If I have a dataframe that has columns that include the same name, is there a way to combine the columns that have the same name with some sort of function (i.e. sum)?
For instance with:
In [186]:
df["NY-WEB01"].head()
Out[186]:
NY-WEB01 NY-WEB01
DateTime
2012-10-18 16:00:00 5.6 2.8
2012-10-18 17:00:00 18.6 12.0
2012-10-18 18:00:00 18.4 12.0
2012-10-18 19:00:00 18.2 12.0
2012-10-18 20:00:00 19.2 12.0
How might I collapse the NY-WEB01 columns (there are a bunch of duplicate columns, not just NY-WEB01) by summing each row where the column name is the same?
I believe this does what you are after:
df.groupby(lambda x:x, axis=1).sum()
Alternatively, between 3% and 15% faster depending on the length of the df:
df.groupby(df.columns, axis=1).sum()
EDIT: To extend this beyond sums, use .agg()
(short for .aggregate()
):
df.groupby(df.columns, axis=1).agg(numpy.max)
You don't need a lambda here, nor do you explicitly have to query df.columns
; groupby
accepts a level
argument you can specify in conjunction with the axis
argument. This is cleaner, IMO.
# Setup
np.random.seed(0)
df = pd.DataFrame(np.random.choice(50, (5, 5)), columns=list('AABBB'))
df
A A B B B
0 44 47 0 3 3
1 39 9 19 21 36
2 23 6 24 24 12
3 1 38 39 23 46
4 24 17 37 25 13
df.groupby(level=0, axis=1).sum()
A B
0 91 6
1 48 76
2 29 60
3 39 108
4 41 75
Handling MultiIndex
es
Another case to consider is when dealing with MultiIndex
columns. Consider
df.columns = pd.MultiIndex.from_arrays([['one']*3 + ['two']*2, df.columns])
df
one two
A A B B B
0 44 47 0 3 3
1 39 9 19 21 36
2 23 6 24 24 12
3 1 38 39 23 46
4 24 17 37 25 13
To perform aggregation across the upper levels, use
df.groupby(level=1, axis=1).sum()
A B
0 91 6
1 48 76
2 29 60
3 39 108
4 41 75
or, if aggregating per upper level only, use
df.groupby(level=[0, 1], axis=1).sum()
one two
A B B
0 91 0 6
1 48 19 57
2 29 24 36
3 39 39 69
4 41 37 38
Dropping Duplicate Columns
If you came here looking to find out how to simply drop duplicate columns (without performing any aggregation), use Index.duplicated
:
df.loc[:,~df.columns.duplicated()]
A B
0 44 0
1 39 19
2 23 24
3 1 39
4 24 37
Or, to keep the last ones, specify keep='last'
(default is 'first'
),
df.loc[:,~df.columns.duplicated(keep='last')]
A B
0 47 3
1 9 36
2 6 12
3 38 46
4 17 13
The groupby
alternatives for the two solutions above are df.groupby(level=0, axis=1).first()
, and ... .last()
, respectively.