How can I find the sum of the perimeter of a matrix?
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Hi, I am wondering if I had a matrix [1,2,3;1,2,3;1,2,3] how I could find the sum of the perimeter. Thanks an advance.
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dpb
2016 年 8 月 5 日
One way amongst any number...
>> M=[1,2,3;1,2,3;1,2,3];
>> I=ones(size(M)); I(2:end-1,2:end-1)=0;
>> S=sum(reshape(M.*I,1,[]))
S =
16
>>
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その他の回答 (3 件)
Stephen23
2016 年 8 月 5 日
編集済み: Stephen23
2016 年 8 月 5 日
In one line is easy and very fast:
>> M = [1,2,3;1,2,3;1,2,3];
>> sum(sum([M([1,end],2:end-1),M(1:end,[1,end]).']))
ans = 16
5 件のコメント
dpb
2018 年 2 月 10 日
[ ... Ricardo Florez Answer moved to comment since was a follow-up question on earlier Answer syntax ... dpb]
Hi. Why is it necessary to write two "sum" in the code: sum(sum(M(...)? Please see below:
the sum(M(:))-sum(sum(M(2:end-1,2:end-1)))
Thank you.
dpb
2018 年 2 月 10 日
編集済み: dpb
2018 年 2 月 11 日
Because sum(M(2:end-1,2:end-1)) is a row vector of the sums of the columns in the submatrix; to get the total have to then add those values. This is taken care of in the first sum by the use of the special syntax (:) that returns all elements of an array/matrix in a column so sum only needs to add the one column vector. There is no syntax in Matlab to apply the secondary indexing operation to the result of the first function call return; there've been many suggest the enhancement.
Sanjay Zamindar
2018 年 6 月 28 日
編集済み: dpb
2018 年 6 月 28 日
function final_sum = peri_sum(A)
size_row = size(A,1)
size_col=size(A,2)
Sum_of_first_row = sum (A(1, [1:1:end])); %base
Sum_of_first_Colmn = sum(A([2:1:size_row],1))
%sum_col= sum(A([2:1:size_col], [1:size_col:size_col]));
Sum_of_last_column= sum(A([2:1:size_row],size_col))
sum_of_last_row = sum(A(size_row,[2:1:size_col-1]))
final_sum = sum_of_last_row+ Sum_of_last_column+Sum_of_first_Colmn+Sum_of_first_row
end
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