python program way to check for type in Python?

Question

def main():
    a,b=numbers(5,1,100)
print("Number of Odd values = " + str(a))
print("Number of Even values = " + str(b))
def numbers(N,A,B):
    even_count,odd_count=0,0
    for i in range(N):
        n=random.randint(A,B)        
        if n%2==0:
            even_count+=1
        else:
            odd_count+=1
    return odd_count, even_count
main()

Need fix this code.i don't know when the number can't go through. print("Number of Odd values = " + str(a)) NameError: name 'a' is not defined

Answer 1

Try this:

import random

def numbers(N,A,B):
    even_count,odd_count=0,0
    for i in range(N):
        n=random.randint(A,B)        
        if n%2==0:
            even_count+=1
        else:
            odd_count+=1
    return odd_count, even_count
    
a,b=numbers(5,1,100) 

print("Number of Odd values = " + str(a))
print("Number of Even values = " + str(b))

Answer 2

you need to make sure the print statements are coming inside the scope of function main() and import module random at the top of the script. below is the formatted version.

def main():
    a,b=numbers(5,1,100)
    
    print("Number of Odd values = " + str(a))
    print("Number of Even values = " + str(b))


def numbers(N,A,B):
    even_count,odd_count=0,0
    for i in range(N):
        n=random.randint(A,B)        
        if n%2==0:
            even_count+=1
        else:
            odd_count+=1
    return odd_count, even_count
main()

Answer 3

I guess you need you to reorder the code,try this:

import random

def numbers(N,A,B):
    even_count, odd_count = 0, 0
    for i in range(N):
        n = random.randint(A, B)
        if n % 2 == 0:
            even_count += 1
        else:
            odd_count += 1
    return odd_count, even_count


def main():
    a, b = numbers(5, 1, 100)
    print("Number of Odd values = " + str(a))
    print("Number of Even values = " + str(b))


main()