使用Python的Dataframe取两列时间值相差一年的所有行方法

data[map(lambda x:datetime.date(x.year-1,x.month,x.day),data['report_date'])==data['date_1y_ago']]
 company_id signal_code_x signal_value_x report_date signal_code_y signal_value_y report_date_last date_1y_ago
0  2154888 r_1002030000   0.62660 2015-09-30 r_1002030000   0.64145  2015-12-31 2014-12-31
1  2154888 r_1002030000   0.64145 2015-12-31 r_1002030000   0.64145  2015-12-31 2014-12-31
2  2154888 r_1002030000   0.60544 2015-03-31 r_1002030000   0.64145  2015-12-31 2014-12-31
3  2154888 r_1002030000   0.54911 2014-12-31 r_1002030000   0.64145  2015-12-31 2014-12-31
4  2154888 r_1002030000   0.61379 2015-06-30 r_1002030000   0.64145  2015-12-31 2014-12-31
5  1702887 r_1002030000   0.62173 2014-03-31 r_1002030000   0.51103  2015-03-31 2014-03-31
6  1702887 r_1002030000   0.55175 2014-12-31 r_1002030000   0.51103  2015-03-31 2014-03-31
7  1702887 r_1002030000   0.51103 2015-03-31 r_1002030000   0.51103  2015-03-31 2014-03-31
8  1702887 r_1002030000   0.58696 2014-06-30 r_1002030000   0.51103  2015-03-31 2014-03-31
9  13484491 r_1002030000   0.17658 2014-12-31 r_1002030000   0.41082  2015-12-31 2014-12-31
10 13484491 r_1002030000   0.41082 2015-12-31 r_1002030000   0.41082  2015-12-31 2014-12-31
11 13484491 r_1002030000   0.39220 2015-09-30 r_1002030000   0.41082  2015-12-31 2014-12-31

company_id signal_code_x signal_value_x report_date signal_code_y signal_value_y report_date_last date_1y_ago
1  2154888 r_1002030000   0.64145 2015-12-31 r_1002030000   0.64145  2015-12-31 2014-12-31
7  1702887 r_1002030000   0.51103 2015-03-31 r_1002030000   0.51103  2015-03-31 2014-03-31
10 13484491 r_1002030000   0.41082 2015-12-31 r_1002030000   0.41082  2015-12-31 2014-12-31