blocks|key|4644342|text|要检查成员资格，请使用in|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4644343|>>>+a+=+['a','b','c','d']
>>>+'a'+in+a
True
>>>+'z'+in+a
False|code-block|syntax|javascript|4644344|entityMap^0|B|2|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

To check membership, use <code>in</code>:

<pre><code>&gt;&gt;&gt; a = ['a','b','c','d']
&gt;&gt;&gt; 'a' in a
True
&gt;&gt;&gt; 'z' in a
False
</code></pre>

blocks|key|2327167|text|any(a)的意思是“a中是否有任何项目是真实的”？结果是True，因为a中的每一项都是真的。(任何非零长度的字符串都是真的，并且a中的每一项都是非零长度的字符串。)|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2327168|然后将结果True与"A"进行比较。True不等于"A"，所以比较的结果当然是False。|2327169|你可能想要这样做：|2327170|"A"+in+a+++#+True|code-block|syntax|javascript|2327171|如果出于某种原因必须使用any()，请尝试：|2327172|any(item+==+"A"+for+item+in+a)|2327173|这种方法的优点是可以很容易地进行不精确的比较(in只会进行精确的比较)。例如：|2327174|any(item.lower()+==+"a"+for+item+in+a)+++#+case-insensitive
any("a"+in+item.lower()+for+item+in+a)+++#+substring+match
any(item.lower().startswith("a")+for+item+in+a)|2327175|entityMap^0|0|6|B|1|T|4|10|1|1T|1|0|5|4|A|3|I|4|P|3|13|5|0|0|0|C|5|0|0|N|2|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]|$9|11|A|12|B|C]|$9|13|A|14|B|C]|$9|15|A|16|B|C]|$9|17|A|18|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|19|8|@$9|1A|A|1B|B|C]|$9|1C|A|1D|B|C]|$9|1E|A|1F|B|C]|$9|1G|A|1H|B|C]|$9|1I|A|1J|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|1K|8|@]|D|@]|E|$]]|$1|J|3|K|5|L|7|1L|8|@]|D|@]|E|$M|N]]|$1|O|3|P|5|6|7|1M|8|@$9|1N|A|1O|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|L|7|1P|8|@]|D|@]|E|$M|N]]|$1|S|3|T|5|6|7|1Q|8|@$9|1R|A|1S|B|C]]|D|@]|E|$]]|$1|U|3|V|5|L|7|1T|8|@]|D|@]|E|$M|N]]|$1|W|3|-4|5|6|7|1U|8|@]|D|@]|E|$]]]|X|$]]

<code>any(a)</code> means "is any item in <code>a</code> truthy"? And the result is <code>True</code> because every item in <code>a</code> is truthy. (Any non-zero-length string is truthy, and every item in <code>a</code> is a non-zero-length string.)

And then you are comparing that result, <code>True</code>, to, say, <code>"A"</code>. <code>True</code> is not equal to <code>"A"</code> so the result of that comparison is, of course, <code>False</code>.

What you probably want to do is something like:

<pre><code>"A" in a # True
</code></pre>

If you must use <code>any()</code> for some reason, try:

<pre><code>any(item == "A" for item in a)
</code></pre>

This approach has the advantage of being able to do imprecise comparisons easily (<code>in</code> will only do exact comparisons). For example:

<pre><code>any(item.lower() == "a" for item in a) # case-insensitive
any("a" in item.lower() for item in a) # substring match
any(item.lower().startswith("a") for item in a)
</code></pre>

blocks|key|1761750|text|问题是，如果可迭代元素的任何一个为True，any()就会返回True，所以只要结果不等于months_list-which中的所有字符串，代码就会一直循环，这可能与您希望发生的情况相反。这里有一种使用它的方法，如果答案与任何字符串匹配，它就会停止或中断循环：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1761751|months_list+=+["January",+"February",+"March",+"April",+"May",+"June",+"July"]

