blocks|key|1076464|text|也许是这样的？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1076465|insert+=+[]
for+k,+v+in+parm.items():
++++if+k+in+check_keys:
++++++++insert.append(k)|code-block|syntax|javascript|1076466|或者这个更短的版本：|1076467|insert+=+[k+for+k,+v+in+parm.items()+if+k+in+check_keys]|1076468|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|$]]

Maybe something like this?

<pre><code>insert = []
for k, v in parm.items():
 if k in check_keys:
 insert.append(k)
</code></pre>

Or this shorter version:

<pre><code>insert = [k for k, v in parm.items() if k in check_keys]
</code></pre>

blocks|key|581191|text|花了些时间但我想我知道你想做什么了。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|581192|如果我是对的，你就不要倒过来查字典，从一个值到一个键。最简单的方法是对反向字典进行验证。|581193|rparm+=+dict(+(value,key)+for+key,value+in+parm.items())|code-block|syntax|javascript|581194|然后对数据进行匹配。|581195|insert+=+[rparm[key]+for+key+in+check_keys]|581196|很抱歉使用了所有这些列表和生成器理解。如果我们不熟悉它们，我可以添加经典的for循环版本。|581197|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|T|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|U|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|V|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

It took some time but i think i figured out what you want to do.

If i am correct you wan't to reverse look up the dictionary from value to key.
The easiest way would be to cerate the reverse dictionary.

<pre><code>rparm = dict( (value,key) for key,value in parm.items())
</code></pre>

Then match the data e.g. 

<pre><code>insert = [rparm[key] for key in check_keys]
</code></pre>

Sorry for using all this list and generator comprehensions. If our are not familiar with them I can add the classic for loop version.

blocks|key|2510166|text|一种可能的方法是反向字典，这可以在线性时间(相对于您的parm的大小)平均。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2510167|一旦字典被反转，搜索的值就变成键，现在很容易访问check_keys列表中的每个键O(1)。|2510168|rev_parm+=+dict((v,+k)+for+k,+v+in+parm.items())
insert+=+[rev_parm[k]+for+k+in+check_keys]|code-block|syntax|javascript|2510169|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

One possible way is the reverse the dictionary, which can be done in linear time (relative to the size of your parm dict) on average. 

Once the dictionary is reversed your searched values become keys 
And are now easily accessible is O(1) per key in your check_keys list.

<pre class="lang-py prettyprint-override"><code>rev_parm = dict((v, k) for k, v in parm.items())
insert = [rev_parm[k] for k in check_keys]
</code></pre>

blocks|key|2510227|text|这不是哈希表/字典的一个很好的用途。给定字典的键，哈希表使您可以快速访问这些值。在这里，您正在尝试访问给定值的字典的键(即。相反的方向)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2510228|这很糟糕，因为：|2510229|太慢了。您需要遍历整个字典才能找到您的值。|unordered-list-item|2510230|结果是不确定的，因为字典是无序的。如果您的输入是parm+=+{'a':+'r',+'b':+'r'}，check_keys+=+[r]，则不确定您将得到[a]还是[b]。|offset|length|style|CODE|2510231|说了这么多，这就解决了你的问题：|2510232|parm+=+{'f':+'w',+'l':+'b',+'b':+'y',+'u':+'o',+'t':+'r',+'r':+'g'}
check_keys+=+['r','b','o','g','w','y']
insert+=+['t','l','u','r','f','b']

insert+=+[]
for+v+in+check_keys:
++++for+k+in+parm:
++++++++if+parm[k]+==+v:
++++++++++++insert.append(k)

print+insert|code-block|syntax|javascript|2510233|当然，假设这是一个更大程序的一部分，我建议您不要使用上面的方法，而是重新评估整个方法。|2510234|entityMap^0|0|0|0|O|R|1G|G|25|3|2A|3|0|0|0|0^^$0|@$1|2|3|4|5|6|7|X|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Y|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Z|8|@]|9|@]|A|$]]|$1|G|3|H|5|F|7|10|8|@$I|11|J|12|K|L]|$I|13|J|14|K|L]|$I|15|J|16|K|L]|$I|17|J|18|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|19|8|@]|9|@]|A|$]]|$1|O|3|P|5|Q|7|1A|8|@]|9|@]|A|$R|S]]|$1|T|3|U|5|6|7|1B|8|@]|9|@]|A|$]]|$1|V|3|-4|5|6|7|1C|8|@]|9|@]|A|$]]]|W|$]]

This is not a great use of a Hash Table/Dictionary. A Hash Table gives you fast access to the values, given the keys of the dictionary. Here, you are trying to access the keys of the dictionary, given the values (ie. the other way around). 

This is bad because:

<ul>
<li>It's slow. You will need to iterate over the entire dictionary to find your values.</li>
<li>The result is non-deterministic, as dictionaries are unordered. If your inputs were instead <code>parm = {'a': 'r', 'b': 'r'}</code>, <code>check_keys = [r]</code>, it is uncertain whether you would get <code>[a]</code>, or <code>[b]</code>. </li>
</ul>

With all that said and done, this should solve your problem:

<pre><code>parm = {'f': 'w', 'l': 'b', 'b': 'y', 'u': 'o', 't': 'r', 'r': 'g'}
check_keys = ['r','b','o','g','w','y']
insert = ['t','l','u','r','f','b']

insert = []
for v in check_keys:
 for k in parm:
 if parm[k] == v:
 insert.append(k)

print insert
</code></pre>

Assuming, of course, that this is part of a larger program, I would suggest that you do NOT use the above approach, but instead re-evaluate your entire approach.

blocks|key|581354|text|您可以使用列表理解来解决这个问题。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|581355|insert+=+[k+for+i+in+check_keys+for+k+in+parm+if+parm[k]+==+i]
#+['t',+'l',+'u',+'r',+'f',+'b']|code-block|syntax|javascript|581356|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|$]]

You could use list comprehension to solve this

<pre><code>insert = [k for i in check_keys for k in parm if parm[k] == i]
# ['t', 'l', 'u', 'r', 'f', 'b']
</code></pre>

let's say I need do this:
I need to match the values in list 'check_keys' against the key-values in dictionary 'parm' and store the matched keys in the same order into a new list 'insert'

<pre><code>parm = {'f': 'w', 'l': 'b', 'b': 'y', 'u': 'o', 't': 'r', 'r': 'g'}
check_keys = ['r','b','o','g','w','y']
insert = ['t','l','u','r','f','b']
</code></pre>

Match a value to key value and get key

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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假设我需要这样做:我需要将列表'check_keys‘中的值与字典'parm’中的键值相匹配，并将匹配的键按相同的顺序存储到一个新的列表'insert‘中。parm = {'f': 'w', 'l': 'b', 'b': 'y', 'u': 'o', 't': 'r', 'r': 'g'}check_keys = ['...

问将值与键值匹配并获取键
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