如何从列表中提取连续的偶数？

Question

In a list, a minimum of 2 even numbers makes a sequence: for example, a list that is [3, 18, 6, 5, 3, 4, 32, 8, 12] has 2 sequences that are even - which are 18, 6 and 4, 32, 8, 12 .

How do I extract the sequences without extracting every even number individually?

def saisie():
    test = False
    while not test:
        N = int(input("donner N: "))
        test = 3<=N<=20
    return N

def remplir(N):
    t=[int]*N
    for i in range(N):
        element = input("enter an element:")
        t.append(element)
    return t

def seq(N, t):
    s=""
    for i in t:
        if t[i] and t[i+=]%2==0:
            s=   

Answer 1

To group items with shared criteria, you can use groupby :

from itertools import groupby

nums = [3, 18, 6, 5, 3, 4, 32, 8, 12]

for is_even, grouper in groupby(nums, key=lambda x: x % 2 == 0):
    if is_even:
        evens = list(grouper)  # grouper objects have no len. Must convert to list
        if len(evens) >= 2:  # only consider sequences of at least 2 even numbers
            print(*evens)

Will give:

18 6
4 32 8 12