Wrap yaml data in single quotes - regex

Question

I would like to wrap all of my YAML data (in a large file) with single quotes. I tried sed, but it did not work:
sed "s/\(.*: \)\(.*\)/\1'\2'/" <data.yml >datanew.yml
This took lines like this:
location_id: 25
street:
text: This is text: it contains colons
And produced lines like:
' location_id: '25
' street: '
' text: This is text: 'it contains colons
... but I would like them to look like:
location_id: '25'
street: ''
text: 'This is text: it contains colons'
Is this possible in sed (or awk or perl or ...)? From my research, it seems like sed may have trouble picking up the first colon since it matches greedy. I am running Ubuntu 14.04.
Additional Information
Note YAML has optional leading whitespace, a token followed by a colon and everything else on the line (which could include one or more additional colons) all of which needs to be wrapped in quotes.
You can test with the above three lines.
More
Thank you all for your suggestions. I am assuming most of them actually work, but not for me. Here is a snapshot from my terminal using one of the suggested patterns. Unfortunately they all fail for me in roughly the same way.
Even more frustrating, when I open the file in vim and run search and replace with that same pattern, it works perfectly. I tried to use that technique for my whole file, but vim wasn't pleased with the 4M lines.
Is my sed somehow broken??

Answer 1

This regex:
^\s*([^:]+)(:\s)(.*?)\s*$
Does what you want. Working Demo
It is easiest to express in Perl.
Given:
$ echo "$tgt"
location_id: 25
street:
text: This is text: it contains colons
In Perl:
$ echo "$tgt" | perl -lne "print if s/^\s*([^:]+)(:\s)(.*?)\s*$/\1\2'\3'/"
location_id: '25'
street: ''
text: 'This is text: it contains colons'

Answer 2

Little change in your sed
sed "s/\([^:]*: \)\(.*\)/\1'\2'/" <data.yml >datanew.yml

Answer 3

Here is an awk you can use:
cat file
Some other data
location_id: 25
street:
awk -v f="'" -F": *" 'NF==2 {$NF=f $NF f}1' file
Some other data
location_id '25'
street ''
It test if line has :, and if then it wraps ' around last filed, empty or not.

Answer 4

The following seems to work for the test cases you provided, as well as some cases I came up with:
sed "s/\([^:]*:\s*\)\(.*\)/\1'\2'/g"
The way it works is by doing a non-greedy match for text up until a colon, then followed by a colon and optional whitespace with [^:]*:\s* . All of that is put into a capture group. The reason I need the "then followed by a colon" is because some lines such as "%YAML 1.1" which appear in the file would match the regex, even though they should not be included. By adding the additional constraint that there be a colon, such lines are excluded from replacement.
The next part is relatively easy, just match any text following the previous capture group. This an be achieved with .* (which also included colons, as you mentioned above in your question).
The sed s command is used to replace the regex matched with the first capture group,\1, which is all the text up until the first colon and optional whitespace, followed by the 2nd capture group \2 which is all the text after the colon and whitespace, in single quotes.
Here's a demo of it:
regex test