Using a function in ian inputdlg

This was the assignment given to me:

The script should request a vector of positive integers entered as a single line in an inputdlg box. The script should create a 1-by-2 vector b where b(1) is the sum of the odd numbers in A and b(2) is the product of the even numbers in A.

An example execution is as follows:

>> script26

The line entered in the inputdlg box is: [2 1 5 2 9 11 7]

b =
  33   4

This I what I have so far:

function[] = script26()

prompt = {'Array Numbers'};
dlg_title = 'Script26';
num_lines = 1;
def = {'2 1 5 2 9 11 7'};

b = inputdlg(prompt,dlg_title,num_lines,def);
b(1) = sum(def(1:2:end))
b(2) = prod(def(2:2:end))
A = [b(1) b(2)]

It won't let me do the sum function. How do I make the function to pull out the odd numbers of an array put into the inputdlg box??

Answer 1

[Geoff Hayes]

Gina - the inputdlg function returns a cell array, so you will have to convert this into the appropriate format before trying to manipulate the data. For example,

 response = inputdlg(prompt,dlg_title,num_lines,def); 
 data     = str2num(char(response));

We convert the cell array to a string using char and then the string of characters to a numeric array using str2num. (Note that how we convert on response and not the default values of def as in your example.)

Now that you have an array of numbers, you will need to iterate over each one and determine if it is odd or even. If the former, then add it to your odd sum. And if the latter, then multiply it against your even product. You won't be able to use sum or product because you won't know the order of the odd or even numbers for every possible input. (For your example, you do know that every other number is odd or even.)

Start with the above and use a for loop to iterate over the data.

[Gina Barlage]

function[] = script26()

prompt = {'Array Numbers'}; dlg_title = 'Script26'; num_lines = 1; def = {'2 1 5 2 9 11 7'};

b = inputdlg(prompt,dlg_title,num_lines,def); data = str2num(char(b)) for ii = data(1:end) if ii == 1:2:max(b) sum_total = sum_total + ii; else sum_total = sum_total; end end for jj = data(1:end) if jj == 0:2:max(b) product = product * jj; else product = product; end end sum product

This is what I now have but it won't let me use the max function. How can I pick out the odd numbers in the array??

[Joseph Cheng]

function[] = script26()
prompt = {'Array Numbers'}; 
dlg_title = 'Script26'; 
num_lines = 1; 
def = {'2 1 5 2 9 11 7'};

 b = inputdlg(prompt,dlg_title,num_lines,def); 
 data = str2num(char(b)) 
 for ii = data(1:end) 
     if ii == 1:2:max(b) 
         sum_total = sum_total + ii; 
     else 
        sum_total = sum_total; 
     end 
  end 

 for jj = data(1:end) 
     if jj == 0:2:max(b) 
     product = product * jj; 
 else 
     product = product; 
 end 
 end 

 sum 
 product

formatted for easier reading

[Joseph Cheng]

you should be using max(data) not max(b) as b is still in the cell formatting.

[Gina Barlage]

when I do 1:2:max(data) it pulls out every odd number between those numbers. How do I only pull out the odd numbers in the array?

[Walter Roberson]

In Joseph's code, the test

if ii == 1:2:max(b)

should be

if any(ii == 1:2:max(b))

and

if jj == 0:2:max(b)

should be

if any(jj == 0:2:max(b))

I would certainly never write the code like that myself, but it is one way of doing things.

[Joseph Cheng]

well it wasn't my code Q('-'Q), i reposted Gina's code since I don't have editing privileges yet. Are you looking for just for it being odd then you should be using mod()==1 to check for odd values. if you're checking for odd values and whether it is within the array then

 if mod(ii,2)==1 & any(ii==data)

check first to see if it is odd, and does the logical operand AND to say also is that number part of the data array (which it should be since ii = data(1:end).) which is kind of redundant since data = data(1:end)

[Gina Barlage]

Thanks so much for all your help!

Answer 2

[Walter Roberson]

Hint:

f = @(x) sin(x).^2;        %does a calculation and returns value
r = rand(1,20);
condition_met = f(r) < 0.33;   %does the calculated value satisfy some condition?
values_true_for = r(condition_met);
values_false_for = r(~condition_met);