How do I get Prime numbers 1-100

Hi Shilpa, Create an exp rule to check prime number and then use below code. 

a!forEach(
  items: enumerate(100)+1,
  expression: if(
    rule!isPrimeNumber(fv!item),
    fv!item,
    {}
  )
)

filter(
  rule!isPrimeNumber(_),
  enumerate(100)+1,
)

Clean..

can you please elaborate

rule!isPrimeNumber(_),

Hi Shilpa ,it can be an expression rule to find out whether a number is prime or not. You would need to create that rule. I would suggest you to create this logic .

Hi Shilpa,

Try this for prime number rule.

rule!isPrimeNumber(_) --> 

load(
local!count: a!forEach(
items: enumerate(100) + 1,
expression: if(mod(ri!number, fv!item) = 0, fv!item, {})
),
if(count(local!count) > 2, {}, ri!number)
)

The idea is to find the number of number by which the given number can be divided.

As the mod() function works with lists, I try the first 100 numbers. The compare operator "=" also works with lists. And the sum() interprets true and false like 1 and 0.

Less than three factors means it is a prime.

sum(mod(ri!number, enumerate(100) + 1) = 0) <= 2

"filter()", for those times when you'd rather have code that's nearly unreadable instead of 4 lines longer

I would only make one change which is to only enumerate up to the number being inspected.

sum(mod(ri!number, enumerate(ri!number) + 1) = 0) <= 2

I would only make one change which is to only enumerate up to the number being inspected.

sum(mod(ri!number, enumerate(ri!number) + 1) = 0) <= 2

I would like to tweak the logic as the below code does not exclude 1 which is not a prime number since 1 has less than one factor, A prime number needs to have exactly two factors.

sum(mod(ri!number, enumerate(ri!number) + 1) = 0) <= 2

Do this instead! 

length(
  wherecontains(true, mod(ri!number, enumerate(ri!number) + 1) = 0)
) = 2

OR

sum(mod(ri!number, enumerate(ri!number) + 1) = 0) = 2

Feel free to correct me if I am wrong.