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How to get a very big file sorted by contents of another variable list in one pass?

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# 1  
Old 10-03-2010
How to get a very big file sorted by contents of another variable list in one pass?
I have two list....

One has 17m (yes that's million) records in it which relate to the path and file name of individual text files.

At the start of the pathname is a two character identifier identifying which category it belongs in as identified by a second file with the two character lookup table.

So for example

Code:
Big File

/BB/26607/B7763/BTHS/name1.txt
/DF/78873/YHH97/H764/name76.txt
/BB/8766/OP764/Y7644/name39.txt

In the second file (usually about 42 different two letter identifiers).

Code:
BB
JH
DF
ND

What I want to do is not just sort the master file but split and prioritise it by the order in the second list and try to do it in one pass as processing 17m records is taking about 7 minutes per pass on the system I'm using.

I can't think of a simple way to do this and am suffering from too much typing today now, so could anyone give me some good ideas please?

Thanks
# 2  
Old 10-03-2010
So you want to output pathnames relating to different categories to separate files? Like those containing "BB" to one file, those containing "JH" to another, etc?
# 3  
Old 10-03-2010
Yes please....

Or drop them into an array with the two character reference being used as the array lookup value i.e.

$NAMEVARIABLE[$TWOCHARVAR]

and then I'll echo that out to individual files

Let me just add....

What happens is that then this list is used to process work on the text files as prioritised by the two character order, the work is broken up and jobs spread to other machines that do the actual processing.
# 4  
Old 10-03-2010
Try:
Code:
perl -ne '/^\/(..)/;open O,">>$1";print O;' bigfile

# 5  
Old 10-03-2010
Assuming you can have 40+ open files (you do not mention your OS)

Code:
#pathname for your two letter filenames
p=/path/to/new/files

awk -F'/' -v p=$p '  FILENAME=="file1" {arr[$0]=sprintf("%s/%s", p, $0) }
                    FILENAME=="file2" {if($1 in arr) { print $0 > arr[$1]}
                                              else {print $0}  } ' file1 file2

This will move data into the separate files. you may need to call ulimit to provide more open file descriptors. Also, any row from file2 without a match in file1 will print to the terminal.
# 6  
Old 10-03-2010
Bartus..

That's very kind and helpful, I should have said I only want to use bash, mainly because I don't have the time to learn perl at the moment and make it a rule to never put code in a script I might need a reference book for later.

If you have any ideas for doing it in bash I'd be grateful (sed, awk, grep all allowed).

Regards

---------- Post updated at 07:20 PM ---------- Previous update was at 07:15 PM ----------

Thanks Jim

I'm using Fedora for the testing and RH live.

Let me see if I can get what you've suggested working, I'm open to as many suggestions as possible though.

Regards
# 7  
Old 10-03-2010
Is this a sort or a sift?
Do you need to sort the content of the 42 files? If so, what is the sort key?

Assuming no sort, sifting the main file into individual files with names based on the data characters 2-3 is relatively easy in a modern shell but it will not be quick with 17 million records.

Code:
For example:
cat filename | while read record
do
     category=`echo "${record}"|cut -c2-3`
     echo "${record}" >> "filename_${category}"
done

You can then do whatever processing you need to do in priority order.

For example:
cat reference_file | while read category
do
         ls -ald "filename_${category}"
done
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