Any() or all()

Question

Iteration over the array with the definition of the parity of the number.
The attachment is a simple option. How can I do this using any () or all () functions ?
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Перебор массива с определением четности числа.
Во вложении простой вариант. Как это сделать с помощью функций any() или all()?
https://code.sololearn.com/cq57LFb3AEwp/?ref=app

Answer 1

here is a code snippet that you can use to see how it works. Input is given as space separated numbers. They will be split and converted to int and then stored in a list. This is the base for any() or all().

inp = [int(i) for i in input('enter some numbers sep. by space: ').split()]
if all(j % 2 == 0 for j in inp):
print('all numbers are Even')
else:
print('not all numbers are Even')

Answer 2

print("NO" if any(filter(lambda k:k%2, b)) else "YES")

Answer 3

print("even" if any(i%2 for i in [2,4,6,8,0,2,4,8,0]) else "odd")

Answer 4

Your code doesn't use an array (or list), the loop is running in range. Did you attach the right code link?

Answer 5

Ok, just FYI, all() or any() needs an iterable object to work with. I don't understand why would you need either one of those functions to check for even or odd number.

Answer 6

Try this I think this is what you want.

print("NO" if any([n % 2 for n in b]) else "YES")

An example is also available in the Python tutorial

https://www.sololearn.com/learn/Python/2456/

Answer 7

Ipang thanks!

Answer 8

Ipang
Ok. It was a code where are you input 10 numbers:)))) and after i look all numbers aret even or not

Answer 9

Ipang
https://code.sololearn.com/c9QhdnSlp7uj/?ref=app

Answer 10

Ipang i want without "for"
Maybe print("yes" if any(b[i]%2) else "no")

Answer 11

No sweat buddy 👌