Issue
Even though I was googling a lot, I couldn't find the solution for my problem.
I have dataframe
filter10 REF
0 NaN 0.00
1 NaN 0.75
2 NaN 1.50
3 NaN 2.25
4 NaN 3.00
5 NaN 3.75
6 NaN 4.50
...
15 2.804688 11.25
16 3.021875 12.00
17 3.578125 12.75
18 3.779688 13.50
...
27 NaN 20.25
28 NaN 21.00
29 NaN 21.75
30 NaN 22.50
31 6.746875 NaN
32 NaN NaN
...
I would like now to add the column df['DIFF']
where function goes through whole column filter10
and when it is the number it finds closest number in REF
column.
And afterwards calculate the difference between them and put it the same row as number in filter10
is.
I would like this output:
filter10 REF DIFF
0 NaN 0.00 NaN
1 NaN 0.75 NaN
2 NaN 1.50 NaN
3 NaN 2.25 NaN
4 NaN 3.00 NaN
5 NaN 3.75 NaN
6 NaN 4.50 NaN
...
15 2.804688 11.25 -0.195312 # 2.804688 - 3 (find closest value in REF) = -0.195312
16 3.021875 12.00 0.021875
17 3.578125 12.75 -0.171875
18 3.779688 13.50 0.029688
...
27 NaN 20.25 NaN
28 NaN 21.00 NaN
29 NaN 21.75 NaN
30 NaN 22.50 NaN
31 6.746875 NaN -0.003125
32 NaN NaN NaN
...
Solution
Use pandas.merge_asof
to find the nearest value:
df['DIFF'] = (pd.merge_asof(df['filter10'].dropna().sort_values().reset_index(),
df[['REF']].dropna().sort_values('REF'),
left_on='filter10', right_on='REF', direction='nearest')
.set_index('index')['REF'].rsub(df['filter10'])
)
Output:
filter10 REF DIFF
0 NaN 0.00 NaN
1 NaN 0.75 NaN
2 NaN 1.50 NaN
3 NaN 2.25 NaN
4 NaN 3.00 NaN
5 NaN 3.75 NaN
6 NaN 4.50 NaN
15 2.804688 11.25 -0.195312
16 3.021875 12.00 0.021875
17 3.578125 12.75 -0.171875
18 3.779688 13.50 0.029688
27 NaN 20.25 NaN
28 NaN 21.00 NaN
29 NaN 21.75 NaN
30 NaN 22.50 NaN
31 6.746875 NaN 2.246875 # likely different due to missing data
32 NaN NaN NaN
Answered By - mozway
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