CODEWITHSUNDEEP
pythonpandasloops
user9639519
user9639519

Reputation:

How to subtract rows in a df based on a value in another column

I am trying to calculate the difference in certain rows based on the values from other columns.

Using the example data frame below, I want to calculate the difference in Time based on the values in the Code column. Specifically, I want to loop through and determine the time difference between B and A. So Time in B - Time in A.

I can do this manually using the iloc function but I was hoping to determine a more efficient way. Especially if I have to repeat this process numerous times.

import pandas as pd
import numpy as np

k = 5
N = 15

d = ({'Time' : np.random.randint(k, k + 100 , size=N),
    'Code' : ['A','x','B','x','A','x','B','x','A','x','B','x','A','x','B']})

df = pd.DataFrame(data=d)

Output:

   Code  Time
0     A    89
1     x    39
2     B    24
3     x    62
4     A    83
5     x    57
6     B    69
7     x    10
8     A    87
9     x    62
10    B    86
11    x    11
12    A    54
13    x    44
14    B    71

Expected Output:

     diff
1    -65
2    -14
3    -1
4     17

Upvotes: 4

Views: 2938

Answers (2)

jezrael
jezrael

Reputation: 863331

First filter by boolean indexing, then subtract by sub with reset_index for default index for align Series a and b and last if want one column DataFrame add to_frame:

a = df.loc[df['Code'] == 'A', 'Time'].reset_index(drop=True)
b = df.loc[df['Code'] == 'B', 'Time'].reset_index(drop=True)

Similar alternative solution:

a = df.loc[df['Code'] == 'A'].reset_index()['Time']
b = df.loc[df['Code'] == 'B'].reset_index()['Time']

c = b.sub(a).to_frame('diff')
print (c)
   diff
0   -65
1   -14
2    -1
3    17

Last for new index start from 1 add rename:

c = b.sub(a).to_frame('diff').rename(lambda x: x + 1)
print (c)
   diff
1   -65
2   -14
3    -1
4    17

Another approach if need count more difference is reshape by unstack:

df = df.set_index(['Code', df.groupby('Code').cumcount() + 1])['Time'].unstack()
print (df)
         1     2     3     4     5     6     7
Code                                          
A     89.0  83.0  87.0  54.0   NaN   NaN   NaN
B     24.0  69.0  86.0  71.0   NaN   NaN   NaN
x     39.0  62.0  57.0  10.0  62.0  11.0  44.0

#last remove `NaN`s rows
c = df.loc['B'].sub(df.loc['A']).dropna()
print (c)
1   -65.0
2   -14.0
3    -1.0
4    17.0
dtype: float64

#subtract with NaNs values - fill_value=0 return non NaNs values
d = df.loc['x'].sub(df.loc['A'], fill_value=0)
print (d)
1   -50.0
2   -21.0
3   -30.0
4   -44.0
5    62.0
6    11.0
7    44.0
dtype: float64

Upvotes: 2

Ami Tavory
Ami Tavory

Reputation: 76366

Assuming your Code is a repeat of 'A', 'x', 'B', 'x', you can just use

>>> (df.Time[df.Code == 'B'].reset_index() - df.Time[df.Code == 'A'].reset_index())[['Time']]
    Time
0   -65
1   -14
2   -1
3   17

But note that the original assumption, that 'A' and 'B' values alternate, seems fragile.

If you want the indexes to run from 1 to 4, as in your question, you can assign the previous to diff, and then use

diff.index += 1
>>> diff
    Time
1   -65
2   -14
3   -1
4   17

Upvotes: 2

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