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Round Robin Distribution of Contents of file to 3 files

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# 1  
Old 07-23-2009
Round Robin Distribution of Contents of file to 3 files
Hi
I need to create a script that distributes in round robin fashion the contents of a file to 3 files. The number of lines in a content of file can vary from 1-n(Each line is just a one letter word).The entire lines needs to get distributed into 3 files ( The order doesnt matter) , at the end of process the number of lines in each file should be more or less the same

eg
Input_file.txt
(Contents - 4 lines)
1
2
3
4

At the end of process

Op_1.txt
1
4

Op_2 .txt
2

Op_3.txt
3


Thanks for the help
# 2  
Old 07-23-2009
You can do something like that :
Code:
awk -v COUNT=3 '{ f=(f%COUNT)+1 ; print > "Op_" f ".txt" }' Input_file.txt

Jean-Pierre.
# 3  
Old 07-23-2009
Thanks
That was really helpful..Thanks
# 4  
Old 08-07-2009
Error Handling on AWK command
How can we do error handling on the awk commands.
Based on earlier threads i am doing this command.Accidently i discovered that if directory /tmp/temp/ is not there it throws out an error ( The directory names are parameterized, this is just an example) , but the return code is still equal zero

ls *.dat | awk -v COUNT=3 -v DIR="/tmp/temp/" '{ f=(f%COUNT)+1 ; print > DIR "_list_" f ".log" }'


Any suggestion ?
# 5  
Old 08-07-2009
Quote:
Originally Posted by police
How can we do error handling on the awk commands.
Based on earlier threads i am doing this command.Accidently i discovered that if directory /tmp/temp/ is not there it throws out an error ( The directory names are parameterized, this is just an example) , but the return code is still equal zero

ls *.dat | awk -v COUNT=3 -v DIR="/tmp/temp/" '{ f=(f%COUNT)+1 ; print > DIR "_list_" f ".log" }'


Any suggestion ?

awk needs an explicit declaration ( exit n ) of its return status ... so assuming you don't have an END statement one way would be:

Code:
$ ls *.dat | awk -v COUNT=3 -v DIR="/tmp/invalid_dir/" 'BEGIN{ if(system("test -d " DIR)>0) exit 2 }{ f=(f%COUNT)+1; print f }'
$ echo $?
2

$ ls *.dat | awk -v COUNT=3 -v DIR="/tmp/valid_dir/"   'BEGIN{ if(system("test -d " DIR)>0) exit 2 }{ f=(f%COUNT)+1; print f }'
1
2
3
1
2
3
$ echo $?
0


otherwise ( END present ):

Code:
ls *.dat | awk -v COUNT=3 -v DIR="/tmp/valid_dir" 'BEGIN{ if(system("test -d " DIR)>0) {e=1; exit 2}}{ f=(f%COUNT)+1; print f }
                                          END  { if(e) exit 2; else print "more actions here"}'

Or it might be easier to test for DIR existence beforehand at the shell level.
# 6  
Old 08-10-2009
Is there any generic way of handling all expection related to AWK .Like $? in Shell
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