Linux – cut the first and last element in CSV

awkcsvlinuxsedtext processing

I have an array in a CSV file as

input.csv
"{1,2,3,4}"
"{1,2,3,4,5,6,7,8,9}"
"{34,6,7,89}"

I want to get the first and the last element of this CSV file as another CSV file

output.csv
    1,4
    1,9
    34,89

I tried with

  cut -d , -f1 -- complement input.csv > output.csv

I know this works for normal CSV.But here I have curly braces and ""
too.

Best Answer

I would do it this way with sed

$ sed -r 's/"\{([0-9]+,).*,([0-9]+)\}"/\1\2/' input
1,4
1,9
34,89

Notes

-r use ERE
\} literal {
([0-9]+,) save some digits followed by a comma for later
\1\2 back reference to saved patterns

(your output is indented, - so maybe you want / \1\2/ or /\t\1\2/ in the replacement - adjust as you like)

Related Solutions

Remove Columns from a CSV File

To remove the fourth column,

$ cut -d, -f4 --complement example.csv > input.csv

Adjust the -f option to match the column number.

If the CSV file is more complicated, you could use some perl and the Text::CSV package,

$ perl -MText::CSV -E '$csv = Text::CSV->new({binary=>1}); 
  while ($row = $csv->getline(STDIN)) 
  {
    print "$row->[0],$row->[1],$row->[2]\n"
  }' < example.csv > input.csv

Shell – Swap first and second columns in a CSV file

Here you go:

awk -F, '{ print $2 "," $1 }' sampleData.csv

Best Answer

Notes

Related Solutions

Remove Columns from a CSV File

Shell – Swap first and second columns in a CSV file

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