如果列表只包含0，我如何在python中打印？

how do i print in python if a list contains only 0s??

如果列表只包含0s，我如何在python中打印？

list1=[0,0,0,0,0,0]
if list1 has all 0s
print("something")

I want the output to be "something"

我希望输出是“东西”

3 个解决方案

#1

Use all():

使用all（）：

if all(item == 0 for item in list1):
   print("something")

Demo:

演示：

>>> list1 = [0,0,0,0,0,0]
>>> all(item == 0 for item in list1)
True

Another alternative will be to use sets, if all the items in list are hashable:

另一种选择是使用集合，如果列表中的所有项目都是可清除的：

>>> set(list1) == {0}
True

But this will create a set in memory and it won't short-circuit like all(), so it is going to be memory inefficient and slow in average cases.

但是这将在内存中创建一个集合并且它不会像all（）那样短路，因此在内存中效率低且在平均情况下会很慢。

>>> list1 = [0,0,0,0,0,0]*1000 + range(1000)
>>> %timeit set(list1) == {0}
1000 loops, best of 3: 292 us per loop
>>> %timeit all(item == 0 for item in list1)
1000 loops, best of 3: 1.04 ms per loop

>>> list1 = range(1000) + [0,0,0,0,0,0]*10
>>> shuffle(list1)
>>> %timeit set(list1) == {0}
10000 loops, best of 3: 61.6 us per loop
>>> %timeit all(item == 0 for item in list1)
1000000 loops, best of 3: 1.3 us per loop

#2

I think a very fast way is to use [].count

我认为一种非常快速的方法是使用[] .count

L.count(0) == len(L)

if the list is HUGE and most not being zeros then all with an iterator may be better, however.

如果列表是巨大的并且大多数不是零，则所有使用迭代器的可能更好。

#3

You can skip the list comprehension / generator expression by doing:

您可以通过执行以下操作跳过列表理解/生成器表达式：

if not any(list1):

#1