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
我们也可以使用.pop:
>>> lst = [23,34,54,45]
>>> remove_element = 23
>>> if remove_element in lst:
... lst.pop(lst.index(remove_element))
...
23
>>> lst
[34, 54, 45]
>>>
其他回答
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:]
这将从数组sys中删除"-v"的所有实例。Argv,如果没有找到实例,则不报错:
while "-v" in sys.argv:
sys.argv.remove('-v')
你可以在一个名为speechToText.py的文件中看到代码的运行:
$ python speechToText.py -v
['speechToText.py']
$ python speechToText.py -x
['speechToText.py', '-x']
$ python speechToText.py -v -v
['speechToText.py']
$ python speechToText.py -v -v -x
['speechToText.py', '-x']
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
通常,如果你告诉Python做一些它不能做的事情,Python会抛出一个异常,所以你必须这样做:
if c in a:
a.remove(c)
or:
try:
a.remove(c)
except ValueError:
pass
异常不一定是坏事,只要它是您所期望的并正确处理的。
我们也可以使用.pop:
>>> lst = [23,34,54,45]
>>> remove_element = 23
>>> if remove_element in lst:
... lst.pop(lst.index(remove_element))
...
23
>>> lst
[34, 54, 45]
>>>