我试图在有空白包围的整数2位数的地方之前分割字符串.最终我希望这可以在Python中工作,但我已经与sed工作了,我无法理解.
我的测试数据如下所示:
13 13 13 13 13 9:07.18 9:12.09 9:15.65
14 14 14 2:04.86 2:05.99 2:06.87 14 4:21.51 4:23.51 4:25.00 14 8:56.28 9:01.09 9:04.58
15 15 57.18 57.61 57.95 15 2:02.61 2:03.72 2:04.58 15 4:17.31 4:19.28 4:20.75 15 8:47.15 8:51.87 8:55.30
16 16 56.34 56.76 57.09 16 2:00.69 2:01.78 2:02.63 16 4:13.75 4:15.69 4:17.14 16 8:39.71 8:44.37 8:47.75
17 25.69 25.85 25.99 17 55.62 56.03 56.36 17 1:59.07 2:00.15 2:00.99 17 4:10.76 4:12.69 4:14.11 17 8:33.73 8:38.34 8:41.68
18 25.43 25.59 25.73 18 55.01 55.42 55.74 18 1:57.74 1:58.81 1:59.63 18 4:08.34 4:10.24 4:11.66 18 8:33.73 8:37.04
19 25.20 25.36 25.49 19 54.50 54.91 55.23 19 1:57.74 1:58.56 19 4:08.34 4:09.74 19 8:33.73
我希望它像这样分开(注意逗号的位置','):
13, 13, 13, 13, 13 9:07.18 9:12.09 9:15.65
14, 14, 14 2:04.86 2:05.99 2:06.87, 14 4:21.51 4:23.51 4:25.00, 14 8:56.28 9:01.09 9:04.58
15, 15 57.18 57.61 57.95, 15 2:02.61 2:03.72 2:04.58, 15 4:17.31 4:19.28 4:20.75, 15 8:47.15 8:51.87 8:55.30
16, 16 56.34 56.76 57.09, 16 2:00.69 2:01.78 2:02.63, 16 4:13.75 4:15.69 4:17.14, 16 8:39.71 8:44.37 8:47.75
17 25.69 25.85 25.99, 17 55.62 56.03 56.36, 17 1:59.07 2:00.15 2:00.99, 17 4:10.76 4:12.69 4:14.11, 17 8:33.73 8:38.34 8:41.68
18 25.43 25.59 25.73, 18 55.01 55.42 55.74, 18 1:57.74 1:58.81 1:59.63, 18 4:08.34 4:10.24 4:11.66, 18 8:33.73 8:37.04
19 25.20 25.36 25.49, 19 54.50 54.91 55.23, 19 1:57.74 1:58.56, 19 4:08.34 4:09.74, 19 8:33.73
上面的数据是相当规律的,因为两位数的整数在[13,19]范围内,但我应该期望的范围是[10,99].
有人可以建议一种方法来执行上述转换吗？我已经使用正则表达式了一段时间,但我不能涵盖所有情况.
先行断言(?=...)可以解决这个问题:
>>> a = """13 13 13 13 13 9:07.18 9:12.09 9:15.65
14 14 14 2:04.86 2:05.99 2:06.87 14 4:21.51 4:23.51 4:25.00 14 8:56.28 9:01.09 9:04.58
15 15 57.18 57.61 57.95 15 2:02.61 2:03.72 2:04.58 15 4:17.31 4:19.28 4:20.75 15 8:47.15 8:51.87 8:55.30
16 16 56.34 56.76 57.09 16 2:00.69 2:01.78 2:02.63 16 4:13.75 4:15.69 4:17.14 16 8:39.71 8:44.37 8:47.75
17 25.69 25.85 25.99 17 55.62 56.03 56.36 17 1:59.07 2:00.15 2:00.99 17 4:10.76 4:12.69 4:14.11 17 8:33.73 8:38.34 8:41.68
18 25.43 25.59 25.73 18 55.01 55.42 55.74 18 1:57.74 1:58.81 1:59.63 18 4:08.34 4:10.24 4:11.66 18 8:33.73 8:37.04
19 25.20 25.36 25.49 19 54.50 54.91 55.23 19 1:57.74 1:58.56 19 4:08.34 4:09.74 19 8:33.73"""
>>> print(re.sub("(\d{2}) (?=\d{2}( |$))","\g<1>, ", a))
13, 13, 13, 13, 13 9:07.18 9:12.09 9:15.65
14, 14, 14 2:04.86 2:05.99 2:06.87, 14 4:21.51 4:23.51 4:25.00, 14 8:56.28 9:01.09 9:04.58
15, 15 57.18 57.61 57.95, 15 2:02.61 2:03.72 2:04.58, 15 4:17.31 4:19.28 4:20.75, 15 8:47.15 8:51.87 8:55.30
16, 16 56.34 56.76 57.09, 16 2:00.69 2:01.78 2:02.63, 16 4:13.75 4:15.69 4:17.14, 16 8:39.71 8:44.37 8:47.75
17 25.69 25.85 25.99, 17 55.62 56.03 56.36, 17 1:59.07 2:00.15 2:00.99, 17 4:10.76 4:12.69 4:14.11, 17 8:33.73 8:38.34 8:41.68
18 25.43 25.59 25.73, 18 55.01 55.42 55.74, 18 1:57.74 1:58.81 1:59.63, 18 4:08.34 4:10.24 4:11.66, 18 8:33.73 8:37.04
19 25.20 25.36 25.49, 19 54.50 54.91 55.23, 19 1:57.74 1:58.56, 19 4:08.34 4:09.74, 19 8:33.73
所以,reg exp.你需要的是(\d{2}) (?=\d{2}( |$)):
(\d{2}) =>在组1中存储2个数字并匹配额外的空格.(?=\d{2}( |$)) =>匹配2个数字和1个空格或EOL,但不要使用它们.这里的关键是,通过不消耗第二个匹配的组,下次应用子功能时将再次处理它.最后,\g<1>,将用相同的数字和附加替换1 ,.