How to split a string of digits into groups of three from right-to-left using only regular expressions?

While I have been able to accomplish this task using a varitey of mixed functions such as regexp() and fliplr(), I am ultimately stumped when trying to resort to only regexp() and/or regexprep(). For example, I am trying to convert the a string similar to the following:

str = '12345678';

such that the resulting output is:

{'12'} {'345'} {'678'}

I am also trying to accomplish this task without the use of loops of any sort.

[Walter Roberson]

Is it permitted to use more than one call to regexp() or regexprep(), or does it need to be just a single call to one or the other?

Is the "execute" group of regexprep() to be permitted?

[Allen]

Single call was preferred, but multiple are acceptable. Was hoping to stay away from the execute groups since that technically uses other functions. However, since I have no experience with trying to generate code using them, I would not have minded seeing a few examples. ;-)

Answer 1

[Stephen23]

With one regexp call (uses a lookaround assertion):

>> str = '12345678';
>> regexp(str, '(^\d{1,2}(?=(\d{3})*$)|\d{3})','match')
ans = 
    '12'    '345'    '678'

With two regexp calls (splits string into two tokens, may have empty cells in the output):

>> tkn = regexp(str,'^(\d{0,2})((\d{3})*)$','tokens','once');
>> out = [tkn(1),regexp(tkn{2},'\d{3}','match')]
out = 
    '12'    '345'    '678'

or with some extra functions to define the regular expression itself:

>> rgx = ['(\d{0,2})',repmat('(\d{3})',1,fix(numel(str)/3))];
>> regexp(str,rgx,'tokens','once')
ans = 
    '12'    '345'    '678'

[Walter Roberson]

I would say that using repmat and fix and so on, would qualify as using a variety of mixed functions, which the user specifically wishes to avoid doing.

[Walter Roberson]

The single-regexp version does not work on '123456' and will produce {'12'} {'456'} instead.

[Stephen23]

"The single-regexp version does not work on '123456' and will produce {'12'} {'456'} instead."

>> regexp('123456', '(^\d{1,2}(?=(\d{3})*$)|\d{3})','match')
ans = 
    '123'    '456'

[Walter Roberson]

You fixed your code after I posted. Before you had

regexp(str,'(^\d{0,2}|\d{3}(?=$|\d{3}))','match')

[Stephen23]

@Walter Roberson: any comments on the corrected code? As far as I can tell, it should work (assuming only digits in the string).

Just out of curiosity, how were you planning to use dynamic regular expressions?

[Walter Roberson]

I was thinking vaguely about recursion ;-)

[Allen]

The first two work perfectly under the limitations. I have been trying to solve that one for quite some time now, but had not been successful with implementing a lookaround assertion to capture more than the first token. Thanks to you both.