How to use sed and wc command to handle whitespace

Question

If I have a CSV file and I want to know the number of columns, I'll use the following command:

  head -1 CSVFile.csv | sed 's/,/\t/g' | wc -w

However, whenever each column has a column name with a space in it, the command doesn't work and gives me a nonsense figure.

What would be the way to edit this command such that it gives me the correct number of columns?

e.g. in my file I could have column name (t - ZK) or (e - 22)

For example my file could be (first 2 row);

 ZZ(v - 1),Tat(t - 1000)
 1.1240128401924,2929292929

Answer 1

Maybe use the last line instead of the first. Change "head" to "tail". That would be a quick, easy solution.

Answer 2

You are piping the sed output to wc -w which would return the number of words in the output. So if a field header contains spaces, those would be considered as different words.

You can use awk:

head -1 CSVFile.csv | awk -F, '{print NF}'

This would return the number of columns in the file (assuming the file is comma-delimited).

Answer 3

Try using awk

awk -F, 'NR==1 {print NF; exit}' CSVFile.csv

---

If you wish to use chain of head, sed and wc

Try using sed replace deliminator as newline \n instead of tab \t and then count number of lines using wc -l instead of counting number of words with wc -w

head -1 CSVFile.csv | sed 's/,/\n/g' | wc -l

Answer 4

perl -ane 'print scalar(@F)-1 if($.==1)' your_file

Answer 5

Assuming there is no "," in header name (like field1,"Surname,name",field3, ...)

sed "1 s/[^,]//g;q" CSVFile.csv | wc -c

Could also be made only in sed but a bit heavy for counting.