Question

我的pandas数据框包含一个包含值或值列表（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，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)

Answer 1

没有那么多有趣/花哨的熊猫用法，但这有效：

import numpy as np
dfIn.loc[:, 'location'] = dfIn.location.apply(np.atleast_1d)
all_locations = np.hstack(dfIn.location)
all_names = np.hstack([[n]*len(l) for n, l in dfIn[['name', 'location']].values])
dfOut = pd.DataFrame({'location':all_locations, 'name':all_names})

它比apply / stack / reindex方法快约40倍。据我所知，该比率几乎适用于所有数据帧大小（没有测试它如何随着每行中列表的大小而缩放）。如果您可以保证所有location个条目都已经是可迭代的，那么您可以删除atleast_1d调用，这样可以提高另外20％的速度。

Answer 2

如果您返回index是地理列表的系列，那么dfIn.apply会将这些系列整理成表格：

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

Answer 3

import pandas as pd


dfIn = pd.DataFrame({
    u'name': ['Tom', 'Jim', 'Claus'],
    u'location': ['Amsterdam', ['Berlin','Paris'], ['Antwerp','Barcelona','Pisa'] ],
})

print(dfIn.explode('location'))

>>>
    name   location
0    Tom  Amsterdam
1    Jim     Berlin
1    Jim      Paris
2  Claus    Antwerp
2  Claus  Barcelona
2  Claus       Pisa

迭代行并扩展pandas数据帧

3 个答案: