blocks|key|4606906|text|我不认为熊猫有什么内置的东西来创建数据的嵌套字典。下面是一些代码，它们一般适用于使用defaultdict的带有MultiIndex的系列|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4606907|嵌套代码遍历MultIndex的每一层，将层添加到字典中，直到最深的层被分配给Series值。|4606908|In++[99]:+from+collections+import+defaultdict

In+[100]:+results+=+defaultdict(lambda:+defaultdict(dict))

In+[101]:+for+index,+value+in+grouped.itertuples():
+++++...:+++++for+i,+key+in+enumerate(index):
+++++...:+++++++++if+i+==+0:
+++++...:+++++++++++++nested+=+results[key]
+++++...:+++++++++elif+i+==+len(index)+-+1:
+++++...:+++++++++++++nested[key]+=+value
+++++...:+++++++++else:
+++++...:+++++++++++++nested+=+nested[key]

In+[102]:+results
Out[102]:+defaultdict(<function+<lambda>+at+0x7ff17c76d1b8>,+{2010:+defaultdict(<type+'dict'>,+{'govnr':+{'pati+mara':+500.0,+'jess+rapp':+80.0},+'mayor':+{'joe+smith':+100.0,+'jay+gould':+12.0}})})

In+[106]:+print+json.dumps(results,+indent=4)
{
++++"2010":+{
++++++++"govnr":+{
++++++++++++"pati+mara":+500.0,+
++++++++++++"jess+rapp":+80.0
++++++++},+
++++++++"mayor":+{
++++++++++++"joe+smith":+100.0,+
++++++++++++"jay+gould":+12.0
++++++++}
++++}
}|code-block|syntax|javascript|4606909|entityMap^0|16|B|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|R|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|S|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|N|$]]

I don't think think there is anything built-in to pandas to create a nested dictionary of the data. Below is some code that should work in general for a series with a MultiIndex, using a <code>defaultdict</code>

The nesting code iterates through each level of the MultIndex, adding layers to the dictionary until the deepest layer is assigned to the Series value.

<pre><code>In [99]: from collections import defaultdict

In [100]: results = defaultdict(lambda: defaultdict(dict))

In [101]: for index, value in grouped.itertuples():
 ...: for i, key in enumerate(index):
 ...: if i == 0:
 ...: nested = results[key]
 ...: elif i == len(index) - 1:
 ...: nested[key] = value
 ...: else:
 ...: nested = nested[key]

In [102]: results
Out[102]: defaultdict(&lt;function &lt;lambda&gt; at 0x7ff17c76d1b8&gt;, {2010: defaultdict(&lt;type 'dict'&gt;, {'govnr': {'pati mara': 500.0, 'jess rapp': 80.0}, 'mayor': {'joe smith': 100.0, 'jay gould': 12.0}})})

In [106]: print json.dumps(results, indent=4)
{
 "2010": {
 "govnr": {
 "pati mara": 500.0, 
 "jess rapp": 80.0
 }, 
 "mayor": {
 "joe smith": 100.0, 
 "jay gould": 12.0
 }
 }
}
</code></pre>

blocks|key|4606955|text|我知道这是一个老问题，但我最近遇到了同样的问题。这是我的解决方案。我从chrisb的例子中借鉴了很多东西(谢谢！)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4606956|这样做的好处是，您可以传递一个lambda，以便从您想要的任何可枚举值以及每个组中获取最终值。|4606957|from+collections+import+defaultdict

def+dict_from_enumerable(enumerable,+final_value,+*groups):
++++d+=+defaultdict(lambda:+defaultdict(dict))
++++group_count+=+len(groups)
++++for+item+in+enumerable:
++++++++nested+=+d
++++++++item_result+=+final_value(item)+if+callable(final_value)+else+item.get(final_value)
++++++++for+i,+group+in+enumerate(groups,+start=1):
++++++++++++group_val+=+str(group(item)+if+callable(group)+else+item.get(group))
++++++++++++if+i+==+group_count:
++++++++++++++++nested[group_val]+=+item_result
++++++++++++else:
++++++++++++++++nested+=+nested[group_val]
++++return+d|code-block|syntax|javascript|4606958|在问题中，您可以像这样调用此函数：|4606959|dict_from_enumerable(grouped.itertuples(),+'amount',+'year',+'office',+'candidate')|4606960|第一个参数也可以是数据数组，甚至不需要pandas。|4606961|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|$]]

I'm aware this is an old question, but I came across the same issue recently. Here's my solution. I borrowed a lot of stuff from chrisb's example (Thank you!).

This has the advantage that you can pass a lambda to get the final value from whatever enumerable you want, as well as for each group.

