将pandas DataFrame列展开为多行

Question

如果我有DataFrame这样的:

pd.DataFrame( {"name" : "John", 
               "days" : [[1, 3, 5, 7]]
              })

给出这个结构:

           days  name
0  [1, 3, 5, 7]  John

如何将其扩展到以下？

   days  name
0     1  John
1     3  John
2     5  John
3     7  John

Answer 1

您可以使用df.itertuples迭代每一行,并使用列表推导将数据重新整形为所需的形式:

import pandas as pd

df = pd.DataFrame( {"name" : ["John", "Eric"], 
               "days" : [[1, 3, 5, 7], [2,4]]})
result = pd.DataFrame([(d, tup.name) for tup in df.itertuples() for d in tup.days])
print(result)

产量

   0     1
0  1  John
1  3  John
2  5  John
3  7  John
4  2  Eric
5  4  Eric

Divakar的解决方案,using_repeat是最快的:

In [48]: %timeit using_repeat(df)
1000 loops, best of 3: 834 µs per loop

In [5]: %timeit using_itertuples(df)
100 loops, best of 3: 3.43 ms per loop

In [7]: %timeit using_apply(df)
1 loop, best of 3: 379 ms per loop

In [8]: %timeit using_append(df)
1 loop, best of 3: 3.59 s per loop

以下是用于上述基准的设置:

import numpy as np
import pandas as pd

N = 10**3
df = pd.DataFrame( {"name" : np.random.choice(list('ABCD'), size=N), 
                    "days" : [np.random.randint(10, size=np.random.randint(5))
                              for i in range(N)]})

def using_itertuples(df):
    return  pd.DataFrame([(d, tup.name) for tup in df.itertuples() for d in tup.days])

def using_repeat(df):
    lens = [len(item) for item in df['days']]
    return pd.DataFrame( {"name" : np.repeat(df['name'].values,lens), 
                          "days" : np.concatenate(df['days'].values)})

def using_apply(df):
    return (df.apply(lambda x: pd.Series(x.days), axis=1)
            .stack()
            .reset_index(level=1, drop=1)
            .to_frame('day')
            .join(df['name']))

def using_append(df):
    df2 = pd.DataFrame(columns = df.columns)
    for i,r in df.iterrows():
        for e in r.days:
            new_r = r.copy()
            new_r.days = e
            df2 = df2.append(new_r)
    return df2

Answer 2

Pandas 0.25 以来的新功能，您可以使用该功能 explode()

https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.explode.html

import pandas as pd
df = pd.DataFrame( {"name" : "John", 
               "days" : [[1, 3, 5, 7]]})

print(df.explode('days'))

印刷

   name days
0  John    1
0  John    3
0  John    5
0  John    7

Answer 3

这是NumPy的一些东西 -

lens = [len(item) for item in df['days']]
df_out = pd.DataFrame( {"name" : np.repeat(df['name'].values,lens), 
               "days" : np.hstack(df['days'])
              })

如指出要快.@unutbu's solution np.concatenate(df['days'].values)np.hstack(df['days'])

它使用循环理解来提取每个'days'元素的长度,这在运行时必须是最小的.

样品运行 -

>>> df
           days  name
0  [1, 3, 5, 7]  John
1        [2, 4]  Eric
>>> lens = [len(item) for item in df['days']]
>>> pd.DataFrame( {"name" : np.repeat(df['name'].values,lens), 
...                "days" : np.hstack(df['days'])
...               })
   days  name
0     1  John
1     3  John
2     5  John
3     7  John
4     2  Eric
5     4  Eric

Answer 4

一种“本地”熊猫解决方案-我们将列拆成一系列，然后根据索引重新加入：

import pandas as pd #import
x2 = x.days.apply(lambda x: pd.Series(x)).unstack() #make an unstackeded series, x2
x.drop('days', axis = 1).join(pd.DataFrame(x2.reset_index(level=0, drop=True))) #drop the days column, join to the x2 series