Python List Comprehension: Find all of the numbers from 1-1000 that have a 3 in them

findThree.py

'''
Find all of the numbers from 1-1000 that have a 3 in them
'''

three = [n for n in range(0,1000) if '3' in str(n)]
print(three)

I have found the following solution without using strings:
numbers=range(1,1001)
r=[i for i in numbers if i%10==3 or (i//10)%10==3 or ((i//10)//10)%10==3 or (((i//10)//10)//10)%10==3]

I think with the string option is better because is the simplest and that's the more legible solution, instead of you can write several conditions to get it without using strings. However, strings are a tool to use to and our goal should be make it easy.

These 3 options work:

x = [i for i in range(1, 1001) if str(i).find("3") != -1]
y = [i for i in range(1, 1001) if str(i).count("3") > 0]
z = [i for i in range(1, 1001) if '3' in str(i)]

print("x ->", x, len(x), "\n")
print("y ->", y, len(y), "\n")
print("z ->", z, len(z))

You could also do it this way:
lst_include3=list(filter(lambda x: '3' in str(x), range(1,1000)))
print(lst_include3)