How can I remove blank spaces in a file that are marked with "." without removing the decimal place in my numbers? (Bash shell)

Question

I have a file that looks like this:

18 29 293.434
12 32 9954.343 
12 343 .
12 45 9493.545

I want to replace the "." on line 3 with "0" using sed.

I can't figure out how to do this without replacing every decimal place in the file.

Any help would be much appreciated.

Can you post expected output ? Seems a mismatch between the title and description... — Gilles Quénot, Jul 23 '20 at 00:28

score 4 · Answer 1 · answered Jul 23 '20 at 00:47

4

With GNU sed:

sed 's/\B\.\B/0/g' file

Output:

18 29 293.434
12 32 9954.343 
12 343 0
12 45 9493.545

See: \B: non-word boundary

answered Jul 23 '20 at 00:47

Cyrus

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For those wanting a posix compliant solution, perhaps `sed -E 's/(^|[[:space:]])[.]([[:space:]]|$)/\10\2/g' file`? – potong Jul 23 '20 at 06:50

score 1 · Answer 2 · answered Jul 23 '20 at 04:12

1

Using any awk in any shell on every UNIX box and assuming you want to replace . with 0 wherever it occurs alone in any field rather than just if it's in the last field of the 3rd line of input:

awk '{for (i=1; i<=NF; i++) if ($i == ".") $i=0} 1' file

answered Jul 23 '20 at 04:12

Ed Morton

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score 0 · Answer 3 · answered Jul 23 '20 at 00:32

0

If I understand well, if the dot is at the end, remove spaces and substitute this dot with 0:

sed '/\.$/{s/ //g;s/\.$/0/}' file
#       ^

$ is an anchor, it means at the end of the line

18 29 293.434
12 32 9954.343
123430
12 45 9493.545

answered Jul 23 '20 at 00:32

Gilles Quénot

173,512
41
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How can I remove blank spaces in a file that are marked with "." without removing the decimal place in my numbers? (Bash shell)

3 Answers3