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Using loop command in UNIX for converting column to row

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# 1  
Old 04-09-2015
Using loop command in UNIX for converting column to row
Dear folks

I have 300 files which one of them are looking like:

Code:
1.SNP
0
0
1
0

I am looking for desire output:

Code:
1.SNP 0 0 1 0

I used this below command to run all of the 300 file at the same time


Code:
for file in *.SNP; do awk '{printf( "%s ", $1 );} END {printf("\n");}' $file > "$(basename "$file" .SNP).gen"; done

But I do not know why all of the *.gen file do not include any information in the *.gen files. Could anyone give me a guide for this problem?
Last edited by Scott; 04-09-2015 at 07:14 PM.. Reason: Code tags, please. We've asked you many-a-time.
# 2  
Old 04-09-2015
There isn't anything immediately obvious wrong with your script...

What operating system are you using?

What shell are you using?

Do you get any output at all when you run your script?

Is a .gen file created for each .SNP file in the directory in which you run this loop?

Pick one of your input files and show us the output (in CODE tags) from the commands:
Code:
ls -l "file.SNP"
od -bc "file.SNP"
awk '{printf( "%s ", $1 );} END {printf("\n");}' "file.SNP" > "$(basename "file.SNP" .SNP).gen"
ls -l "file.gen"
od -bc "file.gen"

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