返回新列中每一行中第一个匹配值的列名

Question

I have a dataframe where the first column is an ID and each other column is a date. Each ID may show the same thing in several columns, may have some leading NaN columns, or may have all NaN columns. I'd like to create a new column with the name of the column where a specific entry first appears.

sample df:

| id_report | req id | 1-Jan | 2-Jan | 3-Jan | 4-Jan |
| --------- | -------------- | ----- | ----- | ----- | ----- |
| 0   | 12345 | NaN | Pend | Pend | Appr |
| 1   | 12346  | NaN | NaN | NaN | NaN |
| 2   | 12347 | NaN | NaN | Pend | Pend |
| 3   | 12348  | NaN | NaN | NaN | Appr |

I've searched and come up with:

id_report["Pend"] = id_report.apply(lambda x: x == "Pend", axis = 1).idxmax(axis = 1)

But this returns "req id" for every row where "Pend" doesn't appear, and I'd like to keep those positions empty.

Desired output:

id_report	req id	1-Jan	2-Jan	3-Jan	4-Jan	Pend
0	12345	NaN	Pend	Pend	Appr	2-Jan
1	12346	NaN	NaN	NaN	NaN	NaN
2	12347	NaN	NaN	Pend	Pend	3-Jan
3	12348	NaN	NaN	NaN	Appr	NaN

Answer 1

You could chain a replace to your current code:

import numpy as np
id_report['Pend'] = (id_report
   .apply(lambda x: x == 'Pend', axis = 1)
   .idxmax(axis = 1)
   .replace('req id', np.nan)
)

	req id	1-Jan	2-Jan	3-Jan	4-Jan	Pend
0	12345	NaN	Pend	Pend	Appr	2-Jan
1	12346	NaN	NaN	NaN	NaN	NaN
2	12347	NaN	NaN	Pend	Pend	3-Jan
3	12348	NaN	NaN	NaN	Appr	NaN