blocks|key|1723949|text|如果您真的想为这个任务编写一个函数，那么您的设计就在正确的轨道上，但是需要对其进行修改，以说明您有一个字典列表。像这样的东西可能会起作用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1723950|def+find_user(userid):
++++for+user_dict+in+big_list_of_user_dictionaries:
++++++++if+user_dict['User_ID']+==+userid:
++++++++++++return+user_dict|code-block|syntax|javascript|1723951|但是，最好创建一个新字典，其中每个键都是userid，每个值都是您的用户信息字典之一。您可以使用Python的字典理解来快速制作这样的词典：|1723952|+user_dict+=+{d['User_ID']+:+d+for+d+in+big_list_of_user_dictionaries}|1723953|然后，您可以通过在user_dict中查找他们的id来查找任何用户的用户信息字典，如下所示：|offset|length|style|CODE|1723954|+print(user_dict['Z000'])|1723955|entityMap^0|0|0|0|0|9|9|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|Y|8|@$M|Z|N|10|O|P]]|9|@]|A|$]]|$1|Q|3|R|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|S|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|T|$]]

If you really want to write a function for this task, your design is on the right track, but needs to be modified to account for the fact that you have a list of dictionaries. Something like this might work:
<pre><code>def find_user(userid):
 for user_dict in big_list_of_user_dictionaries:
 if user_dict['User_ID'] == userid:
 return user_dict
</code></pre>
However, you might be better off creating a new dictionary, where each key is the userid, and each value is one of your user info dictionaries. You could use Python's dictionary comprehensions to make such a dictionary quickly:
<pre><code> user_dict = {d['User_ID'] : d for d in big_list_of_user_dictionaries}
</code></pre>
Then you could find the user info dictionary for any user by looking up their id in the <code>user_dict</code>, like this:
<pre><code> print(user_dict['Z000'])
</code></pre>

blocks|key|2958848|text|您需要迭代列表，并将字典的UserID与输入UserID进行比较：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2958849|def+find_user(val,+lsts):
++++for+d+in+lsts:
++++++++if+val+==+d['User_ID']:
++++++++++++return+d
++++return+"user+not+found"|code-block|syntax|javascript|2958850|然后|2958851|print(find_user('Z000',+lsts))|2958852|版画|2958853|{'User_ID':+'Z000',
+'DOB':+'01.01.1960',
+'State':+'Oregon',
+'Bought':+['P1',+'P2']}|2958854|和|2958855|print(find_user('000',+lsts))|2958856|2958857|user+not+found|2958858|但是，如果您的数据类似于：|2958859|d+=+{+"Data":+[{"User_ID":"Z000"},{"User_ID":"A999"}+]}|2958860|然后，可以将字典中的列表传递给函数，如下所示：|2958861|find_user('Z000',+d['Data'])|2958862|回传|2958863|{'User_ID':+'Z000'}|2958864|entityMap^0|D|6|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1D|8|@$9|1E|A|1F|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|1G|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1H|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|1I|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|1J|8|@]|D|@]|E|$]]|$1|Q|3|R|5|H|7|1K|8|@]|D|@]|E|$I|J]]|$1|S|3|T|5|6|7|1L|8|@]|D|@]|E|$]]|$1|U|3|V|5|H|7|1M|8|@]|D|@]|E|$I|J]]|$1|W|3|P|5|6|7|1N|8|@]|D|@]|E|$]]|$1|X|3|Y|5|H|7|1O|8|@]|D|@]|E|$I|J]]|$1|Z|3|10|5|6|7|1P|8|@]|D|@]|E|$]]|$1|11|3|12|5|H|7|1Q|8|@]|D|@]|E|$I|J]]|$1|13|3|14|5|6|7|1R|8|@]|D|@]|E|$]]|$1|15|3|16|5|H|7|1S|8|@]|D|@]|E|$I|J]]|$1|17|3|18|5|6|7|1T|8|@]|D|@]|E|$]]|$1|19|3|1A|5|H|7|1U|8|@]|D|@]|E|$I|J]]|$1|1B|3|-4|5|6|7|1V|8|@]|D|@]|E|$]]]|1C|$]]

