根据另一列的值向python pandas数据框添加一列

Question

I have some pandas data frame, and I would like to add a column that is the difference of a column, based on the value of a third column. Here is a toy example:

    import pandas as pd
    import numpy as np

     d = {'one' : pd.Series(range(4), index=['a', 'b', 'c', 'd']),
    'two' : pd.Series(range(4), index=['a', 'b', 'c', 'd'])}

    df = pd.DataFrame(d)

    df['three'] = [2,2,3,3]


    four = []
    for i in set(df['three']):
        for j in range(len(df) -1):
            four.append(df[df['three'] == i]['two'][j + 1] - df[df['three']==i]['two'][j])
    four.append(0)

    df['four'] = four

The final column should be [1, 1, 1, Nan], since that is the difference between each of the rows in the 'two' column

This makes more sense in the context of my original code -- my data frame is organized by some IDs, and then by time, and when I take the subset of the data frame by IDs, I'm left with the time series evolution of the variables for each individual ID. However, I keep on either receiving a key error, or attempting to edit a copy of the original data frame. What is the right way to go about this?

Answer 1

You could replace df[df['three'] == i] with a groupby on column three. And perhaps replace ['two'][j + 1] - ['two'][j] with df['two'].shift(-1) - df['two'] .

I think that would be identical to what you are doing now within the nested loop. It depends a bit on what format you want as a result on how you would implement this. One way would be:

df.groupby('three').apply(lambda grp: pd.Series(grp['two'].shift(-1) - grp['two']))

Which would result in:

two    a   b
three       
2      1 NaN
3      1 NaN

The columns names become a bit meaningless after this operation.

Answer 2

如果您要做的只是获取第二列的行之间的差，请使用shift方法。

df['four'] = df.two.shift(-1) - df.two