i 需寻找相当长的水泥(通用采购卡b)编码。
此时此刻,一米仅作假和体格。
the problem is, if the CPV code has been listed in a slightly different format, this algorithm won t find it.
如何最有效地在扼杀中寻找守则的所有不同之处? 这是否只是重新制定每部10 000只聚氯乙烯的编码,并在每种情况下采用经限定的编号?
不同格式的例子如下:
30124120-1
301241201
30124120 - 1
30124120 1
30124120.1
等等。
感谢:
定期表达:
>>> cpv = re.compile(r ([0-9]+[-. ]?[0-9]) )
>>> print cpv.findall( foo 30124120-1 bar 21966823.1 baz )
[ 30124120-1 , 21966823.1 ]
(在数据中与聚氯乙烯相匹配之前进行更新)。)
采用<代码>re的任何功能(定期表述<>)。 See the docs for more info.
您可以定期表达,接受这些守则的若干不同格式,然后使用<代码>re.findall<>或类似于提取信息的内容。 我不敢肯定一点一点,那就是,我没有定期表达这种意愿(尽管你可能看到谷歌有没有吗?)
cpv = re.compile(r (d{8})(?:[ -. /\]*)(d{1}) )
for m in re.finditer(cpv, ex):
cpval,chk = m.groups()
print("{0}-{1}".format(cpval,chk))
适用于抽样数据回报
30124120-1
30124120-1
30124120-1
30124120-1
30124120-1
常规表述可改为
(d{8}) # eight digits
(?: # followed by a sequence which does not get returned
[ -. /\]* # consisting of 0 or more
) # spaces, hyphens, periods, tabs, forward- or backslashes
(d{1}) # followed by one digit, ending at a word boundary
# (ie whitespace or the end of the string)
希望帮助!
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