Swap column names with row values in a pandas DataFrame

user_10770967 0
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I am kind of stucked in a pretty simple task - at least what I thought.

Suppose you have a DataFrame with following shape (So the data frame is given)

df = pd.DataFrame({"col a": [0, 1, 2, 3, "name a"],
                   "col b": [0, 1, 2, 3, "name b"],
                   "col c": [0, 1, 2, 3, "name c"]})

    col a   col b   col c
0       0       0       0
1       1       1       1
2       2       2       2
3       3       3       3
4  name a  name b  name c

If I use the following command

df.columns=df.iloc[4]

the original header disappears. Any suggestion how to achieve my goal. Of course if it would be one sheet I could do it manually, however I have multiple sheets and this need to automatize it

Any kind of support is really appreciated The result should be that way

  name  name  name 
0     0     0     0
1     1     1     1
2     2     2     2
3     3     3     3
4  col a  col b  col c
columnsorting pandas dataframe python
2 Answers

Like in any case where you want to swap 2 values in a program, you need to use an intermediate variable.

You can let python do that for you by taking advantage of the A, B = B, A syntax:

df.columns, df.iloc[4] = df.iloc[4].to_list(), df.columns

output:

  name a name b name c
0      0      0      0
1      1      1      1
2      2      2      2
3      3      3      3
4  col a  col b  col c

ignoring first columns in the swap:

df = df.set_index(['col a', 'col b'])
df.columns, df.iloc[4] = df.iloc[4].to_list(), df.columns
df = df.reset_index()

output:

    col a   col b name c
0       0       0      0
1       1       1      1
2       2       2      2
3       3       3      3
4  name a  name b  col c
user_163434640

It's important to use the .rename method from Pandas instead of just assigning new column names

df = pd.DataFrame({"col a": [0, 1, 2, 3, "name"]})

col_name = df.columns.to_list()[0]
var_name = df.iloc[4].values[0]
df = df.rename(columns={col_name: var_name})
df.iloc[4, :] = col_name

result:

name
0   0
1   1
2   2
3   3
4   col a
user_31303050
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