<pre><code>from collections import defaultdict

def dict_from_enumerable(enumerable, final_value, *groups):
 d = defaultdict(lambda: defaultdict(dict))
 group_count = len(groups)
 for item in enumerable:
 nested = d
 item_result = final_value(item) if callable(final_value) else item.get(final_value)
 for i, group in enumerate(groups, start=1):
 group_val = str(group(item) if callable(group) else item.get(group))
 if i == group_count:
 nested[group_val] = item_result
 else:
 nested = nested[group_val]
 return d
</code></pre>

In the question, you'd call this function like:

<pre><code>dict_from_enumerable(grouped.itertuples(), 'amount', 'year', 'office', 'candidate')
</code></pre>

The first argument can be an array of data as well, not even requiring pandas.

blocks|key|100509|text|我看了一下上面的解决方案，发现它只适用于3层嵌套。此解决方案适用于任意数量的级别。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|100510|import+json
levels+=+len(grouped.index.levels)
dicts+=+[{}+for+i+in+range(levels)]
last_index+=+None

for+index,value+in+grouped.itertuples():

++++if+not+last_index:
++++++++last_index+=+index

++++for+(ii,(i,j))+in+enumerate(zip(index,+last_index)):
++++++++if+not+i+==+j:
++++++++++++ii+=+levels+-+ii+-1
++++++++++++dicts[:ii]+=++[{}+for+_+in+dicts[:ii]]
++++++++++++break

++++for+i,+key+in+enumerate(reversed(index)):
++++++++dicts[i][key]+=+value
++++++++value+=+dicts[i]

++++last_index+=+index


result+=+json.dumps(dicts[-1])|code-block|syntax|javascript|100511|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|$]]

I had a look at the solution above and figured out that it only works for 3 levels of nesting. This solution will work for any number of levels.

<pre><code>import json
levels = len(grouped.index.levels)
dicts = [{} for i in range(levels)]
last_index = None

for index,value in grouped.itertuples():

 if not last_index:
 last_index = index

 for (ii,(i,j)) in enumerate(zip(index, last_index)):
 if not i == j:
 ii = levels - ii -1
 dicts[:ii] = [{} for _ in dicts[:ii]]
 break

 for i, key in enumerate(reversed(index)):
 dicts[i][key] = value
 value = dicts[i]

 last_index = index


result = json.dumps(dicts[-1])
</code></pre>

blocks|key|100535|text|以下是此问题的通用递归解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|100536|def+df_to_dict(df):
++++if+df.ndim+==+1:
++++++++return+df.to_dict()

++++ret+=+{}
++++for+key+in+df.index.get_level_values(0):
++++++++sub_df+=+df.xs(key)
++++++++ret[key]+=+df_to_dict(sub_df)
++++return+ret|code-block|syntax|javascript|100537|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 a generic recursive solution for this problem:

<pre class="lang-py prettyprint-override"><code>def df_to_dict(df):
 if df.ndim == 1:
 return df.to_dict()

 ret = {}
 for key in df.index.get_level_values(0):
 sub_df = df.xs(key)
 ret[key] = df_to_dict(sub_df)
 return ret
</code></pre>

I often use pandas groupby to generate stacked tables. But then I often want to output the resulting nested relations to json. Is there any way to extract a nested json filed from the stacked table it produces? 

Let's say I have a df like: 

<pre><code>year office candidate amount
2010 mayor joe smith 100.00
2010 mayor jay gould 12.00
2010 govnr pati mara 500.00
2010 govnr jess rapp 50.00
2010 govnr jess rapp 30.00
</code></pre>

I can do: 

<pre><code>grouped = df.groupby('year', 'office', 'candidate').sum()

print grouped
 amount
year office candidate 
2010 mayor joe smith 100
 jay gould 12
 govnr pati mara 500
 jess rapp 80
</code></pre>

Beautiful! Of course, what I'd real like to do is get nested json via a command along the lines of grouped.to_json. But that feature isn't available. Any workarounds? 

So, what I really want is something like: 

<pre><code>{"2010": {"mayor": [
 {"joe smith": 100},
 {"jay gould": 12}
 ]
 }, 
 {"govnr": [
 {"pati mara":500}, 
 {"jess rapp": 80}
 ]
 }
}
</code></pre>

Don

pandas groupby to nested json

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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我经常使用pandas groupby来生成堆叠表。但是，我经常希望将得到的嵌套关系输出到json。有没有办法从它产生的堆叠表中提取一个嵌套的json文件？假设我有一个如下的df：year office candidate  amount2010 mayor  joe smith  100.002010 mayor  ...

问pandas groupby to嵌套的json
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