modify value's in CSV with bash-script

Question

I've the following CSV file's:

2012-07-12 15:30:09; 353.2
2012-07-12 15:45:08; 347.4
2012-07-12 16:00:08; 197.6
2012-07-12 16:15:08; 308.2
2012-07-12 16:30:09; 352.6

What I want to do is modify the value in the 2nd column...

What I already can do is extract the value and modify it this way:

#!/bin/bash
cut -d ";" -f2 $1 > .tmp.csv
for num in $(cat .tmp.csv)
    do
        (echo "scale=2;$num/5" | bc -l >> .tmp2.csv)
done
rm .tmp.csv
rm .tmp2.csv

But I need to have column1 in that file too...

I hope one of you can give me a hint, I'm just stuck!

I spent too much time trying to do things like this. If you have python on your system I suggest you try that instead. — Johan Lundberg
– Johan Lundberg, Commented Jul 17, 2012 at 12:50
I've absolutly no py experience, but might be able to work it out with a similar example... — Mirco Schmidt
– Mirco Schmidt, Commented Jul 17, 2012 at 12:51
It'll probably be easiest to use awk, but you'll need to be more specific on what you want to do. — Kevin
– Kevin, Commented Jul 17, 2012 at 12:54
Hi Kevin, I'll divide the value in the 2nd column through 5 and write it back to the csv — Mirco Schmidt
– Mirco Schmidt, Commented Jul 17, 2012 at 12:56

Debaditya · Accepted Answer · 2012-07-17 13:09:43Z

4

From your code, this is what I understood

Input

2012-07-12 15:30:09; 353.2 
2012-07-12 15:45:08; 347.4 
2012-07-12 16:00:08; 197.6 
2012-07-12 16:15:08; 308.2 
2012-07-12 16:30:09; 352.6

Awk code

awk -F ";" '{print $1 ";" $2/5}' input

Output

2012-07-12 15:30:09;70.64
2012-07-12 15:45:08;69.48
2012-07-12 16:00:08;39.52
2012-07-12 16:15:08;61.64
2012-07-12 16:30:09;70.52

answered Jul 17, 2012 at 13:09

Debaditya

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Comments

Steve · Accepted Answer · 2012-07-17 13:10:23Z

3

One way, using awk:

awk '{ $NF = $NF/5 }1' file.txt

Results:

2012-07-12 15:30:09; 70.64
2012-07-12 15:45:08; 69.48
2012-07-12 16:00:08; 39.52
2012-07-12 16:15:08; 61.64
2012-07-12 16:30:09; 70.52

HTH

answered Jul 17, 2012 at 13:10

Steve

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Comments

chepner · Accepted Answer · 2012-07-17 13:30:30Z

3

Here's an almost-pure bash solution, without temp files:

#!/bin/bash

while IFS=$';' read col1 col2; do
    echo "$col1; $(echo "scale=2;$col2/5" | bc -l)"
done

answered Jul 17, 2012 at 13:30

chepner

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Comments

Birei · Accepted Answer · 2012-07-17 12:59:43Z

2

Try with awk:

awk '
    BEGIN {
        ## Split fields with ";".
        FS = OFS = "; "
    }

    {
        $2 = sprintf( "%.2f", $2/5 )
        print $0
    }
' infile

Output:

2012-07-12 15:30:09; 70.64
2012-07-12 15:45:08; 69.48
2012-07-12 16:00:08; 39.52
2012-07-12 16:15:08; 61.64
2012-07-12 16:30:09; 70.52

answered Jul 17, 2012 at 12:59

Birei

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