while+True:
++++answer+=+raw_input("Month?+")
++++if+any(item.lower()+==+answer.lower()+for+item+in+months_list):+
++++++++break
++++print("Sorry,+didn't+recognize+your+answer,+try+again")|code-block|syntax|javascript|1761752|正如其他人所指出的，虽然使用Python的in运算符会更简单，但这种方法仍然会导致线性搜索O(n)被…执行因此，更好(更快)的方法是使用大小写较低的month_names的set，这将利用基于哈希表的查找，O(1)，而不是线性搜索：|BOLD|1761753|months+=+set(month.lower()+for+month+in+("January",+"February",+"March",+"April",
+++++++++++++++++++++++++++++++++++++++++"May",+"June",+"July"))
while+True:
++++answer+=+raw_input("Month?+")
++++if+answer.lower()+in+months:+
++++++++break
++++print("Sorry,+didn't+recognize+your+answer,+try+again")|1761754|进一步细化|1761755|根据所涉及的字符串的性质和比较它们的原因，使用string+casefold()方法而不是lower()进行无大小写的字符串比较可能更好。|1761756|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3/library/stdtypes.html#str.casefold^0|H|4|M|5|V|4|19|B|0|0|L|2|22|B|2E|3|19|4|2V|4|0|0|0|5|0|U|A|19|7|U|A|0|0^^$0|@$1|2|3|4|5|6|7|11|8|@$9|12|A|13|B|C]|$9|14|A|15|B|C]|$9|16|A|17|B|C]|$9|18|A|19|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|1A|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1B|8|@$9|1C|A|1D|B|C]|$9|1E|A|1F|B|C]|$9|1G|A|1H|B|C]|$9|1I|A|1J|B|M]|$9|1K|A|1L|B|M]]|D|@]|E|$]]|$1|N|3|O|5|H|7|1M|8|@]|D|@]|E|$I|J]]|$1|P|3|Q|5|6|7|1N|8|@$9|1O|A|1P|B|M]]|D|@]|E|$]]|$1|R|3|S|5|6|7|1Q|8|@$9|1R|A|1S|B|C]|$9|1T|A|1U|B|C]]|D|@$9|1V|A|1W|1|1X]]|E|$]]|$1|T|3|-4|5|6|7|1Y|8|@]|D|@]|E|$]]]|U|$V|$5|W|X|Y|E|$Z|10]]]]

The problem is that <code>any()</code> returns <code>True</code> if any one of the elements in the iterable is <code>True</code>, so your code keeps looping as long as the answer isn't equal to all the strings in <code>months_list</code>—which is probably the opposite of what you want to happen. Here's a way to use it that stops or breaks-out of the loop if the answer matches any of the strings:
<pre><code>months_list = [&quot;January&quot;, &quot;February&quot;, &quot;March&quot;, &quot;April&quot;, &quot;May&quot;, &quot;June&quot;, &quot;July&quot;]

while True:
 answer = raw_input(&quot;Month? &quot;)
 if any(item.lower() == answer.lower() for item in months_list): 
 break
 print(&quot;Sorry, didn't recognize your answer, try again&quot;)
</code></pre>
As others have pointed out, while it would simpler to use Python's <code>in</code> operator, that way still results in a linear search, O(n), being performed…so an even better (faster) approach would be to use a <code>set</code> of lower-cased <code>month_names</code>, which would utilize a hash table based look-up, O(1), instead of a linear search:
<pre><code>months = set(month.lower() for month in (&quot;January&quot;, &quot;February&quot;, &quot;March&quot;, &quot;April&quot;,
 &quot;May&quot;, &quot;June&quot;, &quot;July&quot;))
while True:
 answer = raw_input(&quot;Month? &quot;)
 if answer.lower() in months: 
 break
 print(&quot;Sorry, didn't recognize your answer, try again&quot;)
</code></pre>
Further refinement
Depending on the nature of strings involved and why you're comparing them, it might be better to use the string <a href="https://docs.python.org/3/library/stdtypes.html#str.casefold" rel="nofollow noreferrer"><code>casefold()</code></a> method instead of <code>lower()</code> to do caseless string comparisons.

Edit: This is what I am trying to do:
I am asking the user to enter a month. then the code will lookup if the month is correct by checking every item in months_list. If not found I want him/her to enter the month again..

Here's the code:

<pre><code>months_list=["January", "February", "March", "April", "May", "June", "July"]
answer=raw_input("Month? \n")
while any(item.lower() != answer.lower() for item in months_list):
 print("Sorry, didn't recognize your answer, try again")
 answer=raw_input("Type in Month\n")
</code></pre>

However this keeps looping regardless if the month is found in the list or not.. I hope this is a good clarifaction.. thank you all in advance

How to check if a variable matches any item in list using the any() function?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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编辑:这是我想要做的:我要求用户输入一个月。然后，代码将通过检查months_list中的每一项来查找月份是否正确。如果找不到，我想让他/她重新输入月份。代码如下：months_list=["January", "February", "March", "April", "May", "June", "July"]an...

问如何使用any()函数检查变量是否与列表中的任何项匹配？
EN

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问如何使用any()函数检查变量是否与列表中的任何项匹配？EN