Replace Single Newlines – Better Methods Using Sed and Awk

awkregular expressionsedtext processing

I am in the habit of writing one line per sentence because I typically compile things to LaTeX, or am writing in some other format where line breaks get ignored. I use a blank line to indicate the start of a new paragraph.

Now, I have a file written in this style which I'd like to just send as plain text. I want to remove all the single linebreaks but leave the double linebreaks intact. This is what I've done:

sed 's/^$/NEWLINE/' file.txt | awk '{printf "%s ",$0}' | sed 's/NEWLINE/\n\n/g' > linebreakfile.txt

This replaces empty lines with some text I am confident doesn't appear in the file: NEWLINE and then it gets rid of all the line breaks with awk (I found that trick on some website) and then it replaces the NEWLINEs with the requisite two linebreaks.

This seems like a long winded way to do a pretty simple thing. Is there a simpler way? Also, if there were a way to replace multiple spaces (which sometimes creep in for some reason) with single spaces, that would be good too.

I use emacs, so if there's some emacs specific trick that's good, but I'd rather see a pure sed or pure awk version.

Best Answer

You can use awk like this:

$ awk ' /^$/ { print; } /./ { printf("%s ", $0); } ' test

Or if you need an extra newline at the end:

$ awk ' /^$/ { print; } /./ { printf("%s ", $0); } END { print ""; } ' test

Or if you want to separate the paragraphs by a newline:

$ awk ' /^$/ { print "\n"; } /./ { printf("%s ", $0); } END { print ""; } ' test

These awk commands make use of actions that are guarded by patterns:

/regex/

END

A following action is only executed if the pattern matches the current line.

And the ^$. characters have special meaning in regular expressions, where ^ matches the beginning of line, $ the end and . an arbitrary character.

Related Solutions

Sed Command – Extracting a Regex Match Without Printing Surrounding Characters

When a regexp contains groups, there may be more than one way to match a string against it: regexps with groups are ambiguous. For example, consider the regexp ^.*$[0-9][0-9]*$$ and the string a12. There are two possibilities:

Match a against .* and 2 against [0-9]*; 1 is matched by [0-9].
Match a1 against .* and the empty string against [0-9]*; 2 is matched by [0-9].

Sed, like all other regexp tools out there, applies the earliest longest match rule: it first tries to match the first variable-length portion against a string that's as long as possible. If it finds a way to match the rest of the string against the rest of the regexp, fine. Otherwise, sed tries the next longest match for the first variable-length portion and tries again.

Here, the match with the longest string first is a1 against .*, so the group only matches 2. If you want the group to start earlier, some regexp engines let you make the .* less greedy, but sed doesn't have such a feature. So you need to remove the ambiguity with some additional anchor. Specify that the leading .* cannot end with a digit, so that the first digit of the group is the first possible match.

If the group of digits cannot be at the beginning of the line:
```
sed -n 's/^.*[^0-9]$[0-9][0-9]*$.*/\1/p'
```
If the group of digits can be at the beginning of the line, and your sed supports the \? operator for optional parts:
```
sed -n 's/^$.*[^0-9]$\?$[0-9][0-9]*$.*/\1/p'
```
If the group of digits can be at the beginning of the line, sticking to standard regexp constructs:
```
sed -n -e 's/^.*[^0-9]$[0-9][0-9]*$.*/\1/p' -e t -e 's/^$[0-9][0-9]*$.*/\1/p'
```

By the way, it's that same earliest longest match rule that makes [0-9]* match the digits after the first one, rather than the subsequent .*.

Note that if there are multiple sequences of digits on a line, your program will always extract the last sequence of digits, again because of the earliest longest match rule applied to the initial .*. If you want to extract the first sequence of digits, you need to specify that what comes before is a sequence of non-digits.

sed -n 's/^[^0-9]*\([0-9][0-9]*\).*$/\1/p'

More generally, to extract the first match of a regexp, you need to compute the negation of that regexp. While this is always theoretically possible, the size of the negation grows exponentially with the size of the regexp you're negating, so this is often impractical.

Consider your other example:

sed -n 's/.*\(CONFIG_[a-zA-Z0-9_]*\).*/\1/p'

This example actually exhibits the same issue, but you don't see it on typical inputs. If you feed it hello CONFIG_FOO_CONFIG_BAR, then the command above prints out CONFIG_BAR, not CONFIG_FOO_CONFIG_BAR.

There's a way to print the first match with sed, but it's a little tricky:

sed -n -e 's/\(CONFIG_[a-zA-Z0-9_]*\).*/\n\1/' -e T -e 's/^.*\n//' -e p

(Assuming your sed supports \n to mean a newline in the s replacement text.) This works because sed looks for the earliest match of the regexp, and we don't try to match what precedes the CONFIG_… bit. Since there is no newline inside the line, we can use it as a temporary marker. The T command says to give up if the preceding s command didn't match.

When you can't figure out how to do something in sed, turn to awk. The following command prints the earliest longest match of a regexp:

awk 'match($0, /[0-9]+/) {print substr($0, RSTART, RLENGTH)}'

And if you feel like keeping it simple, use Perl.

perl -l -ne '/[0-9]+/ && print $&'       # first match
perl -l -ne '/^.*([0-9]+)/ && print $1'  # last match

Sed – How to Remove Multiple Blank Lines from a File

Case 1:

awk '!NF {if (++n <= 2) print; next}; {n=0;print}'

Case 2:

awk '!NF {s = s $0 "\n"; n++; next}
     {if (n>1) printf "%s", s; n=0; s=""; print}
     END {if (n>1) printf "%s", s}'

Best Answer

Related Solutions

Sed Command – Extracting a Regex Match Without Printing Surrounding Characters

Sed – How to Remove Multiple Blank Lines from a File

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