划分下一行的值并在数据框中创建列 [英] divide value of next row and create column in dataframe

问题描述

我有一个类似csv的

id,value
1,100
1,150
1,200
1,250
2,300
2,350
2,400
2,450

我想根据每个唯一ID的值生成一列.

I want to generate one column based on value of each of the unique id.

例如:

• id为1的前2行的值为100、150
• 我正在尝试创建列raise，该列将像这样划分值
• 100/100 = 1，所以带有加高列的第一行将有1
• 150/100 = 1.5，因此带有加高列的第二行将有2
• id 2相同
• 250/250 = 1和300/250 = 1.166

• first 2 rows for id 1 has value 100, 150
• i am trying to create column raise which will divide value like this
• 100/100 = 1 so first row with in raise column will have 1
• 150/100 = 1.5 so second row with in raise column will have 2
• same for id 2
• 250/250 = 1 and 300/250 = 1.166

我取了100，因为它是该id的第一个值，与第二个'id'相同 例如:

I took 100 because its the first value for that id, same for the 2nd 'id' for example:

id,value
1,150
1,100
1,200
1,250

如果是这种情况，则输出应该是

if this was the case then output should be

id,value,raise
1,150,150/150
1,100,100/150
1,200,200/150
1,250,250/150

所以最终我的输出将是

   id  value  raise
0   1    100  1.000
1   1    150  1.500
2   1    200  2.000
3   1    250  2.500
4   2    300  1.000
5   2    350  1.166
6   2    400  1.333
7   2    450  1.500

除了在所有id中使用for循环外，我不知道如何创建它.

I don't know how to create it except using for loop through all the id.

并非所有值都具有相同的间隔，这只是一个示例

Not all the value has same interval it's just an example

推荐答案

由Series划分列. core.groupby.GroupBy.transform.html"rel =" nofollow noreferrer> GroupBy.transform 和

Divide column by Series created by GroupBy.transform with GroupBy.first:

df['raise'] = df['value'].div(df.groupby('id')['value'].transform('first'))
print (df)
   id  value     raise
0   1    100  1.000000
1   1    150  1.500000
2   1    200  2.000000
3   1    250  2.500000
4   2    300  1.000000
5   2    350  1.166667
6   2    400  1.333333
7   2    550  1.833333

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