How can I find the sum of the perimeter of a matrix?
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Alexandra Huff
le 5 Août 2016
Réponse apportée : Ibrahim Abouemira
le 19 Mai 2019
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
le 5 Août 2016
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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Plus de réponses (3)
Stephen23
le 5 Août 2016
Modifié(e) : Stephen23
le 5 Août 2016
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
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dpb
le 10 Fév 2018
[ ... 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
le 10 Fév 2018
Modifié(e) : dpb
le 11 Fév 2018
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
le 28 Juin 2018
Modifié(e) : dpb
le 28 Juin 2018
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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