Python 将元素为字典的列拆分为多个列
我有一个包含字典作为元素的单列的Python 将元素为字典的列拆分为多个列,python,pandas,Python,Pandas,我有一个包含字典作为元素的单列的DataFrame。这是以下代码的结果: dg # is a pandas dataframe with columns ID and VALUE. Many rows contain the same ID def seriesFeatures(series): """This functions receives a series of VALUE for the same ID and extracts tens of complex fe
DataFrame
。这是以下代码的结果:
dg # is a pandas dataframe with columns ID and VALUE. Many rows contain the same ID
def seriesFeatures(series):
"""This functions receives a series of VALUE for the same ID and extracts
tens of complex features from the series, storing them into a dictionary"""
dico = dict()
dico['feature1'] = calculateFeature1
dico['feature2'] = calculateFeature2
# Many more features
dico['feature50'] = calculateFeature50
return dico
grouped = dg.groupby(['ID'])
dh = grouped['VALUE'].agg( { 'all_features' : lambda s: seriesFeatures(s) } )
dh.reset_index()
# Here I get a dh DataFrame of a single column 'all_features' and
# dictionaries stored on its values. The keys are the feature's names
我需要将此'all_features'
列拆分为尽可能多的列(我有太多的行和列,并且我无法更改seriesFeatures
函数),因此输出将是一个包含列ID
,FEATURE1
,FEATURE2
的数据帧,功能3
<代码>功能50。这样做的最佳方式是什么
编辑
一个具体而简单的例子:
dg = pd.DataFrame( [ [1,10] , [1,15] , [1,13] , [2,14] , [2,16] ] , columns=['ID','VALUE'] )
def seriesFeatures(series):
dico = dict()
dico['feature1'] = len(series)
dico['feature2'] = series.sum()
return dico
grouped = dg.groupby(['ID'])
dh = grouped['VALUE'].agg( { 'all_features' : lambda s: seriesFeatures(s) } )
dh.reset_index()
但当我尝试用pd.Series或pd.DataFrame包装它时,它说如果数据是标量值,则必须提供索引。提供index=['feature1','feature2'],我会得到奇怪的结果,例如使用:
dh=grouped['VALUE'].agg({'all_features':lambda s:pd.DataFrame(seriesfeactions,index=['feature1','feature2'])
我认为应该将dict打包成一个系列,然后这将在groupby调用中展开(但随后使用apply
而不是agg
,因为它不再是聚合(标量)结果):
之后,可以将结果重塑为所需的格式
通过您的简单示例,这似乎是可行的:
In [22]: dh = grouped['VALUE'].apply(lambda x: pd.Series(seriesFeatures(x)))
In [23]: dh
Out[23]:
ID
1 feature1 3
feature2 38
2 feature1 2
feature2 30
dtype: int64
In [26]: dh.unstack().reset_index()
Out[26]:
ID feature1 feature2
0 1 3 38
1 2 2 30
谢谢你的示例案例!更新了我的答案。谢谢。我不知道这个
取消堆叠的事情,这似乎是一个很好的解决方案。
In [22]: dh = grouped['VALUE'].apply(lambda x: pd.Series(seriesFeatures(x)))
In [23]: dh
Out[23]:
ID
1 feature1 3
feature2 38
2 feature1 2
feature2 30
dtype: int64
In [26]: dh.unstack().reset_index()
Out[26]:
ID feature1 feature2
0 1 3 38
1 2 2 30