Find and Replace with Spaces using Sed Mac Terminal

Quoted from here:

For POSIX compliance, use the character class [[:space:]] instead of \s, since the latter is a GNU sed extension.

Based on that, I would suggest the following, which, as Jonathan Leffler pointed out, is portable across GNU and BSD implementations.

sed -E 's/[[:space:]]?"[[:space:]]?//g' <path/to/file> 

The -E flag enables extended regular expressions on BSD implementations. On GNU sed it is undocumented, but as discussed here, it enables compatibility with the BSD standard.

Quoted from the manual for BSD sed:

-E Interpret regular expressions as extended (modern) regular expressions rather than basic regular expressions (BRE's).

Applying the above command on a file containing the following single line

"DISH Hartford & New Haven (Hartford)", "206", "FBNHD", " 06028", " East Windsor Hill", "CT", "Hartford County"

it yields

DISH Hartford & New Haven (Hartford),206,FBNHD,06028,East Windsor Hill,CT,Hartford County

This works for me. Is this what you want ?

 sed -e 's|", "|,|g' -e 's|^"||g' -e 's|"$||g' file.csv echo '"DISH Hartford & New Haven (Hartford)", "206", "FBNHD", " 06028", " East Windsor Hill", "CT", "Hartford County"' | sed -e 's|", "|,|g' -e 's|^"||g' -e 's|"$||g' DISH Hartford & New Haven (Hartford),206,FBNHD, 06028, East Windsor Hill,CT,Hartford County 

This should do it:

sed -i 's/\(\s\|\)"\(\|\s\)//g' bs-B2Bformat.csv 

One way is to use python and its csv module:

import csv import sys ## Open file provided as argument. with open(sys.argv[1], 'r') as f: ## Create the csv reader and writer. Avoid to quote fields in output. reader = csv.reader(f, skipinitialspace=True) writer = csv.writer(sys.stdout, quoting=csv.QUOTE_NONE, escapechar='\\') ## Read file line by line, remove leading and trailing white spaces and ## print. for row in reader: row = [field.strip() for field in row] writer.writerow(row) 

Run it like:

python3 script.py csvfile 

That yields:

DISH Hartford & New Haven (Hartford),206,FBNHD,06028,East Windsor Hill,CT,Hartford County 

What all of the current answers seemed to miss:

$ cat bs-B2Bformat.csv | sed s/ " //g sed: 1: "s/": unterminated substitute pattern $ cat bs-B2Bformat.csv |sed s/ \" //g sed: 1: "s/": unterminated substitute pattern $ 

The problem in the above is missing single quotes. It should have been:

$ cat bs-B2Bformat.csv | sed 's/ " //g' ^ ^ 

Without the single quotes, bash splits at the spaces and sends three separate arguments (well at least for the case of \"). sed was seeing its first argument as just s/.

Edit: FYI, single quotes are not required, they just make this case easier. If you want to use double quotes, just escape the one you want to keep for matching:

$ cat bs-B2Bformat.csv | sed "s/ \" //g"