Problem 568. Number of 1s in a binary string

Thats the question:

Find the number of 1s in the given binary string. Example. If the input string is '1100101', the output is 4. If the input string is '0000', the output is 0

I wrote this code, but I don't understand, what is wrong with it.

function y = one(x)
  y = length(find(x))
end

Answer 1

[Les Beckham]

The find() function returns the index of all non-zero elements in the input. ASCII characters (like '0' and '1') are all non-zero (numerically) so your function will just return the length of the input string.

Example:

find('1011010')

ans = 1×7

1 2 3 4 5 6 7

length(find('1011010'))

ans = 7

There are a lot of ways to solve this problem. Look at this documentation page for one way: find characters

[Les Beckham]

Also, if you are just getting started with Matlab, I would highly recommend that you take a couple of hours to go through the free online tutorial: Matlab Onramp

[Mayla]

I have already done Onramp. Maybe I should repeat it again :) Do you have any tipps for other free online tutorials?

[Les Beckham]

Mathworks has several other tutorials in addition to the Onramp.

Check this page that lists some more: Self-Paced Online Courses

Answer 2

[Harald]

Hi,

find is intended for numerical non-0 elements rather than non-'0'. Instead, I would use == to compare to '0', and I would also use sum or nnz rather than length.

Best wishes,

Harald

[Dyuman Joshi]

+1 for using just logical indexing with sum/nnz over length

[Image Analyst]

@Dyuman Joshi my vote registered but yours did not for some reason.

[Mayla]

Thanks for your answers! I found a solution, but I tried

x==1

y=sum(ans)

and it didn't work. Why is that so? And why can't I only use sum(x)?

[Harald]

Hi,

you need to compare to '1' instead of 1.

Also, avoid using ans. Assign to a variable instead:

isOne = x=='1';
y=sum(isOne)
% or, in one line:
y=sum(x == '1')

Best wishes,

Harald

[Dyuman Joshi]

"and it didn't work."

Because you are comparing with a numerical value

x='1011010';

x==1

ans = 1×7 logical array
   0   0   0   0   0   0   0

What you want to do is compare with a character

x=='1'

ans = 1×7 logical array
   1   0   1   1   0   1   0

"And why can't I only use sum(x)?"

Because that will give the sum of the values of x, which is not what the questions asks for.

What you want is, the total number of values i.e. sum where x isequal to '1'. So that's what is done.

[Dyuman Joshi]

@Image Analyst please check again