Как найти минимальное четное число питон

Помогите пожалуйста написать программу на питон, которая будет находить в одномерном массиве минимальное чётное число и максимальное нечётное число.
Программа в основном написана, но не получается найти минимальное чётное число.

n=int(input("Введите количество элементов, которые будут в списке:"))
b=[]
for i in range(0,n):
    a=int(input("Элемент:"))
    b.append(a)
c=[]
d=[]
for i in b:
    if(i%2==0):
        c.append(i)
    else:
        d.append(i)
c.sort()
d.sort()
count1=0
count2=0
for k in c:
    count1=count1-1
for j in d:
    count2=count2+1
print("Минимальное четное число:",c[count1+1])
print("Максимальное нечетное число",d[count2-1])

I’m trying to get the minimum odd number using python. I used lambda, loops and other methods to get minimum odd number but i was not able to get that using functions. here is my code

z= [1,8,-4,-9]

def min_odd(x):
    for i in x:
        if (i%2!=0):
    return min(i)

y = min_odd(z)
print (y)

Can some please tell me what i was missing here.

asked Sep 15, 2018 at 20:14

Try:

min([val for val in z if val % 2 != 0])

answered Sep 15, 2018 at 20:20

It seems your code logics are wrong. First off, you seem to have an indentation error in the return statement. Second off, the min() function requires a collection of items (like an array for example) or a series of arguments to determine the minimum in that series. You can try multiple things.

1. Use another variable to store a temporary minimum. Replace it every time you find a smaller odd value ( for every i in x… if the value is odd and is smaller than the previous odd value, replace it) and have it started with the first odd number you can find.

2. Take all the odd numbers and add them to another array on which you will apply the min function.

Hope this proves useful!

answered Sep 15, 2018 at 20:26

You could pass a generator into the min() function:

def min_odd(iterable):
    return min(i for i in iterable if i % 2)

I didn’t write i % 2 != 0 because any odd number will return 1 which has a Boolean value of True.

I added a parameter to the function that takes the iterable so it can be used for any iterable passed in.

answered Sep 15, 2018 at 20:20

min operates on an iterable. i is not an iterable in your code; it’s the last element of the list.

You can achieve what you want with a filter, though:

min(filter(lambda e: e%2 != 0, x))

answered Sep 15, 2018 at 20:19