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Removing Items

 [1/3] from: kvince1:attbi at: 3-Jul-2002 9:45


Greetings, Simple question (hopefully). I'm reading from a very large quoted-comma delimited text file and writing to a quoted-comma delimited text file. I want to skip over the forty-first element and continue sending the rest of the record to the outbound file. This would be repeated for every record in the inbound file (over three hundred thousand). Inbound: open/direct/lines %InboundTextFile.txt Outbound: open/direct %OutboundTextFile.txt Data: copy/part Inbound [find/skip {","} ** 41st element - part I don't know how to do ** ] Append Outbound Data Any help would be appreciated! Ken

 [2/3] from: greggirwin:mindspring at: 3-Jul-2002 13:57


Hi Ken, << Simple question (hopefully). I'm reading from a very large quoted-comma delimited text file and writing to a quoted-comma delimited text file. I want to skip over the forty-first element and continue sending the rest of the record to the outbound file. This would be repeated for every record in the inbound file (over three hundred thousand). >> REBOL's PARSE function is really good at splitting up data for you, but the rest isn't *quite* so easy. First, you can take the line of input (a string) and parse it into fields, like so:
>> s: {a, b, "c, d, e", f, g}
== {a, b, "c, d, e", f, g}
>> parse s none
== ["a" "b" "c, d, e" "f" "g"] Now you can remove field 41 very easily. I'll remove one item (at field four) in my example here. Just replace 4 with 41 in your case.
>> head remove/part at parse s none 4 1
== ["a" "b" "c, d, e" "g"] Now, it would be nice if REJOIN worked as a mirror of PARSE, but it doesn't, as you can see below. It doesn't know about delimiters or fields that were quoted in the original string.
>> rejoin head remove/part at parse s none 4 1
== "abc, d, eg" To get around that, we can write a function to build a delimited string for us without too much difficulty. I just coded this, without concern for efficiency and very little testing. Caveat emptor. make-dlm-str: func [ "Reduces and joins a block of values." block [block!] "Values to reduce and join" dlm [char! string!] "Delimiter to put between elements" /local mod-blk ][ mod-blk: make block! 2 * length? block foreach item block [ append mod-blk reduce [ either find item dlm [mold item][item] dlm ] ] remove back tail mod-blk rejoin mod-blk ] Now we can do this, which looks OK to me:
>> make-dlm-str head remove/part at parse s none 4 1 ", "
== {a, b, "c, d, e", g} This doesn't preserve the original formatting of course. If that's important, or if you just want to tackle it a different way, you could use PARSE in it's more powerful mode. E.g., as a starting point... fld-sep: #"," valid-chars: complement charset "," field: [ copy data [quoted-string | copy data some valid-chars] (print trim data) [thru fld-sep | to end] ] quoted-string: [{"} thru {"}] rules: [some field to end]
>> s: {a, b, "c, d e", f, g} >> parse s rules
(IANAPG - I Am Not A Parse Guru) so this quick example is likely flawed somehow. HTH! --Gregg

 [3/3] from: lmecir::mbox::vol::cz at: 3-Jul-2002 22:50


Hi Ken, <<Ken>> Simple question (hopefully). I'm reading from a very large quoted-comma delimited text file and writing to a quoted-comma delimited text file. I want to skip over the forty-first element and continue sending the rest of the record to the outbound file. This would be repeated for every record in the inbound file (over three hundred thousand). Inbound: open/direct/lines %InboundTextFile.txt Outbound: open/direct %OutboundTextFile.txt Data: copy/part Inbound [find/skip {","} ** 41st element - part I don't know how to do ** ] Append Outbound Data Any help would be appreciated! <</Ken>> How about: line: first inbound parse/all line [41 [thru #","] copy data to end] etc. -L