基于列列表转置数据框

Iva*_*sky 7

您可以连接explode()，然后将表转回所需的输出！

df = df.explode('cNames').explode('cValues')
df['cValues'] = pd.to_numeric(df['cValues'])
print(df.pivot_table(columns='cNames',index='number',values='cValues'))

输出：

cNames     a     b    c     d
number                       
10       2.0   2.0  2.0   NaN
20      66.0  66.0  NaN  66.0

遗憾的是，explode 的输出是类型，object因此我们必须先将其转换为 ，pd.to_numeric()然后再进行旋转。否则将没有要聚合的数值。

Qua*_*ang 7

一种选择是concat：

pd.concat([pd.Series(x['cValues'], x['cNames'], name=idx) 
               for idx, x in df.iterrows()], 
          axis=1
         ).T.join(df.iloc[:,2:])

或 DataFrame 构造：

pd.DataFrame({idx: dict(zip(x['cNames'], x['cValues']) )
              for idx, x in df.iterrows()
            }).T.join(df.iloc[:,2:])

输出：

      a     b    c     d  number
0   1.0   2.0  3.0   NaN      10
1  55.0  66.0  NaN  77.0      20

---

更新性能按样本数据的运行时间排序

数据框

%%timeit
pd.DataFrame({idx: dict(zip(x['cNames'], x['cValues']) )
              for idx, x in df.iterrows()
            }).T.join(df.iloc[:,2:])
1.29 ms ± 36.8 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

连接：

%%timeit
pd.concat([pd.Series(x['cValues'], x['cNames'], name=idx) 
               for idx, x in df.iterrows()], 
          axis=1
         ).T.join(df.iloc[:,2:])
2.03 ms ± 86.2 µs per loop (mean ± std. dev. of 7 runs, 100 loops each) 

KJDII的新系列

%%timeit
df['series'] = df.apply(lambda x: dict(zip(x['cNames'], x['cValues'])), axis=1)
pd.concat([df['number'], df['series'].apply(pd.Series)], axis=1)

2.09 ms ± 65.2 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

斯科特的申请（pd.Series.explode）

%%timeit
df.apply(pd.Series.explode)\
  .set_index(['number', 'cNames'], append=True)['cValues']\
  .unstack()\
  .reset_index()\
  .drop('level_0', axis=1)

4.9 ms ± 135 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

wwnde 的 set_index.apply(explode)

%%timeit
g=df.set_index('number').apply(lambda x: x.explode()).reset_index()
g['cValues']=g['cValues'].astype(int)
pd.pivot_table(g, index=["number"],values=["cValues"],columns=["cNames"]).droplevel(0, axis=1).reset_index()

7.27 ms ± 162 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

Celius的双爆

%%timeit
df1 = df.explode('cNames').explode('cValues')
df1['cValues'] = pd.to_numeric(df1['cValues'])
df1.pivot_table(columns='cNames',index='number',values='cValues')

9.42 ms ± 189 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

KJD*_*DII 5

import pandas as pd

d = {'cNames': [['a','b','c'], ['a','b','d']], 'cValues': [[1,2,3], 
[55,66,77]], 'number': [10,20]}
df = pd.DataFrame(data=d)

df['series'] = df.apply(lambda x: dict(zip(x['cNames'], x['cValues'])), axis=1)
df = pd.concat([df['number'], df['series'].apply(pd.Series)], axis=1)
print(df)

   number     a     b    c     d
0      10   1.0   2.0  3.0   NaN
1      20  55.0  66.0  NaN  77.0

如果列顺序很重要：

columns = ['a', 'b', 'c', 'd', 'number']
df = df[columns]

      a     b    c     d  number
0   1.0   2.0  3.0   NaN      10
1  55.0  66.0  NaN  77.0      20