group an array according a condition

Question

eric capnu 2017-8-9

1
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https://ww2.mathworks.cn/matlabcentral/answers/352178-group-an-array-according-a-condition

评论： eric capnu 2017-9-1

采纳的回答： Andrei Bobrov

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Hi everyone

i have an array like this

A= 
 10    10.1  12   14   110
 10.1  11    12   15   145
 12    12    12   24   223
 12    11.4  11   23   100
 10    12    20   18   211
 9     23    11   11   98

I want to create a new B array where i get the maximum value of each column according to the first column value

B=
 10.1   11   12    15    145
 12     12   20    24    223
 9      23   11    11    98

i don´t want to use loops. is there a way to do it by using just matlab functions?

thanks in advance

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Answer 1

Andrei Bobrov 2017-8-9

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https://ww2.mathworks.cn/matlabcentral/answers/352178-group-an-array-according-a-condition#answer_277446

编辑：Andrei Bobrov 2017-8-10

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B_table = varfun(@max,array2table(A),'GroupingVariables','A1')

or

B = splitapply(@(x)max(x,[],1),A,findgroups(A(:,1)));

or

g = findgroups(A(:,1)); % or >> [~,~,g] = unique(A(:,1))
[ii,jj] = ndgrid(g,1:size(A,2));
B = accumarray([ii(:),jj(:)],A(:),[],@max);

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John BG 2017-8-10

编辑：John BG 2017-8-10

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and then back to array

    B=table2array(Btable)
    B =
        2.0000    2.0000   10.1000   11.0000   12.0000   15.0000  145.0000
        3.0000    3.0000   12.0000   12.0000   20.0000   24.0000  223.0000
        4.0000    1.0000    9.0000   23.0000   11.0000   11.0000   98.0000

no need for the frequencies

B(:,2)=[]
    B =
        2.0000   10.1000   11.0000   12.0000   15.0000  145.0000
        3.0000   12.0000   12.0000   20.0000   24.0000  223.0000
        4.0000    9.0000   23.0000   11.0000   11.0000   98.0000

eric capnu 2017-9-1

thanks mates. that was very useful... and, what if instead of @max it would be the last element or the first element of the range?

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Answer 2

Stephen23 2017-8-9

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https://ww2.mathworks.cn/matlabcentral/answers/352178-group-an-array-according-a-condition#answer_277437

编辑：Stephen23 2017-8-9

在 MATLAB Online 中打开

You could use accumarray:

A = [
   2   10    10.1  12   14   110
   2   10.1  11    12   15   145
   3   12    12    12   24   223
   3   12    11.4  11   23   100
   3   10    12    20   18   211
   4   9     23    11   11   98];
%
[~,~,X] = unique(A(:,1));
V = 1:size(A,1);
fun = @(r){max(A(r,:),[],1)};
C = accumarray(X,V(:),[],fun);
Z = vertcat(C{:})

produces this:

Z =
0000    10.1000    11.0000    12.0000    15.0000   145.0000
0000    12.0000    12.0000    20.0000    24.0000   223.0000
0000     9.0000    23.0000    11.0000    11.0000    98.0000