Parse cell array into separate cells, with no delimiter

Hello! I am trying to parse the name of a file into separate variables to describe the file, but there are no delimiters in the file name. For example, if my file name is

fileName = {'11171401'}

I want to divide this into:

track = {'1117'}
cycle = {'14'}
segment = {'01'}

I've already chopped off the beginning of the file name using the split function, but I can't seem to figure out how to divide up the rest without a delimiter. Thank you in advance for any help!

[James Tursa]

What are the rules for the name splitting? If it is always 4-2-2 character splits then you could just use indexing.

[Gabrielle Trudeau]

James Tursa, It is a 4-2-2 character split, but I'm not sure how to get around the fact that fileName is saved as a 1x1 cell.

[Gabrielle Trudeau]

I found a solution using 'cell2mat' and 'str2num'!

fileName = {'11171401'};
X = cell2mat(fileName);
track = str2num(X(1:4));
cycle = str2num(X(5:6));
segement = str2num(X(7:8));

[per isakson]

In your solution, track, cycle and segment are all double, not cells containing character vectors as required. Try

fileName = {'11171401'};
track = {fileName{1}(1:4)}
cycle = {fileName{1}(5:6)}
segment = {fileName{1}(7:8)}

[Gabrielle Trudeau]

Thank you!

[Stephen23]

fnm = {'11171401'; '22282512'; '33393623'};
tkn = regexp(fnm,'^(\d{4})(\d\d)(\d\d)$','tokens','once');
tkn = vertcat(tkn{:})

tkn = 3×3 cell array
    {'1117'}    {'14'}    {'01'}
    {'2228'}    {'25'}    {'12'}
    {'3339'}    {'36'}    {'23'}

Answer 1

[Voss]

fileName = {'11171401'; '22282512'; '33393623'};

[track,cycle,segment] = cellfun(@(x)parse_file_name(x),fileName,'UniformOutput',false)

track = 3×1 cell array
    {'1117'}
    {'2228'}
    {'3339'}

cycle = 3×1 cell array
    {'14'}
    {'25'}
    {'36'}

segment = 3×1 cell array
    {'01'}
    {'12'}
    {'23'}

function [t,c,s] = parse_file_name(fn)
t = fn(1:4);
c = fn([5 6]);
s = fn([7 8]);
end

[Gabrielle Trudeau]

Thank you! I was trying to avoid having to write a function, but I'm wondering if this is more efficient (less computationally expensive) than the way I did it...

[Voss]

You're welcome!

The way you did it works only for a scalar cell array fileName. Here's testing it on a non-scalar cell array:

fileName = {'11171401'; '22282512'; '33393623'};
X = cell2mat(fileName)

X = 3×8 char array
    '11171401'
    '22282512'
    '33393623'

% indexing goes down the first column first, giving incorrect results
track = str2num(X(1:4))

track = 1231

cycle = str2num(X(5:6))

cycle = 23

segment = str2num(X(7:8))

segment = 12

Also, your way gives numeric results, but my answer gives cell arrays of character vectors, as specified in the question:

track = {'1117'}
cycle = {'14'}
segment = {'01'}

So since the two methods are really doing completely different things, it's not meaningful to try to compare their relative efficiency.