Finding the indexes of multiple substrings within a larger string.

I’m trying to find the indexes of all two digit pairs in a very long string of numbers, say “c”. I can easily find all occurrences of one string at a time; for example strfind(c, ’00’)…strfind (c, ’01’). But I want a way to do this for all sets one hundred sets; 00 to 99. I tried this:

x=0:99;
dig=sprintf('%02d ',x);
%converts the vector 0to99 into a string with two digits, space between numbers
dub_dig=strsplit(dig);
%splits each pair into cells
dub_dig_str=string(dub_dig);
%converts to a string

How do I get this sequence of strings (dub_dig_str) to work in something like a for loop using the strfind function? When I try this it crashes. I would like to output a matrix of indexes of where each pair occurs, for all pairs.

Thanks

Answer 1

[Stephen23]

idx = regexp(c,'\d\d')     %   no overlaps
idx = regexp(c,'\d(?=\d)') % with overlaps

[Walter Roberson]

A challenge with this is that regexp() will not consider overlaps -- '1234' will find '12' and '34' but not '23' . But the strfind() approach would find the overlaps.

[Steve]

I see, sorry I didn't specify. My goal is to find overlaps. Is there a way to get it work with strfind()?

[Stephen23]

"A challenge with this is that regexp() will not consider overlap"

That is incorrect. Overlaps are very easy too (I updated my answer too):

idx = regexp(c,'\d(?=\d)')

"Is there a way to get it work with strfind()?"

Sure, just call STRFIND 100 times in a loop. Or write STRFIND out 100 times.

Personally, I would not bother when REGEXP is much simpler and more efficient than calling STRFIND multiple times, or fiddling around with ISMEMBER, FIND, UNION, and lots of fiddly indexing.

[Walter Roberson]

oh good idea to use ?= in the context!

[Steve]

Thanks both for your help so far. I attempted playing around with this. My string c is all numbers (no spaces or other characters), so no surprise regexp(c,'\d(?=\d)') finds a pair of digits at all indexes. My goal is to output a separate row of indexes for each pair of numbers (one hundred total, 00to99), stating where each appears in c.

[Stephen23]

"My goal is to output a separate row of indexes for each pair of numbers (one hundred total, 00to99), stating where each appears in c."

Aaah, so you actually want to compare the pairs against another set with a specific order, which is what you were achieving with the loop. Here is an alternative approach:

c = char(randi(+'09',1,123)) % random data

c = '636906240866589674219474419874013401492319709264753858931901195001783553643124423974644494528171633587231581331511128165489'

% Character pairs:
[T,U] = meshgrid('0':'9'); % all pairs
P = cellstr([T(:),U(:)])  % all pairs

P = 100×1 cell array
    {'00'}
    {'01'}
    {'02'}
    {'03'}
    {'04'}
    {'05'}
    {'06'}
    {'07'}
    {'08'}
    {'09'}
    {'10'}
    {'11'}
    {'12'}
    {'13'}
    {'14'}
    {'15'}
    {'16'}
    {'17'}
    {'18'}
    {'19'}
    {'20'}
    {'21'}
    {'22'}
    {'23'}
    {'24'}
    {'25'}
    {'26'}
    {'27'}
    {'28'}
    {'29'}

Q = cellstr(c([1:end-1;2:end]).'); % data pairs
% Find indices of data pairs:
[~,X] = ismember(Q,P);
% Place indices into cell array:
Y = (1:numel(Q)).';
Z = accumarray(X,Y,[100,1],@(a){a})

Z = 100×1 cell array
    {[      64]}
    {4×1 double}
    {0×0 double}
    {0×0 double}
    {0×0 double}
    {0×0 double}
    {[       5]}
    {0×0 double}
    {[       9]}
    {[      44]}
    {0×0 double}
    {3×1 double}
    {2×1 double}
    {2×1 double}
    {[      36]}
    {2×1 double}

Checking the indices of '00' and some random pair:

Z{1}

ans = 64

Z{strcmp(P,'23')}

ans = 3×1

39 80 103

You can probably do something simiar with table operations. Lets try it now:

D = cell2table(Q, 'VariableNames',"Pair");
D.Index = (1:numel(Q)).';
G = groupsummary(D,"Pair",@(a){a})

G = 69×3 table
     Pair     GroupCount     fun1_Index 
    ______    __________    ____________

    {'00'}        1         {[      64]}
    {'01'}        4         {4×1 double}
    {'06'}        1         {[       5]}
    {'08'}        1         {[       9]}
    {'09'}        1         {[      44]}
    {'11'}        3         {3×1 double}
    {'12'}        2         {2×1 double}
    {'13'}        2         {2×1 double}
    {'14'}        1         {[      36]}
    {'15'}        2         {2×1 double}
    {'16'}        2         {2×1 double}
    {'17'}        2         {2×1 double}
    {'19'}        5         {5×1 double}
    {'21'}        1         {[      19]}
    {'23'}        3         {3×1 double}
    {'24'}        2         {2×1 double}

[Steve]

Thank you. This works. I must admit, as a beginner, some of the code looks cryptic (e.g., "@(a){a}", and the output of cells 'Z' is hard to work with mathematically, but I'm sure it's possible. I'm appreciating the tradeoffs between classic numerical functions and string approaches.

Answer 2

[Walter Roberson]

c = 'a91bb48353'

c = 'a91bb48353'

mask = ismember(c, '0':'9');
odd_pair = find(mask(1:2:end-1) & mask(2:2:end)) * 2 - 1

odd_pair = 1×2

7 9

even_pair = find(mask(2:2:end-1) & mask(3:2:end)) * 2

even_pair = 1×3

2 6 8

pair_starts_at = union(odd_pair, even_pair)

pair_starts_at = 1×5

2 6 7 8 9

[Walter Roberson]

c = char(randi([0 9], 1, 30) + '0')

c = '305452612469209463851343808968'

C = c - '0';
odds = C(1:2:end-1) * 10 + C(2:2:end);
evens = C(2:2:end-1) * 10 + C(3:2:end);
odd_idx = (1:numel(odds)) * 2 - 1;
even_idx = (1:numel(evens)) * 2;
indices = accumarray([odds(:); evens(:)] + 1, [odd_idx(:); even_idx(:)], [], @(locs){locs});

populated = find(~cellfun(@isempty, indices));
[num2cell(populated-1), indices(populated)]

ans = 27×2 cell array
    {[ 5]}    {[       2]}
    {[ 8]}    {[      26]}
    {[ 9]}    {[      14]}
    {[12]}    {[       8]}
    {[13]}    {[      21]}
    {[20]}    {[      13]}
    {[24]}    {[       9]}
    {[26]}    {[       6]}
    {[30]}    {[       1]}
    {[34]}    {[      22]}
    {[38]}    {2×1 double}
    {[43]}    {[      23]}
    {[45]}    {[       4]}
    {[46]}    {2×1 double}
    {[51]}    {[      20]}
    {[52]}    {[       5]}

[Steve]

Thank you Walter. This method worked for me as well. Cheers