Pandas column: List of columns in specific order - python-3.x

Question

I have a dataframe, but I'm trying to add a new column which is a list of the column names in order of their values, for each row.
Searching has proved to be difficult, as the search terms have so much in common with doing a column sort overall. Instead, I'm trying to customize the list for each row.
df = pd.DataFrame([
["a",88,3,78,8,40 ],
["b",100,20,29,13,91 ],
["c",77,92,42,72,58 ],
["d",39,53,69,7,40 ],
["e",26,62,77,33,86 ],
["f",94,5,28,96,7 ]
], columns=['id','x1','x2','x3','x4','x5'])
have = df.set_index('id')
+----+-----+----+----+----+----+----------------------------+
| id | x1 | x2 | x3 | x4 | x5 | ordered_cols |
+----+-----+----+----+----+----+----------------------------+
| a | 88 | 3 | 78 | 8 | 40 | ['x2','x4','x5','x3','x1'] |
| b | 100 | 20 | 29 | 13 | 91 | ['x4','x2','x3','x5','x1'] |
| c | 77 | 92 | 42 | 72 | 58 | … |
| d | 39 | 53 | 69 | 7 | 40 | … |
| e | 26 | 62 | 77 | 33 | 86 | … |
| f | 94 | 5 | 28 | 96 | 7 | … |
+----+-----+----+----+----+----+----------------------------+

Answer 1

try stack with sort_values and groupby
assuming your dataframe is called df
df["sorted_cols"] = (
df.stack().sort_values().reset_index(1).groupby(level=0)["level_1"].agg(list)
)
print(df)
x1 x2 x3 x4 x5 sorted_cols
id
a 88 3 78 8 40 [x2, x4, x5, x3, x1]
b 100 20 29 13 91 [x4, x2, x3, x5, x1]
c 77 92 42 72 58 [x3, x5, x4, x1, x2]
d 39 53 69 7 40 [x4, x1, x5, x2, x3]
e 26 62 77 33 86 [x1, x4, x2, x3, x5]
f 94 5 28 96 7 [x2, x5, x3, x1, x4]

Answer 2

The solution by Manakin will be the fastest option, because it is a vectorized.
Use pandas.DataFrame.apply with axis=1, and a list comprehension to sort the column names by the row values.
The list comprehension is from SO: Sorting list based on values from another list, and does not require importing any additional packages.
import pandas as pd
# add the new column
df['ordered_cols'] = df.apply(lambda y: [x for _, x in sorted(zip(y, df.columns))], axis=1)
# display(df)
x1 x2 x3 x4 x5 ordered_cols
id
a 88 3 78 8 40 [x2, x4, x5, x3, x1]
b 100 20 29 13 91 [x4, x2, x3, x5, x1]
c 77 92 42 72 58 [x3, x5, x4, x1, x2]
d 39 53 69 7 40 [x4, x1, x5, x2, x3]
e 26 62 77 33 86 [x1, x4, x2, x3, x5]
f 94 5 28 96 7 [x2, x5, x3, x1, x4]

Answer 3

Here is a simple one line solution using apply and np.argsort :
import numpy as np
have["ordered_cols"] = have.apply(lambda row: have.columns[np.argsort(row.values)].values, axis=1)
have

Answer 4

Hay,
you can try looping over the rows and sorting the values in each row. The code below will do the trick:
ordered_cols = []
for index, row in have.iterrows():
ordered_cols.append(list(have.sort_values(by=index, ascending=True, axis=1).columns))
have['ordered_cols'] = ordered_cols
have
Output:
x1 x2 x3 x4 x5 ordered_cols
id
a 88 3 78 8 40 [x2, x4, x5, x3, x1]
b 100 20 29 13 91 [x4, x2, x3, x5, x1]
c 77 92 42 72 58 [x3, x5, x4, x1, x2]
d 39 53 69 7 40 [x4, x1, x5, x2, x3]
e 26 62 77 33 86 [x1, x4, x2, x3, x5]
f 94 5 28 96 7 [x2, x5, x3, x1, x4]
I hope this was helpful.
Cheers!