When I transpose a dataframe, the headers are considered as "index" by default. But I want it to be a column and not an index. How do I achieve this ?
import pandas as pd dict = {'col-a': [97, 98, 99], 'col-b': [34, 35, 36], 'col-c': [24, 25, 26]} df = pd.DataFrame(dict) print(df.T) 0 1 2 col-a 97 98 99 col-b 34 35 36 col-c 24 25 26 Desired Output:
0 1 2 3 0 col-a 97 98 99 1 col-b 34 35 36 2 col-c 24 25 26 Try T with reset_index:
df=df.T.reset_index() print(df) Or:
df.T.reset_index(inplace=True) print(df) Both Output:
index 0 1 2 0 col-a 97 98 99 1 col-b 34 35 36 2 col-c 24 25 26 If care about column names, add this to the code:
df.columns=range(4) Or:
it=iter(range(4)) df=df.rename(columns=lambda x: next(it)) Or if don't know number of columns:
df.columns=range(len(df.columns)) Or:
it=iter(range(len(df.columns))) df=df.rename(columns=lambda x: next(it)) All Output:
0 1 2 3 0 col-a 97 98 99 1 col-b 34 35 36 2 col-c 24 25 26 Use reset_index and then set default columns names:
df1 = df.T.reset_index() df1.columns = np.arange(len(df1.columns)) print (df1) 0 1 2 3 0 col-a 97 98 99 1 col-b 34 35 36 2 col-c 24 25 26 Another solution:
print (df.rename_axis(0, axis=1).rename(lambda x: x + 1).T.reset_index()) #alternative #print (df.T.rename_axis(0).rename(columns = lambda x: x + 1).reset_index()) 0 1 2 3 0 col-a 97 98 99 1 col-b 34 35 36 2 col-c 24 25 26 