Python 在两个时间序列中查找具有相同值的日期

Python 在两个时间序列中查找具有相同值的日期,python,time-series,pandas,Python,Time Series,Pandas,我有两个以datetime列作为索引的时间序列: 2012-08-10 1.1 2012-08-10 1.1 2012-08-11 1.2 2012-08-11 1.2 2012-08-12 1.8 2012-08-13 1.1 2012-08-14 1.4 2012-08-15 1.3 2012-08-15 1.7 2012-08-16 1.1 2012-08-17 1.6 2012-08-17 1.2 2012-08-1

我有两个以datetime列作为索引的时间序列:

2012-08-10 1.1       2012-08-10 1.1
2012-08-11 1.2       2012-08-11 1.2
2012-08-12 1.8       2012-08-13 1.1
2012-08-14 1.4       2012-08-15 1.3
2012-08-15 1.7       2012-08-16 1.1
2012-08-17 1.6       2012-08-17 1.2
2012-08-18 1.1       2012-08-18 1.1
如何比较它们并获得两个包含日期的时间序列,这两个时间序列中都有日期:

2012-08-10 1.1       2012-08-10 1.1
2012-08-11 1.2       2012-08-11 1.2
2012-08-15 1.7       2012-08-15 1.3
2012-08-17 1.6       2012-08-17 1.2
2012-08-18 1.1       2012-08-18 1.1
使用:

编辑:如果要再次拆分两个文件,请执行以下操作:

$ join in1 in2 | awk '{print $1, $2}' > out1
$ join in1 in2 | awk '{print $1, $3}' > out2
$ cat out1
2012-08-10 1.1
2012-08-11 1.2
2012-08-15 1.7
2012-08-17 1.6
2012-08-18 1.1
$ cat out2
2012-08-10 1.1
2012-08-11 1.2
2012-08-15 1.3
2012-08-17 1.2
2012-08-18 1.1

我喜欢小工具:)

对不起,我是说使用pandas timeseries。不使用bash工具。
$ join in1 in2 | awk '{print $1, $2}' > out1
$ join in1 in2 | awk '{print $1, $3}' > out2
$ cat out1
2012-08-10 1.1
2012-08-11 1.2
2012-08-15 1.7
2012-08-17 1.6
2012-08-18 1.1
$ cat out2
2012-08-10 1.1
2012-08-11 1.2
2012-08-15 1.3
2012-08-17 1.2
2012-08-18 1.1
In [52]: s1.align(s2, join='inner')
Out[52]:
(2012-08-10    1.1
2012-08-11    1.2
2012-08-15    1.7
2012-08-17    1.6
2012-08-18    1.1,
 2012-08-10    1.1
2012-08-11    1.2
2012-08-15    1.3
2012-08-17    1.2
2012-08-18    1.1)