Reputation: 1767
Apologies if this has been asked in a different way, I'm not sure how else to word it to get what I want to do.
Say I have some dataframe with two columns of values:
import pandas as pd
data = {'col1':[0.6, 0.8, 0.9, 0.87, 0.69, 0.88],
'col2':[0.72, 0.91, 0.83, 0.85, 0.96, 0.76]}
df = pd.DataFrame(data)
df
Out[1]:
col1 col2
0 0.60 0.72
1 0.80 0.91
2 0.90 0.83
3 0.87 0.85
4 0.69 0.96
5 0.88 0.76
I want to take column 2 and 'append' it underneath column 1, continuing the index from 6 to 11. I also would like a new 'identifier' column to be created to have the column name to which each datapoint belongs. The closest I can get to this without lots of spaghetti code is the following:
pd.DataFrame(df.stack()).reset_index()
Out[34]:
level_0 level_1 0
0 0 col1 0.60
1 0 col2 0.72
2 1 col1 0.80
3 1 col2 0.91
4 2 col1 0.90
5 2 col2 0.83
6 3 col1 0.87
7 3 col2 0.85
8 4 col1 0.69
9 4 col2 0.96
10 5 col1 0.88
11 5 col2 0.76
I don't want this to be labeled 'level_0', 'level_1', etc. as I have column names in mind that I would like to use.
I'm sure there is an easy and professional way to accomplish this but I don't know enough about it.
Upvotes: 1
Views: 4301
Reputation: 862681
You are really close, not necessary call DataFrame
constructor.
Also for set columns names is possible rename levels of MultiIndex
by rename_axis
:
df = df.stack().rename_axis(('a', 'b')).reset_index(name='val')
print (df)
a b val
0 0 col1 0.60
1 0 col2 0.72
2 1 col1 0.80
3 1 col2 0.91
4 2 col1 0.90
5 2 col2 0.83
6 3 col1 0.87
7 3 col2 0.85
8 4 col1 0.69
9 4 col2 0.96
10 5 col1 0.88
11 5 col2 0.76
Numpy alternative:
a = np.repeat(df.index, len(df.columns))
b = np.tile(df.columns, len(df.index))
c = df.values.ravel()
df = pd.DataFrame(np.column_stack([a,b,c]), columns=['a','b','c'])
print (df)
a b c
0 0 col1 0.6
1 0 col2 0.72
2 1 col1 0.8
3 1 col2 0.91
4 2 col1 0.9
5 2 col2 0.83
6 3 col1 0.87
7 3 col2 0.85
8 4 col1 0.69
9 4 col2 0.96
10 5 col1 0.88
11 5 col2 0.76
Upvotes: 3