use warnings;
use strict;

my $str = "It is very  very cold now";

$str =~ s/(?<! ) (?! )//g;
print $str;
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Prints Itisvery  verycoldnow which I doubt is what you want, but is what you asked for.

or you could use $str =~ s/\b \b//g;

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Perl is Huffman encoded by design.

or you could use $str =~ s/\b \b//g;

...only if all the *non-space* characters are *word* characters:

my $str = '@@@@ $$$ *** )))';
$str =~ s/\b \b//g;
print $str;
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Thanks GrandFather. That almost worked 100%. I have some strings where the is no clear word boundary. Specifically, something like this:

The boy to [ld his fat, her
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In that example, \b is failing to remove the space after to, and before '[', as well as the space after ',', and before her. Would I need to use \B as well?

Update: Ok the question was cleared up while I was writing this. Disreagard this post, the previous answers take care of it.

First, you want to be using s/// for regexes, not tr///.

$row_to_process =~ s/\s//g;

# If you wanted to use tr/// (probably a bit faster here)
$row_to_process =~ tr/ \n\t\r//d;
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Will remove all whitespace from a string. After this, your question is a little confusing. You talk about removing only single spaces. Why would the space between "It" and "is" and the space "cold" and "now" not be removed.

Probably not very efficient, but the best I can come up with at short notice:

Update:See Roy Johnson's comment above on Trix606's similar snippet. Code fixed for single spaces at start and end of string:

what does the $1$2 mean? Doesn't the $ regular expression designate the end of the string?

In the regular expression, yes it does. In the replacement expression $n is the n'th capture string in the regular expression. So for s/(x) (y)/$1$2)/ 'x y' gets replaced with 'xy' for example.

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Perl is Huffman encoded by design.

s/\  (?!\ )//gx ;
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It will remove the last space from any run of spaces. That is why you need both lookahead and lookbehind to make it work properly.

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Caution: Contents may have been coded under pressure.

This solution worked perfectly. Now, do you think you could break it down for me? Forgive me, but I'm having a little trouble understanding it.

Did it though? Consider:

use warnings;
use strict;

my $str1= "It is very  very cold now\n";
my $str2 = $str1;

$str1 =~ s/(?<! ) (?! )//g;
print $str1;

$str2 =~ s/ (?! )//g;
print $str2;
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which prints:

Itisvery  verycoldnow
Itisvery verycoldnow
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Which string is correct? The (?<! ) is a zero width look back negative assertion: it matches if there isn't a space before the current space being matched. The (?! ) is a zero width look ahead negative assertion: it matches if there isn't a space after the current space being matched. Together these ensure that only single spaces are matched: only match a space that doesn't have a space before it and doesn't have a space after it.

You should have a look at perlretut.

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Perl is Huffman encoded by design.