how to split a string

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Mitchie Teotico
Mitchie Teotico on 28 Mar 2020
Answered: DGM on 22 May 2022
i need to create a function named breakline which inputs a string (char, 1 × N) and a linewidth (integer, scalar). The function then splits the string into smaller strings which are smaller than the linewidth.
The output to the function is an array which contains all of the split up lines (cell array, 1 × M).
for example
out=breakupLine(123456789,4)
out =
{
[1,1] = 1234
[1,2] = 5678
[1,3] = 9
}
  1 Comment
Walter Roberson
Walter Roberson on 28 Mar 2020
It is not possible to get the output in exactly that format. The initial part showing the index can only appear if you create the output as all one character vector, contradicting the requirement that it be a cell array.

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Answers (4)

Walter Roberson
Walter Roberson on 28 Mar 2020
If the current vector is shorter than N or exactly N then add the complete current vector to the end of the cell array and return. Otherwise index the first N characters from the current vector and add them to the end of the cell array, and remove N characters from the beginning of the current vector overwriting the current vector and loop back.
  2 Comments
Walter Roberson
Walter Roberson on 29 Mar 2020
ThisRow = RemainingLetters(1:4);
RemainingLetters = RemainingLetters(5:end);

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Nik Niki
Nik Niki on 22 May 2022
Edited: Image Analyst on 22 May 2022
test_data_name="kjashfk, kjasdklasdfas, 3";
test_data_name = split(test_data_name)
test_data_name = 3×1 string array
"kjashfk," "kjasdklasdfas," "3"

Image Analyst
Image Analyst on 22 May 2022
Here is one way:
out=breakupLine('123456789', 4)
out = 1×3 cell array
{'1234'} {'5678'} {'9'}
%=====================================================
function out=breakupLine(str, substringLength)
stringLength = length(str);
loopCounter = 1;
for k = 1 : substringLength : stringLength
index1 = k;
index2 = min(k + substringLength - 1, stringLength);
out{loopCounter} = str(index1 : index2);
loopCounter = loopCounter + 1;
end
end

DGM
DGM on 22 May 2022
To address the OP's particular request:
linew = 4;
teststr = '123456789';
excess = rem(numel(teststr),linew);
output = reshape(teststr(1:end-excess),linew,[]).';
output = [num2cell(output,2); teststr(end-excess+1:end)]
output = 3×1 cell array
{'1234'} {'5678'} {'9' }
I don't know that this is particularly efficient, but it works. It may have advantages at some scale, but I haven't tested that.

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