Sum of each row ?

I have [1 1 0; 0 1 0; 0 0 0;] and I want to write that if average of each row > 0.5 it equals 1 , else 0

for the example above, first row 1+1+0 / 3 > 0.5 then it is 1 ..

for i=1:length(nx);
    s(i)= sum(nx(i,:))/3; 
for n=1:length(nx);

if s(i(n+1))> 0.5;
    s(i(n+1)) == 1;
else
    s(i(n+1))== 0 ;
end    
end
end

[Star Strider]

What functions are you permitted to use in this obvious homework assignment?

[Rena Berman]

(Answers Dev) Restored edit

Answer 1

[Jos (10584)]

Use the matlab way! Use the functions mean, or sum and size. Read the documentation on these functions and pay attention to the dimension argument.

Then take a closer look at logical operators, like >

A = [.9 .7 .4 .6 .2]
B = A > 0.5

Answer 2

[Image Analyst]

Does it have to be a for loop? Why not just do:

rowMeans = mean(nx, 2);
s = zeros(size(nx));
s(rowMeans > 0.5, :) = 1;

You've said both "sum" and "average" - which is it?????. The above is for average. If you want sums, do

rowSums = sum(nx, 2);
s = zeros(size(nx));
s(rowSums > 0.5, :) = 1;

[Image Analyst]

That does not make sense. My code has no requirement on the number of rows. You can have however many rows you want. Just because the number of rows is not specified, that does NOT mean a loop is required.

[Stephen23]

"There are no specific number of rows therefore i am trying to use for loop to see average of each row"

This makes no sense: vectorized code has no restrictions on how many rows it can handle. Why do you think that a loop is required?

[Image Analyst]

Original question:

I have [1 1 0; 0 1 0; 0 0 0;] and I want to write that if average of each row > 0.5 it equals 1 , else 0

for the example above, first row 1+1+0 / 3 > 0.5 then it is 1 ..

for i=1:length(nx);
    s(i)= sum(nx(i,:))/3; 
for n=1:length(nx);

if s(i(n+1))> 0.5;
    s(i(n+1)) == 1;
else
    s(i(n+1))== 0 ;
end    
end
end

Answer 3

[Walter Roberson]

mean(nx, 2) > 0.5      %the 2 means "by rows" here.