summing two columns in a pandas dataframe

Question

I think you've misunderstood some python syntax, the following does two assignments:

In [11]: a = b = 1

In [12]: a
Out[12]: 1

In [13]: b
Out[13]: 1

So in your code it was as if you were doing:

sum = df['budget'] + df['actual']  # a Series
# and
df['variance'] = df['budget'] + df['actual']  # assigned to a column

The latter creates a new column for df:

In [21]: df
Out[21]:
  cluster                 date  budget  actual
0       a  2014-01-01 00:00:00   11000   10000
1       a  2014-02-01 00:00:00    1200    1000
2       a  2014-03-01 00:00:00     200     100
3       b  2014-04-01 00:00:00     200     300
4       b  2014-05-01 00:00:00     400     450
5       c  2014-06-01 00:00:00     700    1000
6       c  2014-07-01 00:00:00    1200    1000
7       c  2014-08-01 00:00:00     200     100
8       c  2014-09-01 00:00:00     200     300

In [22]: df['variance'] = df['budget'] + df['actual']

In [23]: df
Out[23]:
  cluster                 date  budget  actual  variance
0       a  2014-01-01 00:00:00   11000   10000     21000
1       a  2014-02-01 00:00:00    1200    1000      2200
2       a  2014-03-01 00:00:00     200     100       300
3       b  2014-04-01 00:00:00     200     300       500
4       b  2014-05-01 00:00:00     400     450       850
5       c  2014-06-01 00:00:00     700    1000      1700
6       c  2014-07-01 00:00:00    1200    1000      2200
7       c  2014-08-01 00:00:00     200     100       300
8       c  2014-09-01 00:00:00     200     300       500

As an aside, you shouldn't use sum as a variable name as the overrides the built-in sum function.

df['variance'] = df.loc[:,['budget','actual']].sum(axis=1)

Same thing can be done using lambda function. Here I am reading the data from a xlsx file.

import pandas as pd
df = pd.read_excel("data.xlsx", sheet_name = 4)
print df

Output:

  cluster Unnamed: 1      date  budget  actual
0       a 2014-01-01  00:00:00   11000   10000
1       a 2014-02-01  00:00:00    1200    1000
2       a 2014-03-01  00:00:00     200     100
3       b 2014-04-01  00:00:00     200     300
4       b 2014-05-01  00:00:00     400     450
5       c 2014-06-01  00:00:00     700    1000
6       c 2014-07-01  00:00:00    1200    1000
7       c 2014-08-01  00:00:00     200     100
8       c 2014-09-01  00:00:00     200     300

Sum two columns into 3rd new one.

df['variance'] = df.apply(lambda x: x['budget'] + x['actual'], axis=1)
print df

Output:

  cluster Unnamed: 1      date  budget  actual  variance
0       a 2014-01-01  00:00:00   11000   10000     21000
1       a 2014-02-01  00:00:00    1200    1000      2200
2       a 2014-03-01  00:00:00     200     100       300
3       b 2014-04-01  00:00:00     200     300       500
4       b 2014-05-01  00:00:00     400     450       850
5       c 2014-06-01  00:00:00     700    1000      1700
6       c 2014-07-01  00:00:00    1200    1000      2200
7       c 2014-08-01  00:00:00     200     100       300
8       c 2014-09-01  00:00:00     200     300       500

You could also use the .add() function:

 df.loc[:,'variance'] = df.loc[:,'budget'].add(df.loc[:,'actual'])

This is the most elegant solution which follows DRY and work absolutely great.

dataframe_name['col1', 'col2', 'col3'].sum(axis = 1, skipna = True)

Thank you.

If "budget" has any NaN values but you don't want it to sum to NaN then try:

def fun (b, a):
    if math.isnan(b):
        return a
    else:
        return b + a

f = np.vectorize(fun, otypes=[float])

df['variance'] = f(df['budget'], df_Lp['actual'])