set matrix m x n
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I am trying to create a matrix of 200 (rows by 6 (columns). I input by variable, but get a single row with all variable in multiple columns. The objective is to get the max of each row from the 6 columns. Can someone help me out?
h=height(PFset);
for x=1:height(PFset)
A= [XfmrpfCH XfmrpfCT XfmrpfCL XfmrpfCHLUST XfmrpfCHTUST XfmrpfCLTUST];
end
2 件のコメント
Rik
2017 年 12 月 1 日
What are the sizes of the vectors you want to concatenate? Once you have the matrix, just look up the documentation for the max function. You can specify a dimension for it to operate on.
回答 (1 件)
Jan
2017 年 12 月 2 日
The body of the loop does not depend on the loop counter:
for x = 1:height(PFset)
A = [XfmrpfCH XfmrpfCT XfmrpfCL XfmrpfCHLUST XfmrpfCHTUST XfmrpfCLTUST];
end
This is equivalent to:
A = [XfmrpfCH XfmrpfCT XfmrpfCL XfmrpfCHLUST XfmrpfCHTUST XfmrpfCLTUST];
without any loop.
The currently provided information are not enough to guess, what you want instead. Please edit the question and add more details.
5 件のコメント
Rik
2017 年 12 月 2 日
A = [XfmrpfCH', XfmrpfCT', XfmrpfCL', XfmrpfCHLUST', ...
XfmrpfCHTUST', XfmrpfCLTUST'];
Amax = max(A, 2)
Jan
2017 年 12 月 2 日
The reason for loop is to generate multiple row of the six variables.
But this loop does not do this. It creates the same matrix repeatedly, such that it is a waste of time only. Do you understand this?
Currently, I have 200 entries with each variable in separate lists.
Faustino, what are "lists"? Please stay at the standard terms. The question would have been much clearer than. Are "XfmrpfCH" and the other variables [1 x 200] vectors? Then you want either:
A = [XfmrpfCH', XfmrpfCT', XfmrpfCL', XfmrpfCHLUST', ...
XfmrpfCHTUST', XfmrpfCLTUST']
or
A = [XfmrpfCH; XfmrpfCT; XfmrpfCL; XfmrpfCHLUST; ...
XfmrpfCHTUST; XfmrpfCLTUST]
Now either max(A, [], 1) or max(A, [], 2) will find the maximum value in each column or row.
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