Populate df row value based on column header - regex

Question

Appreciate any help. Basically, I have a poor data set and am trying to make it more useful.
Below is a representation
df = pd.DataFrame({'State': ("Texas","California","Florida"),
'Q1 Computer Sales': (100,200,300),
'Q1 Phone Sales': (400,500,600),
'Q1 Backpack Sales': (700,800,900),
'Q2 Computer Sales': (200,200,300),
'Q2 Phone Sales': (500,500,600),
'Q2 Backpack Sales': (800,800,900)})
I would like to have a df that creates separate columns for the Quarters and Sales for the respective state.
I think perhaps regex, str.contains, and loops perhaps?
snapshot below

Answer 1

IIUC, you can use:
df_a = df.set_index('State')
df_a.columns = pd.MultiIndex.from_arrays(zip(*df_a.columns.str.split(' ', n=1)))
df_a.stack(0).reset_index()
Output:
State level_1 Backpack Sales Computer Sales Phone Sales
0 Texas Q1 700 100 400
1 Texas Q2 800 200 500
2 California Q1 800 200 500
3 California Q2 800 200 500
4 Florida Q1 900 300 600
5 Florida Q2 900 300 600
Or we can go further:
df_a = df.set_index('State')
df_a.columns = pd.MultiIndex.from_arrays(zip(*df_a.columns.str.split(' ', n=1)), names=['Quarters','Items'])
df_a = df_a.stack(0).reset_index()
df_a['Quarters'] = df_a['Quarters'].str.extract('(\d+)')
print(df_a)
Output:
Items State Quarters Backpack Sales Computer Sales Phone Sales
0 Texas 1 700 100 400
1 Texas 2 800 200 500
2 California 1 800 200 500
3 California 2 800 200 500
4 Florida 1 900 300 600
5 Florida 2 900 300 600