我想从列表中删除一个值,如果它存在于列表中(它可能不存在)。

a = [1, 2, 3, 4]
b = a.index(6)

del a[b]
print(a)

上面给出了错误:

ValueError: list.index(x): x not in list

所以我必须这样做:

a = [1, 2, 3, 4]

try:
    b = a.index(6)
    del a[b]
except:
    pass

print(a)

难道没有更简单的方法吗?


当前回答

如果你的元素是不同的,那么一个简单的集合差异就可以了。

c = [1,2,3,4,'x',8,6,7,'x',9,'x']
z = list(set(c) - set(['x']))
print z
[1, 2, 3, 4, 6, 7, 8, 9]

其他回答

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]

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')

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')

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']

这就是我的答案,用while和for就可以了

def remove_all(data, value):
    i = j = 0
    while j < len(data):
        if data[j] == value:
            j += 1
            continue
        data[i] = data[j]
        i += 1
        j += 1
    for x in range(j - i):
        data.pop()