Pandas：将 DataFrameGroupBy 对象转换为所需的格式
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Pandas: transforming the DataFrameGroupBy object to desired format
提问by Zhubarb
I have a data frame as follows:
我有一个数据框如下：
import pandas as pd
import numpy as np
df = pd.DataFrame({'id' : range(1,9),
'code' : ['one', 'one', 'two', 'three',
'two', 'three', 'one', 'two'],
'colour': ['black', 'white','white','white',
'black', 'black', 'white', 'white'],
'amount' : np.random.randn(8)}, columns= ['id','code','colour','amount'])
I want to be able to group the ids by codeand colourand then sort them with respect to amount.I know how to groupby():
我希望能够到组idS按code和colour，然后排序他们就amount。我知道如何groupby()：
df.groupby(['code','colour']).head(5)
id code colour amount
code colour
one black 0 1 one black -0.117307
white 1 2 one white 1.653216
6 7 one white 0.817205
three black 5 6 three black 0.567162
white 3 4 three white 0.579074
two black 4 5 two black -1.683988
white 2 3 two white -0.457722
7 8 two white -1.277020
However, my desired output is as below, where I have two columns: 1.code/colourcontains the key strings and 2.id:amountcontains id- amounttuples sorted in descending order wrt amount:
但是，我想要的输出如下，我有两列： 1.code/colour包含关键字符串和 2.id:amount包含id-amount按降序排序的元组 wrt amount：
code/colour id:amount
one/black {1:-0.117307}
one/white {2:1.653216, 7:0.817205}
three/black {6:0.567162}
three/white {4:0.579074}
two/black {5:-1.683988}
two/white {3:-0.457722, 8:-1.277020}
How can I transform the DataFrameGroupByobject displayed above to my desired format? Or, shall I not use groupby()in the first place?
如何DataFrameGroupBy将上面显示的对象转换为我想要的格式？或者，我一开始就不应该使用groupby()吗？
EDIT:Although not in the specified format, the code below kind of gives me the functionality I want:
编辑：虽然不是指定的格式，但下面的代码给了我我想要的功能：
groups = dict(list(df.groupby(['code','colour'])))
groups['one','white']
id code colour amount
1 2 one white 1.331766
6 7 one white 0.808739
How can I reduce the groups to only include the idand amountcolumn?
如何减少组以仅包含id和amount列？
回答by waitingkuo
First, groupby code and colour and then apply a customized function to format id and amount:
首先，groupby 代码和颜色，然后应用自定义函数来格式化 id 和数量：
df = df.groupby(['code', 'colour']).apply(lambda x:x.set_index('id').to_dict('dict')['amount'])
And then modify the index:
然后修改索引：
df.index = ['/'.join(i) for i in df.index]
It will return a series, you can convert it back to DataFrame by:
它将返回一个系列，您可以通过以下方式将其转换回 DataFrame：
df = df.reset_index()
Finally, add the column names by:
最后，通过以下方式添加列名：
df.columns=['code/colour','id:amount']
Result:
结果：
In [105]: df
Out[105]:
code/colour id:amount
0 one/black {1: 0.392264412544}
1 one/white {2: 2.13950686015, 7: -0.393002947047}
2 three/black {6: -2.0766612539}
3 three/white {4: -1.18058561325}
4 two/black {5: -1.51959565941}
5 two/white {8: -1.7659863039, 3: -0.595666853895}
回答by Nipun Batra
Here is an "ugly" way of doing this. First things first- your desired output will not play so well within Pandas since dictis unhashable; so you may lose the real benefit!
这是一种“丑陋”的方法。首先，您想要的输出在 Pandas 中不会表现得很好，因为它dict是不可散列的；所以你可能会失去真正的好处！
od = OrderedDict()
for name, group in df.groupby(['code', 'colour']):
# Convert the group to a dict
temp = group[['id', 'amount']].sort(['amount'], ascending=[0]).to_dict()
# Extract id:amount
temp2 = {temp['id'][key]: temp['amount'][key] for key in temp['amount'].iterkeys()}
od["%s/%s" % (name)] = temp2
This is only a start! Not exactly what you are looking for.
这只是一个开始！不完全是你正在寻找的。
