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Count words in a single column

 
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# 1  
Old 03-27-2015
Count words in a single column
Dear All,

I have set of CSV files (comma separated) and each column have some information in them separated by space. Now I want to count them but have not been successful...

Example data
Quote:
1, 23 56, London Oxford Cambridge
4, 45 65 78, London
desired outcome
Quote:
1, 2, 3
1, 3, 1
I have tried few things including the link below.
Code:
for C in $FILES
do
	name=$(basename $C) 
	grep -o ",.*," $C | wc -w >> result/${name}
done

but this gives me count of everything separated by space Smilie

Can you please help me? thanks
# 2  
Old 03-27-2015
Code:
awk -F, '{for(i=1;i<=NF;i++) printf("%d%s", split($i,a, " "), (i==NF)?ORS:OFS)}' OFS=',' myFile

This User Gave Thanks to vgersh99 For This Post:
A-V
# 3  
Old 03-27-2015
magic ... thank you so much
tired awk but not with split and OFS

now it words perfectly
# 4  
Old 03-27-2015
Bit simplified version of vgersh99's solution

Code:
$ awk -F, '{for(i=1; i<=NF; i++)$i=split($i,a," ")}1' OFS="," file
1,2,3
1,3,1

This User Gave Thanks to Akshay Hegde For This Post:
A-V
# 5  
Old 03-30-2015
Thanks a lot from both of you ...
But both add random 0 at the end of few files ...
the ones I have resided are the ones that do not have values for the last column
e.g. example data
Quote:
1, 23 56, London Oxford Cambridge
4, 45 65 78,
instead of
Quote:
1, 2, 3
1, 3, 0
it is publishing
Quote:
1, 2, 3
1, 3, 0, 0
Can you please help me?
# 6  
Old 03-30-2015
Both proposals deliver the desired output, not the errorneous one you posted. Are you sure there's no hidden control chars in the input file? Did yo run the codes exactly as posted?
This User Gave Thanks to RudiC For This Post:
A-V
# 7  
Old 03-30-2015
With that input, we're seeing the output you seem to want.
Please post the output from the command:
Code:
od -bc file

where file is the input file you fed to either of the awk scripts provided by vgersh99 and Akshay Hegde.
This User Gave Thanks to Don Cragun For This Post:
A-V
 
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