Extracting numbers from mixed string

I have filenames saved as strings such as '2001_06m'. Sometimes the files are inconsistently named as '2001_6m' (missing the zero before the 6) or '2001_06' (missing the m at the end). What code would I use to extract the non-zero integers after underscore in all cases (i.e. output = 6)?

And separately, what code would I use to extract the numbers before the underscore (usually they are 4 digits long, but sometimes 3 digits, i.e. '001' instead of '2001')?

Answer 1

[Jan]

s = '2001_06m';
d = sscanf(s, '%d_%d')

ans =
      2001
         6

Easier and faster than regexp.

[EDITED] If the input is a cell string:

C = {'2001_06m', '002_77q'};
S = sprintf('%s ', C{:});
S(S < '0' | S > '9') = ' ';  % Mask all non-numbers
Num = sscanf(S, '%d %d ', [2, Inf]);

[dpb]

That's true...excepting sscanf isn't vectorized (and I'd forgotten it will return the second value even though it finds the non-convertible character).

[Guillaume]

To be honest, none of the solutions are vectorised. Vectorising strsplit wouldn't be easy either. It wouldn't be too hard to vectorise the regexp solution, but sscanf is certainly more elegant.

Answer 2

[Guillaume]

A possible regexp version would be:

str2double(regexp(filename, '(\d+)_(\d+)', 'tokens', 'once'))

edit: following the discussion in Jan's answer, a vectorised version for when filenames is a cell array of char arrays or a string array:

tokens = regexp(filenames, '(\d+)_(\d+)', 'tokens', 'once');
str2double(vertcat(tokens{:}))

Note that the vertcat call will fail if a filename does not match the pattern.

Answer 3

[Stephen23]

Fully vectorized, one line, and more efficient than regexp and/or str2double:

>> C = {'2001_06m','2001_7m','2001_08'};
>> sscanf(sprintf('%sm',C{:}),'%*d_%d%*[m]') % second number
ans =
   6
   7
   8
>> sscanf(sprintf('%sm',C{:}),'%d_%*d%*[m]') % first number
ans =
   2001
   2001
   2001
>> sscanf(sprintf('%sm',C{:}),'%d_%d%*[m]') % both numbers
ans =
   2001
      6
   2001
      7
   2001
      8

"...usually they are 4 digits long, but sometimes 3 digits, i.e. '001' instead of '2001'"

My answer works regardless of the number of digits in the numbers.

Answer 4

[dpb]

v=str2double(strip(splitstr(s,'_'),'m'));      % chuckles...

In reality, write a regexp expression is the more general solution but I have to spend too much time figuring out the syntax and am too impatient... :)