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


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.


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
  • Going with this approach, with a small twist - sedding the commas to spaces then using wc -w to get the word count. – An̲̳̳drew Apr 29 '11 at 18:42

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)

Found this solution:

cat file.csv | sed 's/,/ /g' | wc -w

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.


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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