与python的数据争用

我正努力在熊猫数据框上进行数据争论。这个问题我已经解决了两天了

我有一个数据框，看起来像这样：

['a','e']
['b','f']
['c','g']
['d','h']
['a','i']
['b','j']
['c','k']
['d','l']

我需要把它变成这样

['a',['e','i']]
['b',['f','j']]
['c',['g','k']]
['d',['h','l']]

因此，基本上是围绕第一列旋转原始数据帧，然后从第二列创建字符串列表

---

谢谢

您可以使用

groupy/agg

：

import pandas as pd

data = [['a','e'], ['b','f'], ['c','g'], ['d','h'], ['a','i'], ['b','j'], 
        ['c','k'], ['d','l']]

df = pd.DataFrame(data, columns=['first', 'second'])
print(df.groupby(['first']).agg(lambda x: x.tolist()))

屈服

       second
first        
a      [e, i]
b      [f, j]
c      [g, k]
d      [h, l]

---

您可以使用

groupy/agg

：

import pandas as pd

data = [['a','e'], ['b','f'], ['c','g'], ['d','h'], ['a','i'], ['b','j'], 
        ['c','k'], ['d','l']]

df = pd.DataFrame(data, columns=['first', 'second'])
print(df.groupby(['first']).agg(lambda x: x.tolist()))

屈服

       second
first        
a      [e, i]
b      [f, j]
c      [g, k]
d      [h, l]

---

您可以使用

groupy/agg

：

import pandas as pd

data = [['a','e'], ['b','f'], ['c','g'], ['d','h'], ['a','i'], ['b','j'], 
        ['c','k'], ['d','l']]

df = pd.DataFrame(data, columns=['first', 'second'])
print(df.groupby(['first']).agg(lambda x: x.tolist()))

屈服

       second
first        
a      [e, i]
b      [f, j]
c      [g, k]
d      [h, l]

---

您可以使用

groupy/agg

：

import pandas as pd

data = [['a','e'], ['b','f'], ['c','g'], ['d','h'], ['a','i'], ['b','j'], 
        ['c','k'], ['d','l']]

df = pd.DataFrame(data, columns=['first', 'second'])
print(df.groupby(['first']).agg(lambda x: x.tolist()))

屈服

       second
first        
a      [e, i]
b      [f, j]
c      [g, k]
d      [h, l]