一个更好的RE？ [英] A better RE?
问题描述
我想要一个与21MAR06 31APR06 1236，
匹配的字符串，其中最后一部分是天数（1-7），即它可以包含
数字1-7，按顺序，每个只有一个，至少一个
数字。我想把它当成三组。我在考虑
r"（\\\\ [AZ] \\\）（\\\\ [[[[[[[[[[[[[[[[[[[[dd d）（1？2？3？4？5？6？7？）
但是即使第三组是空的也会匹配，
对吗？
这个人有没有好的而不是过于复杂的RE？
P.S.我知道现在你有两个问题回复......
I want an re that matches strings like "21MAR06 31APR06 1236",
where the last part is day numbers (1-7), i.e it can contain
the numbers 1-7, in order, only one of each, and at least one
digit. I want it as three groups. I was thinking of
r"(\d\d[A-Z]\d\d) (\d\d[A-Z]\d\d) (1?2?3?4?5?6?7?)"
but that will match even if the third group is empty,
right? Does anyone have good and not overly complex RE for
this?
P.S. I know the "now you have two problems reply..."
推荐答案
Magnus Lycka写道：
Magnus Lycka wrote:
我想要一个与21MAR06 31APR06 1236等字符串匹配的重新编号，其中最后一部分是日期编号（1-7），即它可以包含数字1-7，in订单，每个只有一个，至少有一个数字。我想把它当成三组。我想的是
r"（\d\ [AZ] \\\）（\\\\ [[[[[[[[[[[[[[[[[[[dd ？2？3？4？5？6？7？）
但即使第三组是空的，它也会匹配，对吧？有没有人对这个有好的而不是过于复杂的RE？
I want an re that matches strings like "21MAR06 31APR06 1236",
where the last part is day numbers (1-7), i.e it can contain
the numbers 1-7, in order, only one of each, and at least one
digit. I want it as three groups. I was thinking of
r"(\d\d[A-Z]\d\d) (\d\d[A-Z]\d\d) (1?2?3?4?5?6?7?)"
but that will match even if the third group is empty,
right? Does anyone have good and not overly complex RE for
this?
怎么样（未经测试）
r" ;（\d \ [AZ] {3} \\\）（\\\\ [AZ] {3} \\\）（？= [1234567]）（1 ？2？3？4？5？6？7？）
其中{3}表示需要三份以前的RE部分，并且
（？= [1234567]）表示要求1-7中的至少一个，但如果匹配，则不要向前移动
。
< ; / F>
how about (untested)
r"(\d\d[A-Z]{3}\d\d) (\d\d[A-Z]{3}\d\d) (?=[1234567])(1?2?3?4?5?6?7?)"
where {3} means require three copies of the previous RE part, and
(?=[1234567]) means require at least one of 1-7, but don''t move
forward if it matches.
</F>
Magnus Lycka写道：
Magnus Lycka wrote:
我想要一个与21MAR06 31APR06 1236等字符串匹配的重新编号，
其中最后一部分是天数（1-7），即它可以包含数字1-7，按顺序，每个只有一个，至少有一个数字。我想把它当成三组。我想的是
r"（\d\ [AZ] \\\）（\\\\ [[[[[[[[[[[[[[[[[[[dd ？2？3？4？5？6？7？）
但即使第三组是空的，它也会匹配，对吧？对于
这个有没有好的而不是过于复杂的RE？
P.S。我知道现在你有两个问题回复......
I want an re that matches strings like "21MAR06 31APR06 1236",
where the last part is day numbers (1-7), i.e it can contain
the numbers 1-7, in order, only one of each, and at least one
digit. I want it as three groups. I was thinking of
r"(\d\d[A-Z]\d\d) (\d\d[A-Z]\d\d) (1?2?3?4?5?6?7?)"
but that will match even if the third group is empty,
right? Does anyone have good and not overly complex RE for
this?
P.S. I know the "now you have two problems reply..."
txt = 21MAR06 31APR06 1236
m =''（？：JAN | FEB | MAR | APR | MAI | JUN | JUL | AUG | SEP | OCT | NOV | DE Z）''
＃非捕获组（：？）
p = re.compile（r"（\\\ %% s \\\\））（\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ （？= [1234567]）（1？2？3？4？5？6？7？）％（m，m））
p.match（txt）.group（1 ）
''21MAR06''
p.match（txt）.group（2）
''31APR06''
p.match（ txt）.group（3）
txt = "21MAR06 31APR06 1236" m = ''(?:JAN|FEB|MAR|APR|MAI|JUN|JUL|AUG|SEP|OCT|NOV|DE Z)'' # non capturing group (:?)
p = re.compile(r"(\d\d%s\d\d) (\d\d%s\d\d) (?=[1234567])(1?2?3?4?5?6?7?)" % (m,m))
p.match(txt).group(1) ''21MAR06''
p.match(txt).group(2) ''31APR06''
p.match(txt).group(3)
1236
1236
Magnus Lycka写道：
Magnus Lycka wrote:
我想要一个匹配像21MAR06 31APR06 1236这样的字符串的re，其中最后一部分是日期编号（1-7），即它可以包含
数字1-7，按顺序，每个只有一个，至少有一个数字。我想把它当成三组。我想的是
r"（\d\ [AZ] \\\）（\\\\ [[[[[[[[[[[[[[[[[[[dd ？2？3？4？5？6？7？）
但即使第三组是空的，它也会匹配，对吧？对于
这个人有没有好的而不是过于复杂的RE？
最简单：
I want an re that matches strings like "21MAR06 31APR06 1236",
where the last part is day numbers (1-7), i.e it can contain
the numbers 1-7, in order, only one of each, and at least one
digit. I want it as three groups. I was thinking of
r"(\d\d[A-Z]\d\d) (\d\d[A-Z]\d\d) (1?2?3?4?5?6?7?)"
but that will match even if the third group is empty,
right? Does anyone have good and not overly complex RE for
this?
Simplest:
exp = r"（\d {2} [AZ] {3} \d {2}）（\d {2} [AZ] {3} \d {2}）（\d +）"
re.match （exp，s）.groups（）
exp = r"(\d{2}[A-Z]{3}\d{2}) (\d{2}[A-Z]{3}\d{2}) (\d+)"
re.match(exp, s).groups()
（''21MAR06''，''31APR06''，''1236''）br />
但这可能会给你误报，取决于真实数据。
如果你想尽可能严格，这会变得有点有毛的。
PS我知道现在你有两个问题回复......
(''21MAR06'', ''31APR06'', ''1236'')
but this could give you false positive, depending on the real data.
If you want to be as strict as possible, this becomes a little bit hairy.
P.S. I know the "now you have two problems reply..."
！ - ）
- < br $> b $ b bruno desthuilliers
python -c" print''@''。join（[''。''。join（[w [:: - 1] for w在p.split（''。''）]）中'/ o **** @ xiludom.gro'中的
p。分裂（''@''）]）"
!-)
--
bruno desthuilliers
python -c "print ''@''.join([''.''.join([w[::-1] for w in p.split(''.'')]) for
p in ''o****@xiludom.gro''.split(''@'')])"
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