如何计算pandas中一行中所有元素的加权和？

Question

我有一个包含多列的pandas数据框.我想weighted_sum从行中的值和另一个列向量数据帧创建一个新列weight

weighted_sum 应具有以下值:

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

我找到了这个功能sum(axis=1),但它不会让我倍增weight.

编辑:我改变了一点.

weight 看起来像这样:

     0
col1 0.5
col2 0.3
col3 0.2

df 看起来像这样:

col1 col2 col3
1.0  2.2  3.5
6.1  0.4  1.2

df*weight返回一个充满Nan值的数据帧.

Answer 1

问题在于,您将帧与具有不同行索引的不同大小的帧相乘.这是解决方案:

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

您可以访问该列:

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

或者dot用来取回另一个DataFrame

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

把它们放在一起:

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

以下是timeit每种方法的s,使用更大的方法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

广泛的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

因此,dot更快,更易读.

注意:如果您的任何数据包含NaNs,那么您不应该使用dot您应该使用乘法和和方法.dot无法处理NaNs,因为它只是一个薄的包装numpy.dot()(不处理NaNs).

Answer 2

假设权重是每列的一系列权重,您可以乘以并得到总和:

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

这种方法的好处是它可以处理您不想称量的列:

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