# How to find the mean of a column in a matrix using for loop?

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Hunter Nau on 19 Apr 2022
Commented: KSSV on 19 Apr 2022
For part B I have to use for loop to find the average of each column. I am getting the correct output but I am getting 5 of the same outputs. For part C I need to find the mean of all the temperatures, but I keep getting answers from 1:5.
%% Part A
TempSenVal=[58.3 59.2 60.2 63.0 63.3; 62.5 63.8 64.9 65.2 65.8;...
63.2 64.2 65.3 67.8 67.9; 64.0 65.1 67.8 68.2 69.1; 65.5 66.0 67.9 68.9 70.2]
%% Part B
for k=TempSenVal
mean(TempSenVal);
end
%% Part C
n=length(TempSenVal);
for k=TempSenVal
mean(n);
end

KSSV on 19 Apr 2022
TempSenVal=[58.3 59.2 60.2 63.0 63.3; 62.5 63.8 64.9 65.2 65.8;...
63.2 64.2 65.3 67.8 67.9; 64.0 65.1 67.8 68.2 69.1; 65.5 66.0 67.9 68.9 70.2] ;
[m,n] = size(TempSenVal) ;
M = zeros(1,n) ;
for i = 1:n
M(i) = mean(TempSenVal(:,i)) ;
end
% just use mean
M
M = 1×5
62.7000 63.6600 65.2200 66.6200 67.2600
mean(TempSenVal)
ans = 1×5
62.7000 63.6600 65.2200 66.6200 67.2600
##### 2 CommentsShowHide 1 older comment
KSSV on 19 Apr 2022
mean(TempSenVal(:))

R2022a

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