Extracting numbers from mixed string

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Al_G
Al_G 2018-10-7
编辑： Stephen23 2018-10-8
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')?

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采纳的回答

Jan
Jan 2018-10-7
编辑：Jan 2018-10-8
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]);
  2 个评论
dpb
dpb 2018-10-8
编辑：dpb 2018-10-8
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
Guillaume 2018-10-8
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.

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更多回答（3 个）

Guillaume
Guillaume 2018-10-7
编辑：Guillaume 2018-10-8
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.
Stephen23
Stephen23 2018-10-8
编辑：Stephen23 2018-10-8
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.
dpb
dpb 2018-10-7
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... :)

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