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An̲̳̳drew
Asked: 2011-04-30 10:12:33 +0800 CST

Counting number of items in a single-row CSV file?

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I have a file containing a long comma-delimited list of numbers, like this:

2,8,42,75,101

What's the simplest command (from a Unix shell) to get the count of numbers in this file? In the example above, that would be 5.

scripting csv unix-shell

5 Answers

  1. Best Answer

    Easiest is probably to just count the commas:

    sed 's/[^,]//g' yourfile.csv | wc -c

    Normally you'd add one to get the number of elements, but if there's a newline there it's counted too. Convenient in this case, I guess.

    Also with awk:

    awk -F, '{print NF}' yourfile.csv
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  2. as per here: http://www.cyberciti.biz/faq/finding-bash-shell-array-length-elements/

    this isnt too difficult

    if you can define your list of elements in an array like so:

    ArrayName=("element 1" "element 2" "element 3")

    then its as simple as:

    echo ${#ArrayName[@]}

    however you have a csv so this may be better: http://www.thelinuxblog.com/working-with-csv-files-in-bash/

    echo $(cat file.csv | sed ‘s/, /_ /g’ | tr ‘,’ ‘\n’ | wc -l)
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  3. Found this solution:

    cat file.csv | sed 's/,/ /g' | wc -w
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  4. If you wanted to use python you could do:

    str = given_csv_line_as_string
if str[-1] == ',': str = str[:-1]
print len(given_csv_line_as_string.split(','))

    Note that this will return the correct length regardless if there is a hanging comma.

    I'm sure there is something similar and more sysadmin-y in perl but I don't use perl.

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  5. Or use grep

    grep -o "," | wc -l

    Actually to get this correct you would need to add one to the result

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