blocks|key|2823707|text|你为什么不能这样把它们储存起来？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2823708|{
+0+:+[
++++[6],
++++[1,4,5,6]
++++...],
+1+:+...
}|code-block|syntax|javascript|2823709|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|$]]

Why can't you store them like this?

<pre><code>{
 0 : [
 [6],
 [1,4,5,6]
 ...],
 1 : ...
}
</code></pre>

blocks|key|1110940|text|这个数据的格式有点奇怪，但这可能会起作用。不过，我建议将这些数据纳入比这更好的结构。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1110941|def+get_values(data,+key=0):
+++++out+=+[]
+++++for+item+in+data:
+++++++++if+item.get(key):
+++++++++++++out.extend(item[key])
+++++return+out|code-block|syntax|javascript|1110942|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|$]]

Well the format of that data is kind of strange, but this will probably work. I would suggest putting that data into a better structure than that though. 

<pre><code>def get_values(data, key=0):
 out = []
 for item in data:
 if item.get(key):
 out.extend(item[key])
 return out
</code></pre>

blocks|key|1110960|text|您有一个相当奇怪的数据结构，但在编写数据结构时，可以使用以下代码对其进行修改：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1110961|list_of_dicts+=+[...]
output+=+dict()

for+entry+in+list_of_dicts:
++++for+key+in+entry:
++++++++if+output.has_key(key):
++++++++++++output[key].append(entry[key])
++++++++else:
++++++++++++output[key]+=+entry[key]|code-block|syntax|javascript|1110962|注意，这段代码并不是假定列表中的每个dict都只有一个键，因为在一般情况下，每个dict可能包含多个条目。此外，代码并不认为每个list元素都是唯一的，在这种情况下，一个集合可能更合适。|1110963|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|$]]

You have a rather strange data structure, but as it is written, it can be modified with the following code:

<pre><code>list_of_dicts = [...]
output = dict()

for entry in list_of_dicts:
 for key in entry:
 if output.has_key(key):
 output[key].append(entry[key])
 else:
 output[key] = entry[key]
</code></pre>

Note that this code does not assume every dict in the list to have only a single key, as in the general case it could be possible for each dict to contain multiple entries. Further, the code does not assume each list element to be unique, in which case a set might have been more appropriate.

blocks|key|1175886|text|使用dict.setdefault()函数(对于所有值)的解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1175887|d+=+{}
for+i+in+l:
++++k+=+list(i.keys())[0]
++++d.setdefault(k,+[])
++++d[k]+=+d[k]+%2B+i[k]

print(d)|code-block|syntax|javascript|1175888|产出：|1175889|{0:+[6,+1,+4,+5,+6,+3],+2:+[1,+13,+15,+8,+19,+8,+1,+23,+8,+26,+23,+23,+23,+34,+39,+23,+34,+1,+1,+42,+23,+34,+24,+15,+8],+3:+[5,+6,+8,+9,+0,+1,+5,+6,+13,+17,+18,+12,+2,+6,+9,+13,+15,+16,+4,+5,+6,+17,+20,+21,+22,+5,+12,+17,+20,+21,+26,+27,+2,+8,+9,+22,+26,+28,+4,+5,+6,+20,+21,+8,+1,+13,+32,+33,+16,+13,+9,+17,+20,+21,+27,+28,+32,+35,+36,+6,+9,+16,+8,+37,+38,+7,+9,+26,+27,+28,+29,+9,+22,+1,+34,+15,+25,+29,+1,+14,+20,+40,+14,+13,+41,+5,+32,+35,+36,+37,+38,+6,+21,+42]}|1175890|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3/library/stdtypes.html#dict.setdefault^0|2|H|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|Z|8|@]|9|@]|C|$]]|$1|K|3|L|5|F|7|10|8|@]|9|@]|C|$G|H]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]]]

