我有一个pandas数据帧,其中包含值或值列表(长度不等)的列.我想“扩展”行,因此列表中的每个值都会成为列中的单个值.一个例子说明了一切:
dfIn = pd.DataFrame({u'name': ['Tom', 'Jim', 'Claus'],
u'location': ['Amsterdam', ['Berlin','Paris'], ['Antwerp','Barcelona','Pisa'] ]})
location name
0 Amsterdam Tom
1 [Berlin, Paris] Jim
2 [Antwerp, Barcelona, Pisa] Claus
我想变成:
dfOut = pd.DataFrame({u'name': ['Tom', 'Jim', 'Jim', 'Claus','Claus','Claus'],
u'location': ['Amsterdam', 'Berlin','Paris', 'Antwerp','Barcelona','Pisa']})
location name
0 Amsterdam Tom
1 Berlin Jim
2 Paris Jim
3 Antwerp Claus
4 Barcelona Claus
5 Pisa Claus
我首先尝试使用apply但据我所知,它不可能返回多个系列. iterrows似乎是诀窍.但是下面的代码给了我一个空数据框……
def duplicator(series):
if type(series['location']) == list:
for location in series['location']:
subSeries = series
subSeries['location'] = location
dfOut.append(subSeries)
else:
dfOut.append(series)
for index, row in dfIn.iterrows():
duplicator(row)
解决方法:
如果返回索引为位置列表的系列,则dfIn.apply会将这些系列整理到表中:
import pandas as pd
dfIn = pd.DataFrame({u'name': ['Tom', 'Jim', 'Claus'],
u'location': ['Amsterdam', ['Berlin','Paris'],
['Antwerp','Barcelona','Pisa'] ]})
def expand(row):
locations = row['location'] if isinstance(row['location'], list) else [row['location']]
s = pd.Series(row['name'], index=list(set(locations)))
return s
In [156]: dfIn.apply(expand, axis=1)
Out[156]:
Amsterdam Antwerp Barcelona Berlin Paris Pisa
0 Tom NaN NaN NaN NaN NaN
1 NaN NaN NaN Jim Jim NaN
2 NaN Claus Claus NaN NaN Claus
然后,您可以堆叠此DataFrame以获取:
In [157]: dfIn.apply(expand, axis=1).stack()
Out[157]:
0 Amsterdam Tom
1 Berlin Jim
Paris Jim
2 Antwerp Claus
Barcelona Claus
Pisa Claus
dtype: object
这是一个系列,而你想要一个DataFrame.使用reset_index进行一点按摩可以获得所需的结果:
dfOut = dfIn.apply(expand, axis=1).stack()
dfOut = dfOut.to_frame().reset_index(level=1, drop=False)
dfOut.columns = ['location', 'name']
dfOut.reset_index(drop=True, inplace=True)
print(dfOut)
产量
location name
0 Amsterdam Tom
1 Berlin Jim
2 Paris Jim
3 Amsterdam Claus
4 Antwerp Claus
5 Barcelona Claus
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