How to compute weighted sum of all elements in a row in pandas?

Question

I have a pandas data frame with multiple columns. I want to create a new column weighted_sum from the values in the row and another column vector dataframe weight

weighted_sum should have the following value:

row[weighted_sum] = row[col0]*weight[0] + row[col1]*weight[1] + row[col2]*weight[2] + ...

I found the function sum(axis=1), but it doesn't let me multiply with weight.

Edit: I changed things a bit.

weight looks like this:

     0
col1 0.5
col2 0.3
col3 0.2

df looks like this:

col1 col2 col3
1.0  2.2  3.5
6.1  0.4  1.2

df*weight returns a dataframe full of Nan values.

Answer 1

The problem is that you're multiplying a frame with a frame of a different size with a different row index. Here's the solution:

In [121]: df = DataFrame([[1,2.2,3.5],[6.1,0.4,1.2]], columns=list('abc'))

In [122]: weight = DataFrame(Series([0.5, 0.3, 0.2], index=list('abc'), name=0))

In [123]: df
Out[123]:
           a          b          c
0       1.00       2.20       3.50
1       6.10       0.40       1.20

In [124]: weight
Out[124]:
           0
a       0.50
b       0.30
c       0.20

In [125]: df * weight
Out[125]:
           0          a          b          c
0        nan        nan        nan        nan
1        nan        nan        nan        nan
a        nan        nan        nan        nan
b        nan        nan        nan        nan
c        nan        nan        nan        nan

You can either access the column:

In [126]: df * weight[0]
Out[126]:
           a          b          c
0       0.50       0.66       0.70
1       3.05       0.12       0.24

In [128]: (df * weight[0]).sum(1)
Out[128]:
0         1.86
1         3.41
dtype: float64

Or use dot to get back another DataFrame

In [127]: df.dot(weight)
Out[127]:
           0
0       1.86
1       3.41

To bring it all together:

In [130]: df['weighted_sum'] = df.dot(weight)

In [131]: df
Out[131]:
           a          b          c  weighted_sum
0       1.00       2.20       3.50          1.86
1       6.10       0.40       1.20          3.41

Here are the timeits of each method, using a larger DataFrame.

In [145]: df = DataFrame(randn(10000000, 3), columns=list('abc'))
weight
In [146]: weight = DataFrame(Series([0.5, 0.3, 0.2], index=list('abc'), name=0))

In [147]: timeit df.dot(weight)
10 loops, best of 3: 57.5 ms per loop

In [148]: timeit (df * weight[0]).sum(1)
10 loops, best of 3: 125 ms per loop

For a wide DataFrame:

In [162]: df = DataFrame(randn(10000, 1000))

In [163]: weight = DataFrame(randn(1000, 1))

In [164]: timeit df.dot(weight)
100 loops, best of 3: 5.14 ms per loop

In [165]: timeit (df * weight[0]).sum(1)
10 loops, best of 3: 41.8 ms per loop

So, dot is faster and more readable.

NOTE: If any of your data contain NaNs then you should not use dot you should use the multiply-and-sum method. dot cannot handle NaNs since it is just a thin wrapper around numpy.dot() (which doesn't handle NaNs).

Answer 2

Assuming weights is a Series of weights for each columns, you can just multiply and do the sum:

In [11]: df = pd.DataFrame([[1, 2, 3], [4, 5, 6]], columns=['a', 'b', 'c'])

In [12]: weights = pd.Series([7, 8, 9], index=['a', 'b', 'c'])

In [13]: (df * weights)
Out[13]: 
    a   b   c
0   7  16  27
1  28  40  54

In [14]: (df * weights).sum(1)
Out[14]: 
0     50
1    122
dtype: int64

The benefit of this approach is it takes care of columns which you don't want to weigh:

In [21]: weights = pd.Series([7, 8], index=['a', 'b'])

In [22]: (df * weights)
Out[22]: 
    a   b   c
0   7  16 NaN
1  28  40 NaN

In [23]: (df * weights).sum(1)
Out[23]: 
0    23
1    68
dtype: float64