在Openrefine中匹配不同的日期

Question

I am trying to use the value.match command in OpenRefine 2.6 for splitting the information presents in a column into (at least) 2 columns. The data are, however, quite messed up.
I have sometimes full dates:

May 30, 1949

Sometimes full dates are combined with other dates and attributes:

May 30, 1949, published 1979
May 30, 1949 and 1951, published 1979
May 30, 1949, printed 1980
May 30, 1949, print executed 1988
May 30, 1949, prints executed 1988
published 1940

Sometimes you have timespan:

1905-05 OR 1905-1906

Sometimes only the year

1905

Sometimes year with attributes

August or September 1908

Doesn't seems to follow any specific schema or order.
I would like to extract (at least)ca start and end date year, in order to have two columns:

-----------------------  
|start_date | end_date|  
|1905       | 1906    |   
-----------------------  

without the rest of the attributes.

I can find the last date using
value.match(/.*(\\d{4}).*?/)[0]
and the first one with
value.match(/.*^(\\d{4}).*?/)[0]
but I got some trouble with the two formulas.
The latter cannot match anything in case of:
May 30, 1949 and 1951, published 1979
while in the case of:
Paris, winter 1911-12 The latter formula cannot match anything and the former formula match 1911

Anyone know how I can resolve the problem?
I would need a solution that take the first date as start_date and final date as end_date, or better (don't know if it is possible) earliest date as start_date and latest date as end_date. Moreover, I would be glad to have some clue about how to extract other information, such as if published or printed or executed is present in the text -> copy date to a new column name “execution”. should be something like create a new column if(value.match("string1|string2|string3" + (\\d{4}), "perform the operation", do nothing)

Answer 1

value.match() is a very useful but sometimes tricky function. To extract a pattern from a text, I prefer to use Python/Jython's regular expressions :

import re

pattern = re.compile(r"\d{4}")

return pattern.findall(value)

From there, you can create a string with all the years concatenated:

return ",".join(pattern.findall(value))

Or select only the first:

return pattern.findall(value)[0]

Or the last:

return pattern.findall(value)[-1]

etc.

Same thing for your sub-question:

import re

pattern = re.compile(r"(published|printed|executed)\s+(\d+)")

return pattern.findall(value)[0][1]

Or :

import re

pattern = re.compile(r"(published|printed|executed)\s+(\d+)")

m = re.search(pattern, value)

return m.group(2)

Example:

Answer 2

Here is a regex which will extract start_date and end_date in named groups :

If there is only one date, then it consider it's the start_date :

((?<start_date>\\d{4}).*?)?(?<end_date>\\d{4}|(?<=-)\\d{2})?$

Demo