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?
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
一些最简单的基准测试方法:
import random
from copy import copy
sample = random.sample(range(100000), 10000)
remove = random.sample(range(100000), 1000)
%%timeit
sample1 = copy(sample)
remove1 = copy(remove)
for i in reversed(sample1):
if i in remove1:
sample1.remove(i)
# 271 ms ± 16 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
# remove all instances
%%timeit
sample1 = copy(sample)
remove1 = copy(remove)
filtered = list(filter(lambda x: x not in remove1, sample1))
# 280 ms ± 18.9 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
# remove all instances
%%timeit
sample1 = copy(sample)
remove1 = copy(remove)
filtered = [ele for ele in sample1 if ele not in remove1]
# 293 ms ± 72.1 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
# remove all instances
%%timeit
sample1 = copy(sample)
remove1 = copy(remove)
for val in remove1:
if val in sample1:
sample1.remove(val)
# 558 ms ± 40.7 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
# only remove first occurrence
%%timeit
sample1 = copy(sample)
remove1 = copy(remove)
for val in remove1:
try:
sample1.remove(val)
except:
pass
# 609 ms ± 11.7 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
# only remove first occurrence
其他回答
在列表中查找一个值,然后删除该索引(如果它存在的话),使用list的remove方法更容易完成:
>>> a = [1, 2, 3, 4]
>>> try:
... a.remove(6)
... except ValueError:
... pass
...
>>> print a
[1, 2, 3, 4]
>>> try:
... a.remove(3)
... except ValueError:
... pass
...
>>> print a
[1, 2, 4]
如果你经常这样做,你可以把它包在一个函数中:
def remove_if_exists(L, value):
try:
L.remove(value)
except ValueError:
pass
如果你知道要删除什么值,这里有一个简单的方法(就像我能想到的一样简单):
a = [0, 1, 1, 0, 1, 2, 1, 3, 1, 4]
while a.count(1) > 0:
a.remove(1)
你会得到 [0,0,2,3,4]
你可以这样做
a=[1,2,3,4]
if 6 in a:
a.remove(6)
但以上需要在list a中搜索2次6,所以尝试except会更快
try:
a.remove(6)
except:
pass
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')
list1=[1,2,3,3,4,5,6,1,3,4,5]
n=int(input('enter number'))
while n in list1:
list1.remove(n)
print(list1)