Python 在不使用 `not` 命令的情况下检查列表是否为空
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Check if list is empty without using the `not` command
提问by user2240288
How can I find out if a list is empty without using the not command?
Here is what I tried:
如何在不使用 not 命令的情况下确定列表是否为空？
这是我尝试过的：
if list3[0] == []:
print("No matches found")
else:
print(list3)
I am very much a beginner so excuse me if I do dumb mistakes.
我是一个初学者，所以如果我犯了愚蠢的错误，请原谅。
采纳答案by John Kugelman
In order of preference:
按优先顺序：
# Good
if not list3:
# Okay
if len(list3) == 0:
# Ugly
if list3 == []:
# Silly
try:
next(iter(list3))
# list has elements
except StopIteration:
# list is empty
If you have both an if and an else you might also re-order the cases:
如果您同时拥有 if 和 else，您还可以重新排序这些案例：
if list3:
# list has elements
else:
# list is empty
回答by Ali Afshar
Check its length.
检查其长度。
l = []
print len(l) == 0
回答by dawg
You find out if a list is empty by testing the 'truth' of it:
您可以通过测试列表的“真相”来确定列表是否为空：
>>> bool([])
False
>>> bool([0])
True
While in the second case 0is False, but the list [0]is True because it contains something. (If you want to test a list for containing all falsey things, use allor any: any(e for e in li)is True if any item in liis truthy.)
虽然在第二种情况下0是 False，但列表[0]是 True，因为它包含一些东西。（如果要测试包含所有虚假内容的列表，请使用all或any: any(e for e in li)is True 如果其中的任何项目li为真。）
This results in this idiom:
这导致了这个习语：
if li:
# li has something in it
else:
# optional else -- li does not have something
if not li:
# react to li being empty
# optional else...
According to PEP 8, this is the proper way:
根据PEP 8，这是正确的方法：
? For sequences, (strings, lists, tuples), use the fact that empty sequences are false.
Yes: if not seq: if seq: No: if len(seq) if not len(seq)
? 对于序列（字符串、列表、元组），使用空序列为假的事实。
Yes: if not seq: if seq: No: if len(seq) if not len(seq)
You test if a list has a specific index existing by using try:
您可以使用以下命令测试列表是否存在特定索引try：
>>> try:
... li[3]=6
... except IndexError:
... print 'no bueno'
...
no bueno
So you may want to reverse the order of your code to this:
因此，您可能希望将代码的顺序颠倒为：
if list3:
print list3
else:
print "No matches found"
回答by Thanakron Tandavas
this is how you would do that
这就是你会怎么做
if len(list3) == 0:
print("No matches found")
回答by Kartik
Python provides an inbuilt any() function to check whether an iterable is empty or not:
Python 提供了一个内置的 any() 函数来检查可迭代对象是否为空：
>>>list=[]
>>>any(list)
False
The function returns True if the iterable contains a 'True' value, and False otherwise.
如果可迭代对象包含“真”值，则该函数返回 True，否则返回 False。
However, note that the list [0] also returns False with any().
但是，请注意列表 [0] 也使用 any() 返回 False。
