要删除一个元素的第一次出现，使用list.remove:

>>> xs = ['a', 'b', 'c', 'd']
>>> xs.remove('b')
>>> print(xs)
['a', 'c', 'd']

要移除所有出现的元素，使用列表推导式:

>>> xs = ['a', 'b', 'c', 'd', 'b', 'b', 'b', 'b']
>>> xs = [x for x in xs if x != 'b']
>>> print(xs)
['a', 'c', 'd']

正如许多其他答案所述，list.remove()将工作，但如果项不在列表中则抛出ValueError。在python 3.4+中，有一个有趣的方法来处理这个问题，使用suppress contextmanager:

from contextlib import suppress
with suppress(ValueError):
    a.remove('b')

以下是如何做到这一点(不需要理解列表):

def remove_all(seq, value):
    pos = 0
    for item in seq:
        if item != value:
           seq[pos] = item
           pos += 1
    del seq[pos:]

arr = [1, 1, 3, 4, 5, 2, 4, 3]

# to remove first occurence of that element, suppose 3 in this example
arr.remove(3)

# to remove all occurences of that element, again suppose 3
# use something called list comprehension
new_arr = [element for element in arr if element!=3]

# if you want to delete a position use "pop" function, suppose 
# position 4 
# the pop function also returns a value
removed_element = arr.pop(4)

# u can also use "del" to delete a position
del arr[4]

这个例子是快速的，并且会从列表中删除一个值的所有实例:

a = [1,2,3,1,2,3,4]
while True:
    try:
        a.remove(3)
    except:
        break
print a
>>> [1, 2, 1, 2, 4]

一句话:

a.remove('b') if 'b' in a else None

有时它很有用。

更简单:

if 'b' in a: a.remove('b')