Python：确定列表的所有项目是否为同一项目
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Python: determine if all items of a list are the same item
提问by Einar
In some of my code I put a series of objects in a list and I build an additional list out of their attributes, which is a string. I need to determine if all the items in this second list have the exact same value, without knowing beforehand which value it is, and return a bool so that I can do different things in my code depending on the result.
在我的一些代码中，我将一系列对象放在一个列表中，并根据它们的属性构建了一个附加列表，它是一个字符串。我需要确定第二个列表中的所有项目是否具有完全相同的值，而无需事先知道它是哪个值，并返回一个 bool 以便我可以根据结果在我的代码中执行不同的操作。
I can't know the names of the properties beforehand, that is why I'm trying to make something as generic as possible.
我无法事先知道属性的名称，这就是为什么我要尝试尽可能通用的原因。
To make the example clear, an ideal function, called "all_same" would work like this:
为了使示例清晰，一个名为“all_same”的理想函数将如下工作：
>>> property_list = ["one", "one", "one"]
>>> all_same(property_list)
True
>>> property_list = ["one", "one", "two"]
>>> all_same(property_list)
False
I was thinking of making a list of unique elements and then check if its length is 1, but I'm not sure if it's the most elegant solution out there.
我正在考虑制作一个唯一元素列表，然后检查它的长度是否为 1，但我不确定它是否是最优雅的解决方案。
采纳答案by FogleBird
def all_same(items):
return all(x == items[0] for x in items)
Example:
例子：
>>> def all_same(items):
... return all(x == items[0] for x in items)
...
>>> property_list = ["one", "one", "one"]
>>> all_same(property_list)
True
>>> property_list = ["one", "one", "two"]
>>> all_same(property_list)
False
>>> all_same([])
True
回答by wheaties
You could cheat and use set:
你可以欺骗和使用set：
def all_same( items ):
return len( set( items ) ) == 1 #== len( items )
or you could use:
或者你可以使用：
def all_same( items ):
return all( map(lambda x: x == items[0], items ) )
or if you're dealing with an iterable instead of a list:
或者如果您正在处理可迭代而不是列表：
def all_same( iterable ):
it_copy = tee( iterable, 1 )
return len( set( it_copy) ) == 1
回答by FogleBird
I originally interpreted you to be testing identity ("the same item"), but you're really testing equality("same value"). (If you were testing identity, use isinstead of ==.)
我最初将您解释为测试身份（“相同项目”），但您实际上是在测试相等性（“相同值”）。（如果您正在测试身份，请使用is而不是 ==。）
def all_same(items):
it = iter(items)
for first in it:
break
else:
return True # empty case, note all([]) == True
return all(x == first for x in it)
The above works on any iterable, not just lists, otherwise you could use:
以上适用于任何可迭代对象，而不仅仅是列表，否则您可以使用：
def all_same(L):
return all(x == L[0] for x in L)
(But, IMHO, you might as well use the general version—it works perfectly fine on lists.)
（但是，恕我直言，您不妨使用通用版本——它在列表上运行得非常好。）
回答by tokland
This works both for sequences and iterables:
这适用于序列和可迭代对象：
def all_same(items):
it = iter(items)
first = next(it, None)
return all(x == first for x in it)
回答by Jonas K
This is likely to be faster if you know values are in a list.
如果您知道值在列表中，这可能会更快。
def all_same(values):
return values.count(values[0]) == len(values)
回答by rezakamalifard
Best way to do this is to use Python sets.You need to define all_samelike this:
最好的方法是使用 Python 集合。你需要这样定义all_same：
def all_same(items):
return len(set(items)) < 2
Test:
测试：
>>> def all_same(items):
... return len(set(items)) < 2
...
>>>
>>> property_list = ["one", "one", "one"]
>>> all_same(property_list)
True
>>> property_list = ["one", "one", "two"]
>>> all_same(property_list)
False
>>> property_list = []
>>> all_same(property_list)
True
回答by Landon Powell
I created this snippet of code for that same issue after thinking it over. I'm not exactly sure if it works for every scenario though.
在仔细考虑之后，我为同一问题创建了这段代码。不过，我不确定它是否适用于所有场景。
def all_same(list):
list[0]*len(list) == list
