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
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]
其他回答
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
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。
另一种可能是使用set而不是list,如果set适用于您的应用程序。
IE中如果你的数据不是有序的,并且没有副本,那么
my_set=set([3,4,2])
my_set.discard(1)
是没有错误的。
通常,列表只是一个方便的容器,用于存放实际上是无序的项目。有些问题询问如何从列表中删除一个元素的所有出现。如果你一开始就不想被骗,一套也很方便。
my_set.add(3)
不改变上面的my_set。
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]
这是一个效率较低的解决方案,但它仍然有效:
A =[] //这是你的列表
B //需要删除的元素
counter = a.count(b)
while counter > 0:
if b in a:
a.remove(b)
counter -= 1
print(a)