CODEWITHSUNDEEP
shellcsv
jms1980
jms1980

Reputation: 1075

Extracting data from a particular column number in csv file

I have a csv file that looks like this:

arruba,jamaica, bermuda, bahama, keylargo, montigo, kokomo
80,70,90,85,86,89,83
77,75,88,87,83,85,77
76,77,83,86,84,86,84

I want to have a shell script set up so that I can extract out the data so that I can categorize data by columns.

I know that the line of code:

IFS="," read -ra arr <"vis.degrib"
for ((i=0 ; i<${#arr[@]} ; i++));do
     ifname=`printf "%s\n" "${arr[i]}"`
     echo "$ifname"
done

will print out the individual column components for the first row. How do I also do this again for subsequent rows?

Thank you for your time.

Upvotes: 0

Views: 56

Answers (1)

karakfa
karakfa

Reputation: 67507

I'm extrapolating from the OP

 awk -F, 'NR==1{for(i=1;i<=NF;i++) {gsub(/^ +/,"",$i);print $i}}' vis.degrib

will print

arruba
jamaica
bermuda
bahama
keylargo
montigo
kokomo

note that there is trimming of the space from the beginning of each field. If you remove the condition NR==1, the same will be done for all rows. Was this your request? Please comment...

Perhaps you want to convert columnar format to row based format (transpose)? There are many ways, this awk script will do

 awk -F, -v OFS=, '{sep=(NR==1)?"":OFS} {for(i=1;i<=NF;i++) a[i]=a[i] sep $i} END{for(i=1;i<=NF;i++) print a[i]}' vis.degrib

will print

arruba,80,77,76
jamaica,70,75,77
 bermuda,90,88,83
 bahama,85,87,86
 keylargo,86,83,84
 montigo,89,85,86
 kokomo,83,77,84

you can again trim the space from the beginning of the labels as shown above.

Another approach without awk.

 tr ',' '\n' <vis.degrib | pr -4ts,

will generate the same

 arruba,80,77,76
 jamaica,70,75,77
  bermuda,90,88,83
  bahama,85,87,86
  keylargo,86,83,84
  montigo,89,85,86
  kokomo,83,77,84

4 is the number of rows in the original file.

Upvotes: 1

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