列表推导式:

def remove_values_from_list(the_list, val):
   return [value for value in the_list if value != val]

x = [1, 2, 3, 4, 2, 2, 3]
x = remove_values_from_list(x, 2)
print x
# [1, 3, 4, 3]

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))

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]

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

很多答案都很好。如果你是python初学者，这里有一个简单的方法，以防你肯定想使用remove()方法。

rawlist = [8, 1, 8, 5, 8, 2, 8, 9, 8, 4]

ele_remove = 8

for el in rawlist:
    if el == ele_remove:
        rawlist.remove(ele_remove)

对于太大的列表，它可能会慢一些。