# Finding the indexes of multiple substrings within a larger string.

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Steve il 24 Mar 2023
Commentato: Steve il 1 Apr 2023
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
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### Risposta accettata

Stephen23 il 24 Mar 2023
Modificato: Stephen23 il 24 Mar 2023
idx = regexp(c,'\d\d') % no overlaps
idx = regexp(c,'\d(?=\d)') % with overlaps
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Stephen23 il 26 Mar 2023
Modificato: Stephen23 il 27 Mar 2023
"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 il 1 Apr 2023
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.

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### Più risposte (1)

Walter Roberson il 24 Mar 2023
c = 'a91bb48353'
c = 'a91bb48353'
odd_pair = 1×2
7 9
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
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Walter Roberson il 26 Mar 2023
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 il 1 Apr 2023
Thank you Walter. This method worked for me as well. Cheers

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