concat方法可以实现对象在轴向的的粘合或者堆叠。
In [55]: s1 = pd.Series([0, 1], index=['a', 'b'])
In [56]: s2 = pd.Series([2, 3, 4], index=['c', 'd', 'e'])
In [57]: s3 = pd.Series([5, 6], index=['f', 'g'])
In [58]: pd.concat([s1, s2, s3]) # 要以列表的方式提供参数
Out[58]:
a 0
b 1
c 2
d 3
e 4
f 5
g 6
dtype: int64
In [59]: pd.concat([s1, s2, s3], axis=1) # 横向堆叠，但出现警告信息
C:\ProgramData\Anaconda3\Scripts\ipython:1: FutureWarning: Sorting because non-concatenation axis is not aligned. A future version
of pandas will change to not sort by default.
......
Out[59]:
0 1 2
a 0.0 NaN NaN
b 1.0 NaN NaN
c NaN 2.0 NaN
d NaN 3.0 NaN
e NaN 4.0 NaN
f NaN NaN 5.0
g NaN NaN 6.0
In [60]: pd.concat([s1, s2, s3], axis=1,sort=True) # 按人家的要求做
Out[60]:
0 1 2
a 0.0 NaN NaN
b 1.0 NaN NaN
c NaN 2.0 NaN
d NaN 3.0 NaN
e NaN 4.0 NaN
f NaN NaN 5.0
g NaN NaN 6.0
对于DataFrame，默认情况下都是按行往下合并的，当然也可以设置axis参数：
In [66]: df1 = pd.DataFrame(np.arange(6).reshape(3, 2), index=['a', 'b', 'c'],
...: columns=['one', 'two'])
...:
In [67]: df2 = pd.DataFrame(5 + np.arange(4).reshape(2, 2), index=['a', 'c'],
...: columns=['three', 'four'])
...:
In [68]: df1
Out[68]:
one two
a 0 1
b 2 3
c 4 5
In [69]: df2
Out[69]:
three four
a 5 6
c 7 8
In [71]: pd.concat([df1, df2], sort=True)
Out[71]:
four one three two
a NaN 0.0 NaN 1.0
b NaN 2.0 NaN 3.0
c NaN 4.0 NaN 5.0
a 6.0 NaN 5.0 NaN
c 8.0 NaN 7.0 NaN
In [72]: pd.concat([df1, df2], axis=1, sort=True)
Out[72]:
one two three four
a 0 1 5.0 6.0
b 2 3 NaN NaN
c 4 5 7.0 8.0
有这么种场景，某个对象里缺失的值，拿另外一个对象的相应位置的值来填补。在Numpy层面，可以这么做：
In [74]: a = pd.Series([np.nan, 2.5, 0, 3.5, 4.5, np.nan],
...: index=['f', 'e', 'd', 'c', 'b', 'a'])
In [75]: b = pd.Series([0, np.nan, 2.1, np.nan, np.nan, 5], index=list('abcdef'))
In [76]: a
Out[76]:
f NaN
e 2.5
d 0.0
c 3.5
b 4.5
a NaN
dtype: float64
In [77]: b
Out[77]:
a 0.0
b NaN
c 2.1
d NaN
e NaN
f 5.0
dtype: float64
In [78]: np.where(pd.isnull(a), b, a)
Out[78]: array([0. , 2.5, 0. , 3.5, 4.5, 5. ])
np.where(pd.isnull(a), b, a)，这一句里，首先去pd.isnull(a)种判断元素，如果是True，从b里拿数据，否则从a里拿，得到最终结果。
实际上，Pandas为这种场景提供了一个专门的combine_first方法：
In [80]: b.combine_first(a) Out[80]: a 0.0 b 4.5 c 2.1 d 0.0 e 2.5 f 5.0 dtype: float64
对于DataFrame对象，combine_first逐列做相同的操作，因此你可以认为它是根据你传入的对象来‘修补’调用对象的缺失值。
In [81]: df1 = pd.DataFrame({'a': [1., np.nan, 5., np.nan],
...: 'b': [np.nan, 2., np.nan, 6.],
...: 'c': range(2, 18, 4)})
...:
In [82]: df2 = pd.DataFrame({'a': [5., 4., np.nan, 3., 7.],
...: 'b': [np.nan, 3., 4., 6., 8.]})
...:
In [83]: df1
Out[83]:
a b c
0 1.0 NaN 2
1 NaN 2.0 6
2 5.0 NaN 10
3 NaN 6.0 14
In [84]: df2
Out[84]:
a b
0 5.0 NaN
1 4.0 3.0
2 NaN 4.0
3 3.0 6.0
4 7.0 8.0
In [85]: df1.combine_first(df2)
Out[85]:
a b c
0 1.0 NaN 2.0
1 4.0 2.0 6.0
2 5.0 4.0 10.0
3 3.0 6.0 14.0
4 7.0 8.0 NaN
np.where要求两个对象遵循广播机制，combine_first没有这一要求。
水平堆叠