You want to iterate over the list and compare the <code>UserID</code> of a dictionary with an input UserID:
<pre><code>def find_user(val, lsts):
 for d in lsts:
 if val == d['User_ID']:
 return d
 return &quot;user not found&quot;
</code></pre>
Then
<pre><code>print(find_user('Z000', lsts))
</code></pre>
prints
<pre><code>{'User_ID': 'Z000',
 'DOB': '01.01.1960',
 'State': 'Oregon',
 'Bought': ['P1', 'P2']}
</code></pre>
and
<pre><code>print(find_user('000', lsts))
</code></pre>
prints
<pre><code>user not found
</code></pre>
If however, your data is like:
<pre><code>d = { &quot;Data&quot;: [{&quot;User_ID&quot;:&quot;Z000&quot;},{&quot;User_ID&quot;:&quot;A999&quot;} ]}
</code></pre>
Then you can pass the list inside the dictionary to the function like:
<pre><code>find_user('Z000', d['Data'])
</code></pre>
which returns
<pre><code>{'User_ID': 'Z000'}
</code></pre>

blocks|key|3270111|text|希望这段代码对你有用。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3270112|+def+find_user(val):
+++++++for+dict_key+in+l:
++++++++++if+dict_key["User_ID"]+==+val:
+++++++++++++return+dict_key
+++++++else:
++++++++++return+"User+Not+Found"
++++
++++print(find_user("Z000"))|code-block|syntax|javascript|3270113|这是你所有字典的目录。|3270114|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Hope this code works for you.
<pre><code> def find_user(val):
 for dict_key in l:
 if dict_key[&quot;User_ID&quot;] == val:
 return dict_key
 else:
 return &quot;User Not Found&quot;
 
 print(find_user(&quot;Z000&quot;))
</code></pre>
here l is the list that stores all your dictionaries.

I'm trying to write a function that takes a value for a given key (User_ID) and return the full dictionary for that value. I understand that this can probably be achieved without writing a function but as a beginner I'm trying to build my knowledge with functions.
My data is a list of dictionaries and looks like this:
<pre><code>[
 {
 &quot;User_ID&quot;:&quot;Z000&quot;,
 &quot;DOB&quot;:&quot;01.01.1960&quot;,
 &quot;State&quot;:&quot;Oregon&quot;,
 &quot;Bought&quot;:[&quot;P1&quot;,&quot;P2&quot;]
 },
 {
 &quot;User_ID&quot;:&quot;A999&quot;,
 &quot;DOB&quot;:&quot;01.01.1980&quot;,
 &quot;State&quot;:&quot;Texas&quot;,
 &quot;Bought&quot;:[&quot;P5&quot;,&quot;P9&quot;]
 }
]
</code></pre>
I wrote the following function but I realized that this is would only work for a dictionary but I have a list of dictionaries. How can I make it to take the <code>User_ID</code> value and return the full dictionary including the <code>User_ID</code>, <code>DOB</code>, <code>State</code> and <code>Bought</code>.
<pre><code>def find_user(val):
 for key, value in dict_1.items():
 if val == key:
 return value
 
 return &quot;user not found&quot;
</code></pre>

Python function that takes value from a dict and return full dictionary for the value

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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我试图编写一个函数，该函数接受给定键(User_ID)的值，并返回该值的完整字典。我知道这可能不需要编写函数就能实现，但作为初学者，我正在尝试用函数来构建我的知识。我的数据是字典列表，如下所示：[   {      "User_ID":"Z000",      "DOB":"01.01.1960",      "Sta...

问Python函数，该函数从dict获取值，并为该值返回完整字典。
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问Python函数，该函数从dict获取值，并为该值返回完整字典。EN