You can solve this with a

1. FASTEST was the pl/perl method which I placed last on this list because it requires pl/perl, and is likely not needed for most workloads.

2. FAST A string function, such as one the pattern explained below

   length(str) - length(*replace(str, replaceStr))
     / length(replaceStr)
   

3. Something that converts from a string to an array.
4. SLOW Something that converts from a string to a table.

Possible solutions

String

Using length and regexp_replace

Most RDBMS's provide some method to calculate substring occurrences like this,

SELECT length(data) - length(replace(data, '2', ''))
  / length('2')
FROM foo;

This method isn't applicable here because without an anchor, we can't be sure if we're replacing the substring inside of something-space-delimited. As an example, the above replaces the 2 in 329. We can remedy that by using regexp_replace to anchor the substring.

SELECT length(data) - length(regexp_replace(data, '\m2\M', '', 'g'))
  / length('2')
FROM foo;

Because we're not splitting on simple spaces (' ') though we could with more complexity, we may also want to accommodate sub-strings of different lengths like in this question. This is why we explicitly include / length('2'). That reduces to a no-op, but if we we're search for something that was longer than one character, it'd be required.

SELECT length(data) - length(regexp_replace(data, '\m42\M', '', 'g'))
  / length('42')
FROM foo;

Using ARRAY[]

Splitting into an ARRAY[]

Here we have to subtract one match splits a string into two fragments, and thus the occurrences is one less than the fragment counts: this xyx split on y, produces {'x', 'x'} and we want the length to be 1 corresponding to the occurrences of y.

SELECT array_length(x, 1) - 1
FROM foo
CROSS JOIN LATERAL regexp_split_to_array(data, '\m2\M') AS t(x);
-- un-anchored version for reference.
-- CROSS JOIN LATERAL string_to_array(data, '2') AS t(x);

Alternatively, we can use string_to_array to separate something that's space-delimited and then count the matches,

SELECT id, array_length(array_positions(x, '2'), 1)
FROM foo
CROSS JOIN LATERAL string_to_array(data, ' ') AS t(x);

Using a TABLE

Splitting into a table with regexp_split_to_table

Here we split the regex into a table.. In this method you're using GROUP BY and count().

SELECT id, x
FROM foo
CROSS JOIN LATERAL regexp_split_to_table(data, ' ')
  AS t(x);

 id | x 
----+---
  1 | 1
  1 | 2
  1 | 2
  1 | 2
  1 | 2
  1 | 2
  1 | 2
(7 rows)

And, from there you can run regular SQL on it.

SELECT id, x, count(*)
FROM (
    SELECT id, x
    FROM foo
    CROSS JOIN LATERAL regexp_split_to_table(data, ' ')
      AS t(x)
) AS t(id,x)
GROUP BY id, x;

 id | x | count 
----+---+-------
  1 | 1 |     1
  1 | 2 |     6

Using regex_matches

Here we get away from splitting and instead use the \m, and \M anchors for word-boundaries.

SELECT count(*)
FROM foo
CROSS JOIN LATERAL regexp_matches(data, '\m2\M', 'g');

Procedural Languages

Perl

This method turned out to the be the fastest overall,

CREATE LANGUAGE plperl

CREATE FUNCTION count_occurances(inputStr text, regex text)
RETURNS smallint
AS $BODY$
  scalar @{[ $_[0] =~ m/$_[1]/g ]} 
$BODY$
LANGUAGE plperl
IMMUTABLE;

Summary and Performance Implications

Following the same format of data, performance implications can be obtained with

CREATE TABLE foo
AS
  SELECT
    1 AS id,
    array_to_string(
      ARRAY(SELECT trunc(random()*100+1)::int % 100 FROM generate_series(1,5000) AS t(x)),
      ' '
    ) AS data
;

Under these constraints, I found that the procedural method with plperl was the fastest. Next I found the following as being the fastest native method,

length(str) - regexp_replace(str, replacement, g)
  / length(replacement)

Keep in mind the tried and true method of string replacement outside of needing to anchor the string is still the fastest and most efficient native method however clunky it may be,

length(str) - replace(str, replacement)
  / length(replacement)

That said, the ARRAY[] method is massively faster than splitting to a table.