Cody Problem 14. Find the numeric mean of the prime numbers in a matrix.

I found difficulties with the 14th Problem on Cody here is my solution:

function out = meanOfPrimes(in)
[a,b]=size(in)
sum = 0;
for i = 1:a
    for j = 1:b
        if isprime(in(i,j))
            in(i,j) = str2num(in(i,j))
            
            sum = sum + in(i,j);
            
        end
    end
end

out = sum/2;
end

Unfortunately it doesn't work, can anyone help me with that!

Thanks in advance.

[Geoff Hayes]

Hidd_1 - why do you do the following

 in(i,j) = str2num(in(i,j))

Isn't the in array already numeric? Won't this fail if already numeric? What is your intention here?

Also, try to avoid creating variables whose name matches those of in-built MATLAB functions. In this case, that would be sum.

Answer 1

[David Hill]

Input is not a string.

function out = meanOfPrimes(in)
[a,b]=size(in)
sum = 0;
c=0;
for i = 1:a
    for j = 1:b
        if isprime(in(i,j))
            %in(i,j) = str2num(in(i,j))
            sum = sum + in(i,j);
            c=c+1;%need to keep track of how many
            
        end
    end
end
out = sum/c;%mean is sum of primes/number of primes
end

Alternatively, linear index into matrix

function out = meanOfPrimes(in)
sum = 0;
c=0;
for i = 1:numel(in)
    if isprime(in(i))
       sum = sum + in(i);
       c=c+1;
    end
end
out = sum/c;

Or better yet, no loop

out=mean(a(isprime(a)));

Answer 2

[the cyclist]

After you correct the problem that @Geoff Hayes pointed out in his comment, you still have a mathematical one.

out = sum/2

is not the correct way to calculate the mean value from the sum of the prime values.

Is that enough of a hint?

Also, I'll just point out that you did not need to use the for loops here. You could have applied isprime to the whole input array at once, harnessing the vectorized nature of MATLAB.

[Hidd_1]

Thanks I solved the problem! Here is my solution:

function out = meanOfPrimes(in)
[a,b]=size(in);
s = 0;
k=0;
for i = 1:a
    for j = 1:b
        if isprime(in(i,j))
            s = s + in(i,j);
            k = k+1;
            
        end
    end
end
out = s/k
end

[s] I wonder how can I solve it using only isprime? [/s]

@David HillThanks!

Answer 3

[Hernia Baby]

As Geoff Hayes says, variable 'in' is NOT string type.

You can check the data type with class function.

in = [8 3 5 9];
class(in)

ans =

'double'

ｰｰｰｰｰｰｰｰｰｰｰｰｰｰ

Therefore, it will work well when omitting str2num.

clear,clc;

in = [8 3 5 9];

[a,b]=size(in);
sum = 0;

for i = 1:a
    for j = 1:b
        if isprime(in(i,j))
            in(i,j) = in(i,j);
            sum = sum + in(i,j);
            
        end
    end
end
out = sum/2

out =

4

Answer 4

[Abhinav]

function out = meanOfPrimes(in)
    
    primeTF = isprime(in);
    primeTFIndex = find(primeTF);
    x = [];
    
    for i = 1:length(primeTFIndex)
        x = [x, in(primeTFIndex(i))];
    end
    
    out = mean(x);
end