The solution using <a href="https://docs.python.org/3/library/stdtypes.html#dict.setdefault" rel="nofollow noreferrer">dict.setdefault()</a> function(for all values):

<pre><code>d = {}
for i in l:
 k = list(i.keys())[0]
 d.setdefault(k, [])
 d[k] = d[k] + i[k]

print(d)
</code></pre>

The output:

<pre><code>{0: [6, 1, 4, 5, 6, 3], 2: [1, 13, 15, 8, 19, 8, 1, 23, 8, 26, 23, 23, 23, 34, 39, 23, 34, 1, 1, 42, 23, 34, 24, 15, 8], 3: [5, 6, 8, 9, 0, 1, 5, 6, 13, 17, 18, 12, 2, 6, 9, 13, 15, 16, 4, 5, 6, 17, 20, 21, 22, 5, 12, 17, 20, 21, 26, 27, 2, 8, 9, 22, 26, 28, 4, 5, 6, 20, 21, 8, 1, 13, 32, 33, 16, 13, 9, 17, 20, 21, 27, 28, 32, 35, 36, 6, 9, 16, 8, 37, 38, 7, 9, 26, 27, 28, 29, 9, 22, 1, 34, 15, 25, 29, 1, 14, 20, 40, 14, 13, 41, 5, 32, 35, 36, 37, 38, 6, 21, 42]}
</code></pre>

blocks|key|1175892|text|下面是一个很好的功能：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1175893|from+itertools+import+chain

def+find(index):
++++return+list(chain.from_iterable([i[index]+for+i+in+d+if+i.keys()==[index]]))|code-block|syntax|javascript|1175894|它提供了以下输出：|1175895|>>>+find(0)
[6,+1,+4,+5,+6,+3]
>>>+find(2)
[1,+13,+15,+8,+19,+8,+1,+23,+8,+26,+23,+23,+23,+34,+39,+23,+34,+1,+1,+42,+23,+34,+24,+15,+8]
>>>+|1175896|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|$]]

Here is a nice function to do it:

<pre><code>from itertools import chain

def find(index):
 return list(chain.from_iterable([i[index] for i in d if i.keys()==[index]]))
</code></pre>

Which gives the following output:

<pre><code>&gt;&gt;&gt; find(0)
[6, 1, 4, 5, 6, 3]
&gt;&gt;&gt; find(2)
[1, 13, 15, 8, 19, 8, 1, 23, 8, 26, 23, 23, 23, 34, 39, 23, 34, 1, 1, 42, 23, 34, 24, 15, 8]
&gt;&gt;&gt; 
</code></pre>

blocks|key|1110999|text|下面是您的问题的简单逻辑代码:希望您在此之后找到自己的出路；只需在python(3)中完美地工作。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1111000|f+=+[{0:+[6]},
++++{0:+[1,+4,+5,+6]},
++++{0:+[3]},
++++{2:+[1,+13,+15]},
++++{2:+[8,+19]},
++++{2:+[8]},
++++{2:+[1,+23]},
++++{2:+[8,+26]},
++++{2:+[23]},
++++{2:+[23]},
++++{2:+[23,+34]},
++++{2:+[39]},
++++{2:+[23,+34,+1]},
++++{2:+[1,+42]},
++++{2:+[23,+34]},
++++{2:+[24]},
++++{2:+[15,+8]},
++++{3:+[5,+6,+8,+9]},
++++{3:+[0]},
++++{3:+[1]},
++++{3:+[5,+6,+13,+17,+18]},
++++{3:+[12]},
++++{3:+[2,+6,+9,+13,+15,+16]}]

++++c+=+int+(input("Enter+the+number+that+you+want+to+look+for+"))
++++z+=+[]
++++for+x+in+f:

++++++++for+keys+,+values+in+x.items():

++++++++++++if+keys==c+:
+++++++++++++++++z+=+z+%2B+values
++++print+(z)|code-block|syntax|javascript|1111001|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|$]]

