how to split a string

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Mitchie Teotico
Mitchie Teotico 2020-3-28
回答： DGM 2022-5-22
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 个评论
Walter Roberson
Walter Roberson 2020-3-28
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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回答（4 个）

Walter Roberson
Walter Roberson 2020-3-28
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 个评论
Mitchie Teotico
Mitchie Teotico 2020-3-29
could you show an example
Walter Roberson
Walter Roberson 2020-3-29
ThisRow = RemainingLetters(1:4);
RemainingLetters = RemainingLetters(5:end);

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Nik Niki
Nik Niki 2022-5-22
编辑：Image Analyst 2022-5-22
Ran in:
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 2022-5-22
Ran in:
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 2022-5-22
Ran in:
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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