replace long string of variables

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Cristian Martin 2022-6-1

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1731855-replace-long-string-of-variables

评论： Cristian Martin 2022-6-1

Hi,

I have a script like below. The problem is that I must to create until index50, how can I write the script more easier and faster?

index1= strcmp(a(:,1), 'mama') & strcmp(a(:,3), '1');

val1 = unique(index1);

val_unica1 = nnz(val1);

index2= strcmp(a(:,1), 'mama') & strcmp(a(:,3), '2');

val2 = unique(index2);

val_unica2 = nnz(val2);

index3= strcmp(a(:,1), 'mama') & strcmp(a(:,3), '3');

val3 = unique(index3);

val_unica3 = nnz(val3);

total= sum(val_unica1) + sum(val_unica2) + sum(val_unica3);

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Stephen23 2022-6-1

编辑：Stephen23 2022-6-1

在 MATLAB Online 中打开

Very interesting code. Because STRCMP returns a boolean array:

index1= strcmp(a(:,1), 'mama') & strcmp(a(:,3), '1');

this line:

val1 = unique(index1);

will return one of [], FALSE, TRUE, or [FALSE,TRUE]. You then count how many times TRUE occurs:

val_unica1 = nnz(val1);

which will return either 0 or 1. It would be simpler and more efficient to use ANY.

Cristian Martin 2022-6-1

编辑：Cristian Martin 2022-6-1

Yes, because I want for the below matrix to set an edit text for counting a sum of unique entry for mama 1, mama2, mama3 and so on

a = {'mama', '5', '1';

'mama', '4', '1';

'mama', '6', '1';

'mama', '7', '2';

'mama', '7', '1';

'mama', '7', '3';

'mama', '8', '1'};

disp(total);

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Answer 1

Steven Lord 2022-6-1

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1731855-replace-long-string-of-variables#answer_976225

在 MATLAB Online 中打开

a = {'mama', '5', '1';
    'mama', '4', '1';
    'mama', '6', '1';
    'mama', '7', '2';
    'mama', '7', '2';
    'mama', '7', '1';
    'mama', '7', '3';
    'mama', '7', '3';
    'mama', '8', '1'};
rowsWithMama = strcmp(a(:, 1), 'mama')
rowsWithMama = 9×1 logical array
   1
   1
   1
   1
   1
   1
   1
   1
   1
rowsWith1 = strcmp(a(:, 3), '1')
rowsWith1 = 9×1 logical array
   1
   1
   1
   0
   0
   1
   0
   0
   1
nnz(rowsWithMama & rowsWith1)
ans = 5

If you have multiple values you want to test for in column 3 consider using ismember.

rowsWith2Or3 = ismember(a(:, 3), {'2', '3'})
rowsWith2Or3 = 9×1 logical array
   0
   0
   0
   1
   1
   0
   1
   1
   0

or, using string to make creation of the list of search terms easier by avoiding needing to manually create '2', '3', '4', ... '50':

s = string(2:3)
s = 1×2 string array
    "2"    "3"
rowsWith2Or3 = ismember(a(:, 3), s)
rowsWith2Or3 = 9×1 logical array
   0
   0
   0
   1
   1
   0
   1
   1
   0