pandas 的算术运算和数据对齐

pandas 还有一个重要的功能，就是他可以对不同索引的对象进行算数运算。
对象相加， 如果存在不同的索引对，则结果的索引就是该索引对的并集。

先来个例子

Series

In [33]: s1 = Series([7.3, -2.5, 3.4, 1.5], index=['a', 'c', 'd', 'e'])

In [34]: s2 = Series([-2.1, 3.6, -1.5, 4, 3.1], index=['a', 'c', 'e', 'f', 'g'])

In [35]: s1
Out[35]:
a    7.3
c   -2.5
d    3.4
e    1.5
dtype: float64

In [36]: s2
Out[36]:
a   -2.1
c    3.6
e   -1.5
f    4.0
g    3.1
dtype: float64

In [37]: s1 + s2
Out[37]:
a    5.2
c    1.1
d    NaN
e    0.0
f    NaN
g    NaN
dtype: float64

生成值

In [38]: s3 = Series([-2.1, 3.6, -1.5, 4, 3.1], index=['a', 'c', 'e', 'f', 'g'])

In [39]: s1 + s2 + s3
Out[39]:
a    3.1
c    4.7
d    NaN
e   -1.5
f    NaN
g    NaN
dtype: float64
也就是说NaN值不会变

DataFrame

add   用于加法（+）方法
sub   用于减法（-）方法
div   用于除法（/）方法
mul   用于乘法（*）方法

In [45]: df1 = DataFrame(np.arange(9.).reshape((3,3)), columns=list('bcd'), index=['Ohio', "Texas", "Colorado"])

In [46]: df2 = DataFrame(np.arange(12.).reshape((4,3)), columns=list('bde'), index=["Uhah", 'Ohio', "Texas", "Oregon"])

　　

In [47]: df1 + df2
Out[47]:
            b   c     d   e
Colorado  NaN NaN   NaN NaN
Ohio      3.0 NaN   6.0 NaN
Oregon    NaN NaN   NaN NaN
Texas     9.0 NaN  12.0 NaN
Uhah      NaN NaN   NaN NaN

那么可以使用add方法，传入df2一个fill_valued参数
In [8]: df1.add(df2, fill_value=0)
Out[8]:
            b    c     d     e
Colorado  6.0  7.0   8.0   NaN
Ohio      3.0  1.0   6.0   5.0
Oregon    9.0  NaN  10.0  11.0
Texas     9.0  4.0  12.0   8.0
Uhah      0.0  NaN   1.0   2.0

DataFrame和Series之间的运算

Series

In [40]: arr = np.arange(12.).reshape((3, 4))

In [41]: arr
Out[41]:
array([[  0.,   1.,   2.,   3.],
       [  4.,   5.,   6.,   7.],
       [  8.,   9.,  10.,  11.]])

In [42]: arr[0]
Out[42]: array([ 0.,  1.,  2.,  3.])

In [43]: arr - arr[0]
Out[43]:
array([[ 0.,  0.,  0.,  0.],
       [ 4.,  4.,  4.,  4.],
       [ 8.,  8.,  8.,  8.]])

DataFrame

In [44]: frame = DataFrame(np.arange(12.).reshape((4,3)), columns=list('bde'), index=["Uhah", 'Ohio', "Texas", "Oregon"])

In [45]: series = frame.ix[0]

In [46]: frame - series
Out[46]:
          b    d    e
Uhah    0.0  0.0  0.0
Ohio    3.0  3.0  3.0
Texas   6.0  6.0  6.0
Oregon  9.0  9.0  9.0

注意：如果某个索引值在DataFrame的列或Series的索引中找不到， 则参与运算的两个对象就会被重新索引以形成并集

In [47]: series2 = Series(range(3), index=['b', 'e', 'f'])

In [48]: frame + series2
Out[48]:
          b   d     e   f
Uhah    0.0 NaN   3.0 NaN
Ohio    3.0 NaN   6.0 NaN
Texas   6.0 NaN   9.0 NaN
Oregon  9.0 NaN  12.0 NaN

如果希望列在行上广播，必须使用算术运算方法

In [63]: frame.sub(series, axis=0)
Out[63]:
          b    d    e
Uhah   -1.0  0.0  1.0
Ohio   -1.0  0.0  1.0
Texas  -1.0  0.0  1.0
Oregon -1.0  0.0  1.0