Menu
Writing some test cases and my mind wanders, assuming there is a better way to write something like this. I have a list, its numbers transition from all odd values to all even, doesn't matter where. I need to assert this is the case, here's what I came up with:
values = [1, 3, 5, 7, 5, 3, 5, 3, 5, 7, 4, 6, 8, 4, 2, 2, 8, 6]
# find all the indexes of odd and even values
odds = [i for (i, v) in enumerate(values) if v % 2 == 1]
evens = [i for (i, v) in enumerate(values) if v % 2 == 0]
# indexes should be a continuous sequence: 0, 1, 2, 3 ... n
assert odds + evens == range(evens[-1] + 1)
Seems like a long way to go. Suggestions on how this could be reduced?
808
Example 1: count Even and Odd numbers from given list using for loop Iterate each element in the list using for loop and check if num % 2 == 0, the condition to check even numbers. If the condition satisfies, then increase even count else increase odd count.
def is_list_even() returns true if all integers in the list are even and false otherwise. def is_list_odd() returns true if all integers in the list are odd and false otherwise.
num = int (input (“Enter any number to test whether it is odd or even: “) if (num % 2) == 0: print (“The number is even”) else: print (“The provided number is odd”) Output: Enter any number to test whether it is odd or even: 887 887 is odd. The program above only accepts integers as input.
Python Code:num = int(input("Enter a number: ")) mod = num % 2 if mod > 0: print("This is an odd number. ") else: print("This is an even number. ")
A possible solution is to consider that you allow only
odd->odd
odd->even
even->even
in other words the only forbidden transition is
even->odd
and this translates to
(0, 1) not in ((x%2, y%2) for x, y in zip(values, values[1:]))
[x for x in values if x % 2 == 1] + [x for x in values if x % 2 == 0] == values
This is only true, if values starts with all of it's own odd values, followed by all of its even values.
37
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With
