Finding min and max values in a structure

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L'O.G.
L'O.G. 2022 年 11 月 28 日
コメント済み: Voss 2022 年 11 月 28 日
For a 1x1 structure with multiple fields, each field consisting of a cell of numbers, how can you find the minimum and maximum values over the entire structure?

回答 (3 件)

David Hill
David Hill 2022 年 11 月 28 日
d=struct2cell(yourStruct);
m=0;M=0;
for k=1:length(d)
m=min([m,d{k}]);
M=max([M,d{k}]);
end
  1 件のコメント
L'O.G.
L'O.G. 2022 年 11 月 28 日
I get an error for both lines in the loop:
`Invalid data type. First argument must be numeric or logical.`

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Matt J
Matt J 2022 年 11 月 28 日
編集済み: Matt J 2022 年 11 月 28 日
each field consisting of a cell of numbers
If you really do mean a cell array, that seems like an inadvisable organization of the data. You should probably have each field be a numeric matrix.
yourStruct.a=[1,2];
yourStruct.b=[3 4;
5 6];
Max = max( structfun(@(z)max(z,[],'all'), yourStruct) )
Max = 6
Min = min( structfun(@(z)min(z,[],'all'), yourStruct) )
Min = 1

Voss
Voss 2022 年 11 月 28 日
編集済み: Voss 2022 年 11 月 28 日
"a 1x1 structure with multiple fields, each field consisting of a cell of numbers"
yourStruct.a={1,2};
yourStruct.b={3 4;
5 6};
Max = max( structfun(@(z)max([z{:}],[],'all'), yourStruct) )
Max = 6
Min = min( structfun(@(z)min([z{:}],[],'all'), yourStruct) )
Min = 1
Another way:
d = struct2cell(yourStruct);
m = NaN;
M = NaN;
for k = 1:numel(d)
m = min([m,d{k}{:}]);
M = max([M,d{k}{:}]);
end
disp(M);
6
disp(m);
1
  1 件のコメント
Voss
Voss 2022 年 11 月 28 日
@Sean de Wolski: This answer is not a copy of the other answers. This answer uses the methods proposed by the other answers, adapted to work for a structure of cell arrays of numbers (as opposed to a structure of numeric arrays, which the other answers work for), which is the description given in the question, and which neither other answer works for.

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