您可以使用DataFrame.from_records
,但首先需要從col
列的值創建嵌套list
:
df = pd.DataFrame({'col':[['A',1,3,4,'Null'],['B',4,5,6,'Null'],['C',7,8,9,'Null']]})
print (df)
col
0 [A, 1, 3, 4, Null]
1 [B, 4, 5, 6, Null]
2 [C, 7, 8, 9, Null]
print (df.col.values.tolist())
[['A', 1, 3, 4, 'Null'], ['B', 4, 5, 6, 'Null'], ['C', 7, 8, 9, 'Null']]
df1 = pd.DataFrame.from_records(df.col.values.tolist(),
columns=['colA','colB','colC','colD','colE'])
print(df1)
colA colB colC colD colE
0 A 1 3 4 Null
1 B 4 5 6 Null
2 C 7 8 9 Null
如果不需要指定列名:
df1 = pd.DataFrame.from_records(df.col.values.tolist())
print(df1)
0 1 2 3 4
0 A 1 3 4 Null
1 B 4 5 6 Null
2 C 7 8 9 Null
時序:
#len(df) = 4k
df = pd.concat([df]*1000).reset_index(drop=True)
In [80]: %timeit pd.DataFrame(df['col'].apply(pd.Series).values, columns=['colA','colB','colC','colD','colE'])
1 loop, best of 3: 753 ms per loop
In [81]: %timeit pd.DataFrame.from_records(df.col.values.tolist(), columns=['colA','colB','colC','colD','colE'])
100 loops, best of 3: 3.73 ms per loop