标签归档:inclusion

如何检查以下所有项目是否都在列表中?

问题:如何检查以下所有项目是否都在列表中?

我发现存在一个有关的问题,即如何查找列表中是否至少有一项:如何检查列表中
是否有以下一项?

但是,找到列表中是否存在所有项的最佳方式是什么?

搜索文档后,我发现此解决方案:

>>> l = ['a', 'b', 'c']
>>> set(['a', 'b']) <= set(l)
True
>>> set(['a', 'x']) <= set(l)
False

其他解决方案是这样的:

>>> l = ['a', 'b', 'c']
>>> all(x in l for x in ['a', 'b'])
True
>>> all(x in l for x in ['a', 'x'])
False

但是在这里您必须进行更多的键入。

还有其他解决方案吗?

I found, that there is related question, about how to find if at least one item exists in a list:
How to check if one of the following items is in a list?

But what is the best and pythonic way to find whether all items exists in a list?

Searching through the docs I found this solution:

>>> l = ['a', 'b', 'c']
>>> set(['a', 'b']) <= set(l)
True
>>> set(['a', 'x']) <= set(l)
False

Other solution would be this:

>>> l = ['a', 'b', 'c']
>>> all(x in l for x in ['a', 'b'])
True
>>> all(x in l for x in ['a', 'x'])
False

But here you must do more typing.

Is there any other solutions?


回答 0

<=Python 这样的运算符通常不会被覆盖以表示与“小于或等于”明显不同的东西。对于标准库来说,这样做是不寻常的-对我来说,它听起来像是旧版API。

使用等效的且更明确命名的方法set.issubset。注意,您不需要将参数转换为集合;如有需要,它将为您完成。

set(['a', 'b']).issubset(['a', 'b', 'c'])

Operators like <= in Python are generally not overriden to mean something significantly different than “less than or equal to”. It’s unusual for the standard library does this–it smells like legacy API to me.

Use the equivalent and more clearly-named method, set.issubset. Note that you don’t need to convert the argument to a set; it’ll do that for you if needed.

set(['a', 'b']).issubset(['a', 'b', 'c'])

回答 1

我可能会set以以下方式使用:

set(l).issuperset(set(['a','b'])) 

或反过来:

set(['a','b']).issubset(set(l)) 

我觉得它更具可读性,但可能会过分杀人。集合对于计算集合之间的并集/交叉点/差异特别有用,但是在这种情况下,它可能不是最佳选择。

I would probably use set in the following manner :

set(l).issuperset(set(['a','b'])) 

or the other way round :

set(['a','b']).issubset(set(l)) 

I find it a bit more readable, but it may be over-kill. Sets are particularly useful to compute union/intersection/differences between collections, but it may not be the best option in this situation …


回答 2

我喜欢这两个,因为它们看起来最合乎逻辑,后者更短并且可能最快(此处使用set文字语法显示,该文字语法已 反向移植到Python 2.7):

all(x in {'a', 'b', 'c'} for x in ['a', 'b'])
#   or
{'a', 'b'}.issubset({'a', 'b', 'c'})

I like these two because they seem the most logical, the latter being shorter and probably fastest (shown here using set literal syntax which has been backported to Python 2.7):

all(x in {'a', 'b', 'c'} for x in ['a', 'b'])
#   or
{'a', 'b'}.issubset({'a', 'b', 'c'})

回答 3

如果您的列表包含这样的重复项,该怎么办:

v1 = ['s', 'h', 'e', 'e', 'p']
v2 = ['s', 's', 'h']

集不包含重复项。因此,以下行返回True。

set(v2).issubset(v1)

要计算重复项,可以使用以下代码:

v1 = sorted(v1)
v2 = sorted(v2)


def is_subseq(v2, v1):
    """Check whether v2 is a subsequence of v1."""
    it = iter(v1)
    return all(c in it for c in v2) 

因此,以下行返回False。

is_subseq(v2, v1)

What if your lists contain duplicates like this:

v1 = ['s', 'h', 'e', 'e', 'p']
v2 = ['s', 's', 'h']

Sets do not contain duplicates. So, the following line returns True.

set(v2).issubset(v1)

