以:
开头import pandas as pd
lis1= [['apples'],['bananas','oranges','cinnamon'],['pears','juice']]
lis2= [['john'],['stacy'],['ron']]
pd.DataFrame({'fruits':lis1,'users':lis2})
fruits users
0 [apples] [john]
1 [bananas, oranges, cinnamon] [stacy]
2 [pears, juice] [ron]
我想结尾为:
lis3= ['apples','bananas','oranges','cinnamon','pears','juice']
lis4= ['john','stacy','stacy','stacy','ron','ron']
pd.DataFrame({'fruits': lis3, 'users':lis4})
fruits users
0 apples john
1 bananas stacy
2 oranges stacy
3 cinnamon stacy
4 pears ron
5 juice ron
首先,我需要创建一个新的数据框,其中每个项目都位于其自己的行中。其次,名称变量需要根据“水果”的数量重复其自身。因此,在示例中,John拥有一个水果,而Stacy有5个水果-因此在用户名下,Stacy必须重复5次。
答案 0 :(得分:3)
itertools
from itertools import chain, product, starmap
pd.DataFrame(
[*chain(*starmap(product, zip(df.fruits, df.users)))],
columns=df.columns
)
fruits users
0 apples john
1 bananas stacy
2 oranges stacy
3 cinnamon stacy
4 pears ron
5 juice ron
如果您只有2列,这也可以使用
pd.DataFrame(
[*chain(*starmap(product, zip(*map(df.get, df))))],
columns=df.columns
)
generator
def f(z):
for A, B in z:
for a in A:
for b in B:
yield (a, b)
pd.DataFrame([*f(zip(df.fruits, df.users))], columns=df.columns)
fruits users
0 apples john
1 bananas stacy
2 oranges stacy
3 cinnamon stacy
4 pears ron
5 juice ron
答案 1 :(得分:2)
假设lis1
和lis2
具有相同数量的元素,则可以在压缩列表后通过列表理解来做到这一点。
pd.DataFrame(
[{'fruit':F, 'users':U} for (f, u) in zip(lis1, lis2) for F in f for U in u]
)
以下代码产生以下输出:
fruit users
0 apples john
1 bananas stacy
2 oranges stacy
3 cinnamon stacy
4 pears ron
5 juice ron
答案 2 :(得分:1)
这是一个具有大量堆叠和卸载功能的解决方案:
开始于:
>>> df
fruits users
0 [apples] [john]
1 [bananas, oranges, cinnamon] [stacy]
2 [pears, juice] [ron]
使用:
final = (df.stack().apply(pd.Series)
.stack(0).unstack(1)
.ffill()
.reset_index(drop=True))
>>> final
fruits users
0 apples john
1 bananas stacy
2 oranges stacy
3 cinnamon stacy
4 pears ron
5 juice ron