blocks|key|1850215|text|你找到了吗？：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1850216|print+([x+for+x+in+cons++if+sorted(list(x))+==+sorted(list(alphabet))])|code-block|syntax|javascript|1850217|或者如果需要找到不完全的巧合|1850218|print+([x+for+x+in+cons++if+''.join(sorted(list(alphabet)))+in+''.join(sorted(list(x)))])|1850219|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

did you find it?:

<pre><code>print ([x for x in cons if sorted(list(x)) == sorted(list(alphabet))])
</code></pre>

or if need to find not complete coincidence

<pre><code>print ([x for x in cons if ''.join(sorted(list(alphabet))) in ''.join(sorted(list(x)))])
</code></pre>

blocks|key|2635650|text|cons=['qwe','wer','ert','rty','tyu','yui','uio','iop','asd','sdf','dfg',
++++++'fgh','ghj','hjk','jkl','zxc','xcv','cvb','vbn','bnm']|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2635651|UPDATE+(基于OP注释)如果需要cons中的字符串为alphabet的精确匹配子字符串，则此代码只需检查来自cons的字符串是否为alphabet的子字符串(精确匹配)，如果是，则包含它：|unstyled|offset|length|style|BOLD|CODE|2635652|print([sub+for+sub+in+cons+if+sub+in+alphabet])|2635653|对于情况：alphabet+=+'qwer'，ouput+=+['qwe',+'wer']|2635654|但是，基于OP，该测试检查cons字符串中的所有字母是否存在于alphabet中(不一定按确切的顺序和频率)。这使用all()方法，如果传递的所有参数都是True，则返回True。|2635655|print([sub+for+sub+in+cons+if+all(let+in+alphabet+for+let+in+sub)])|2635656|说明：let+in+alphabet根据每个字母let是否在用户输入字符串alphabet中返回真假。这是对cons，for+let+in+sub子字符串中的每个字母执行的。因此，用户输入的每个字母都将被选中，并且只有当True字符串中的所有字母都在字母表中时，才能计算为sub。如果此检查计算结果为True，则结果中包含该子字符串并打印该子字符串。如果来自con的子字符串与用户输入alphabet的长度不同，则此函数可以工作。|2635657|如果字母=+'er'，输出=+[]+(没有字符串包含alphabet中的所有字母)|2635658|如果字母=+'qwer'，输出=+['qwe',+'wer']|2635659|entityMap^0|0|0|6|J|4|T|8|1L|4|1W|8|0|0|5|8|G|6|V|E|0|D|4|V|8|1M|5|25|4|2D|4|0|0|3|F|O|3|11|8|1I|4|1N|E|32|4|3S|3|4Z|3|5C|8|0|6|4|F|2|Q|8|0|6|6|H|E|0^^$0|@$1|2|3|4|5|6|7|11|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|12|8|@$G|13|H|14|I|J]|$G|15|H|16|I|K]|$G|17|H|18|I|K]|$G|19|H|1A|I|K]|$G|1B|H|1C|I|K]]|9|@]|A|$]]|$1|L|3|M|5|6|7|1D|8|@]|9|@]|A|$B|C]]|$1|N|3|O|5|F|7|1E|8|@$G|1F|H|1G|I|K]|$G|1H|H|1I|I|K]|$G|1J|H|1K|I|K]]|9|@]|A|$]]|$1|P|3|Q|5|F|7|1L|8|@$G|1M|H|1N|I|K]|$G|1O|H|1P|I|K]|$G|1Q|H|1R|I|K]|$G|1S|H|1T|I|K]|$G|1U|H|1V|I|K]]|9|@]|A|$]]|$1|R|3|S|5|6|7|1W|8|@]|9|@]|A|$B|C]]|$1|T|3|U|5|F|7|1X|8|@$G|1Y|H|1Z|I|K]|$G|20|H|21|I|K]|$G|22|H|23|I|K]|$G|24|H|25|I|K]|$G|26|H|27|I|K]|$G|28|H|29|I|K]|$G|2A|H|2B|I|K]|$G|2C|H|2D|I|K]|$G|2E|H|2F|I|K]]|9|@]|A|$]]|$1|V|3|W|5|F|7|2G|8|@$G|2H|H|2I|I|K]|$G|2J|H|2K|I|K]|$G|2L|H|2M|I|K]]|9|@]|A|$]]|$1|X|3|Y|5|F|7|2N|8|@$G|2O|H|2P|I|K]|$G|2Q|H|2R|I|K]]|9|@]|A|$]]|$1|Z|3|-4|5|F|7|2S|8|@]|9|@]|A|$]]]|10|$]]

<pre><code>cons=['qwe','wer','ert','rty','tyu','yui','uio','iop','asd','sdf','dfg',
 'fgh','ghj','hjk','jkl','zxc','xcv','cvb','vbn','bnm']
</code></pre>

UPDATE (based on OP comments) If you need the strings in <code>cons</code> to be exact match substrings of <code>alphabet</code>, then this code simply checks if the string from <code>cons</code> is a substring of <code>alphabet</code> (exact match) and includes it if so:

<pre><code>print([sub for sub in cons if sub in alphabet])
</code></pre>

for case: <code>alphabet</code> = <code>'qwer'</code>, ouput = <code>['qwe', 'wer']</code>

However, based on the OP, this test checks if all the letters in the strings of <code>cons</code> exist in the <code>alphabet</code> (not necessarily in exact order and frequency). This uses the <code>all()</code> method which returns <code>True</code> if all arguments passed are <code>True</code>.