To count for duplicates, you can use the code:

v1 = sorted(v1)
v2 = sorted(v2)


def is_subseq(v2, v1):
    """Check whether v2 is a subsequence of v1."""
    it = iter(v1)
    return all(c in it for c in v2) 

So, the following line returns False.

is_subseq(v2, v1)

回答 4

这就是我在网上搜索的内容,但不幸的是,我在python解释器上进行实验时发现不是在线的。

>>> case  = "caseCamel"
>>> label = "Case Camel"
>>> list  = ["apple", "banana"]
>>>
>>> (case or label) in list
False
>>> list = ["apple", "caseCamel"]
>>> (case or label) in list
True
>>> (case and label) in list
False
>>> list = ["case", "caseCamel", "Case Camel"]
>>> (case and label) in list
True
>>>

如果您有一个完整的变量列表 sublist variable

>>>
>>> list  = ["case", "caseCamel", "Case Camel"]
>>> label = "Case Camel"
>>> case  = "caseCamel"
>>>
>>> sublist = ["unique banana", "very unique banana"]
>>>
>>> # example for if any (at least one) item contained in superset (or statement)
...
>>> next((True for item in sublist if next((True for x in list if x == item), False)), False)
False
>>>
>>> sublist[0] = label
>>>
>>> next((True for item in sublist if next((True for x in list if x == item), False)), False)
True
>>>
>>> # example for whether a subset (all items) contained in superset (and statement)
...
>>> # a bit of demorgan's law
...
>>> next((False for item in sublist if item not in list), True)
False
>>>
>>> sublist[1] = case
>>>
>>> next((False for item in sublist if item not in list), True)
True
>>>
>>> next((True for item in sublist if next((True for x in list if x == item), False)), False)
True
>>>
>>>

This was what I was searching online but unfortunately found not online but while experimenting on python interpreter.

>>> case  = "caseCamel"
>>> label = "Case Camel"
>>> list  = ["apple", "banana"]
>>>
>>> (case or label) in list
False
>>> list = ["apple", "caseCamel"]
>>> (case or label) in list
True
>>> (case and label) in list
False
>>> list = ["case", "caseCamel", "Case Camel"]
>>> (case and label) in list
True
>>>

and if you have a looong list of variables held in a sublist variable

>>>
>>> list  = ["case", "caseCamel", "Case Camel"]
>>> label = "Case Camel"
>>> case  = "caseCamel"
>>>
>>> sublist = ["unique banana", "very unique banana"]
>>>
>>> # example for if any (at least one) item contained in superset (or statement)
...
>>> next((True for item in sublist if next((True for x in list if x == item), False)), False)
False
>>>
>>> sublist[0] = label
>>>
>>> next((True for item in sublist if next((True for x in list if x == item), False)), False)
True
>>>
>>> # example for whether a subset (all items) contained in superset (and statement)
...
>>> # a bit of demorgan's law
...
>>> next((False for item in sublist if item not in list), True)
False
>>>
>>> sublist[1] = case
>>>
>>> next((False for item in sublist if item not in list), True)
True
>>>
>>> next((True for item in sublist if next((True for x in list if x == item), False)), False)
True
>>>
>>>

回答 5

如何使用lambda表达式执行此操作的示例如下:

issublist = lambda x, y: 0 in [_ in x for _ in y]

An example of how to do this using a lambda expression would be:

issublist = lambda x, y: 0 in [_ in x for _ in y]

回答 6

不是OP的情况,但是-对于任何想要在字典中声明交集 并由于不良谷歌搜索而最终到这里的人(例如我)-您需要与dict.items

>>> a = {'key': 'value'}
>>> b = {'key': 'value', 'extra_key': 'extra_value'}
>>> all(item in a.items() for item in b.items())
True
>>> all(item in b.items() for item in a.items())
False

那是因为dict.items返回键/值对的元组,就像Python中的任何对象一样,它们可以互换地比较

Not OP’s case, but – for anyone who wants to assert intersection in dicts and ended up here due to poor googling (e.g. me) – you need to work with dict.items:

>>> a = {'key': 'value'}
>>> b = {'key': 'value', 'extra_key': 'extra_value'}
>>> all(item in a.items() for item in b.items())
True
>>> all(item in b.items() for item in a.items())
False

That’s because dict.items returns tuples of key/value pairs, and much like any object in Python, they’re interchangeably comparable