Here is the simple logical code for your problem: Hope you will find your own way out after this; Will just work perfectly in python(3)

<pre><code>f = [{0: [6]},
 {0: [1, 4, 5, 6]},
 {0: [3]},
 {2: [1, 13, 15]},
 {2: [8, 19]},
 {2: [8]},
 {2: [1, 23]},
 {2: [8, 26]},
 {2: [23]},
 {2: [23]},
 {2: [23, 34]},
 {2: [39]},
 {2: [23, 34, 1]},
 {2: [1, 42]},
 {2: [23, 34]},
 {2: [24]},
 {2: [15, 8]},
 {3: [5, 6, 8, 9]},
 {3: [0]},
 {3: [1]},
 {3: [5, 6, 13, 17, 18]},
 {3: [12]},
 {3: [2, 6, 9, 13, 15, 16]}]

 c = int (input("Enter the number that you want to look for "))
 z = []
 for x in f:

 for keys , values in x.items():

 if keys==c :
 z = z + values
 print (z)
</code></pre>

blocks|key|1172014|text|d+=+{}
for+i+in+l:
++++k,v+=+list(i.items())[0]
++++d.setdefault(k,+[]).extend(v)

print(d)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1172015|(k，v)+(0,3))|unstyled|1172016|(k，v)+(2,1，13，15)|1172017|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|F|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>d = {}
for i in l:
 k,v = list(i.items())[0]
 d.setdefault(k, []).extend(v)

print(d)
</code></pre>

<h1>(k,v) (0,[3]))</h1>

<h1>(k,v) (2,[1, 13, 15])</h1>

I have a list in dictionary form looking like this:

<pre><code>[{0: [6]},
 {0: [1, 4, 5, 6]},
 {0: [3]},
 {2: [1, 13, 15]},
 {2: [8, 19]},
 {2: [8]},
 {2: [1, 23]},
 {2: [8, 26]},
 {2: [23]},
 {2: [23]},
 {2: [23, 34]},
 {2: [39]},
 {2: [23, 34, 1]},
 {2: [1, 42]},
 {2: [23, 34]},
 {2: [24]},
 {2: [15, 8]},
 {3: [5, 6, 8, 9]},
 {3: [0]},
 {3: [1]},
 {3: [5, 6, 13, 17, 18]},
 {3: [12]},
 {3: [2, 6, 9, 13, 15, 16]},
 {3: [4, 5, 6, 17, 20, 21, 22]},
 {3: [5, 12, 17, 20, 21, 26, 27]},
 {3: [2, 8, 9, 22, 26, 28]},
 {3: [4, 5, 6, 20, 21, 8, 1, 13, 32, 33]},
 {3: [16, 13]},
 {3: [9, 17, 20, 21, 27, 28, 32, 35, 36]},
 {3: [6, 9, 16, 8, 37, 38]},
 {3: [7, 9, 26, 27, 28, 29]},
 {3: [9, 22, 1, 34]},
 {3: [15, 25]},
 {3: [29, 1]},
 {3: [14, 20, 40]},
 {3: [14, 13, 41]},
 {3: [5, 32, 35, 36, 37, 38]},
 {3: [6, 21, 42]}]
</code></pre>

Goes on like this for a while. The first numbers are really important.
Now I want to find values '0', '2' and '3' and look which values are connected to this.

What I want ultimately is to look up the different values and save all the values connected in a list.

For 0:

<pre><code>0: [6,1, 4, 5, 6,3]
</code></pre>

I can't really find out how to do it.

Search value in list dictionary

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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我有一个字典形式的列表，如下所示：[{0: [6]}, {0: [1, 4, 5, 6]}, {0: [3]}, {2: [1, 13, 15]}, {2: [8, 19]}, {2: [8]}, {2: [1, 23]}, {2: [8, 26]}, {2: [23]}, {2: [23]}, {2: [23, 34...

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