<pre><code>print([sub for sub in cons if all(let in alphabet for let in sub)])
</code></pre>

Explanation: <code>let in alphabet</code> returns True or False based on whether each letter, <code>let</code>, is in the user input string, <code>alphabet</code>. This is executed for each letter in the sub string of <code>cons</code>, <code>for let in sub</code>. Therefore each letter from the user input will be checked and only evaluate to <code>True</code> if all letters from the <code>sub</code> string are in the alphabet. If this check evaluates to True, than the substring is included in the results and printed. This function works if the substrings from <code>con</code> are of different lengths from the user input, <code>alphabet</code>.

If alphabet = <code>'er'</code>, output = <code>[]</code> (no strings have all letters in <code>alphabet</code>)

If alphabet = <code>'qwer'</code>, output = <code>['qwe', 'wer']</code>

blocks|key|1850334|text|如果您不关心列表中输出字符串的顺序，那么当然可以这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1850335|inputs+=+['qwe',+'wer',+'ert',+'rty',+'tyu',+'yui',+'uio',+'iop',+'asd',+'sdf',
++++++++++'dfg',+'fgh',+'ghj',+'hjk',+'jkl',+'zxc',+'xcv',+'cvb',+'vbn',+'bnm']

def+string_intersections(input_string):

++++def+intersect(string):
++++++++input_letters+=+set(input_string)
++++++++string_letters+=+set(string)
++++++++has_intersection+=+input_letters+==+string_letters
++++++++return+has_intersection

++++result+=+list(filter(intersect,+inputs))
++++return+result

string_intersections("boo")|code-block|syntax|javascript|1850336|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

In the case that you don't care about the order of the output strings in your list, then you could of course do:

<pre><code>inputs = ['qwe', 'wer', 'ert', 'rty', 'tyu', 'yui', 'uio', 'iop', 'asd', 'sdf',
 'dfg', 'fgh', 'ghj', 'hjk', 'jkl', 'zxc', 'xcv', 'cvb', 'vbn', 'bnm']

def string_intersections(input_string):

 def intersect(string):
 input_letters = set(input_string)
 string_letters = set(string)
 has_intersection = input_letters == string_letters
 return has_intersection

 result = list(filter(intersect, inputs))
 return result

string_intersections("boo")
</code></pre>

blocks|key|2635744|text|尝试使用set|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2635745|cons=['qwe','wer','ert','rty','tyu','yui','uio','iop','asd','sdf','dfg',
++++++'fgh','ghj','hjk','jkl','zxc','xcv','cvb','vbn','bnm']
alphabet+=+"weq"
print([c+for+c+in+cons+if+set(alphabet).issubset(set(c))])+#+output+will+be+['qwe']|code-block|syntax|javascript|2635746|entityMap^0|4|3|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|$]]

try with <code>set</code>:

<pre><code>cons=['qwe','wer','ert','rty','tyu','yui','uio','iop','asd','sdf','dfg',
 'fgh','ghj','hjk','jkl','zxc','xcv','cvb','vbn','bnm']
alphabet = "weq"
print([c for c in cons if set(alphabet).issubset(set(c))]) # output will be ['qwe']
</code></pre>

So here i have two lists

<pre><code>cons=['qwe','wer','ert','rty','tyu','yui','uio','iop','asd','sdf','dfg',
 'fgh','ghj','hjk','jkl','zxc','xcv','cvb','vbn','bnm']

print([i for e in alphabet for i in cons if e in i])
</code></pre>

Alphabet is a user input

If the user were to type, for example, qwe, then the shell would print <code>'qwe','wer','ert','rty'</code> (all items that have at least one letter similar) BUT i only want it to print the items from cons if of all items from the input are similar to that. E.g, print 'qwe'

How do i do this?

EDIT: to make this more clear, if i were to input 'qwer', i would like the outputs to be 'qwe' and 'wer'

Trying to see if one items in a list are in another

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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所以这里我有两个列表cons=['qwe','wer','ert','rty','tyu','yui','uio','iop','asd','sdf','dfg',      'fgh','ghj','hjk','jkl','zxc','xcv','cvb','vbn','bnm']print([i for e in ...

问试图查看列表中的某一项是否在另一项中
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问试图查看列表中的某一项是否在另一项中EN