如何创建一个不被 3 整除的 5 个下一个数字的列表

Question

dear community, I started learning Python and want to create a list of 5 next (after the given number) numbers that are not divided by 3 (modulo). Here is my code:

def main(N):
    
#print a list of 5 next numbers that are not divided by 3
    l = list()
    for i in range(int(N), 150):
        if len(l) < 5:
            if i%3==0:
                l.append(i+1)
    return [l]

at the end I get fe for the print(main(99)) :

>>> [100, 103, 106, 109, 112]

I also tried to put i+=1 after if statement but in the end I got even weirder result:

[101, 101, 102, 104, 104]

What do I do wrong and how can I fix it? Thanks!

Answer 1

I would write a generator that yields numbers not divisible by 3 with a base number and count as arguments:

def three_no_factor(n, cnt):
    x = 1
    while x <= cnt:
        if n % 3 != 0:
            yield n
            x += 1
        n += 1

Then use that to get the list of 5 numbers not divisible by 3:

>>> [x for x in three_no_factor(100,5)]
[100, 101, 103, 104, 106]

Answer 2

Unless there's a specific reason to cap the values at 150, you can just let the loop run until the list is full.

def next_nontriples(N):
# return a list of 5 next numbers that are not divisible by 3
    ls = list()
    nx = N + 1
    while len(ls) < 5:
        if nx%3 != 0:
            ls.append(nx)
        nx += 1
    return ls

Using the != operator is the key. Note that I'm returning the resulting list, whereas you return a list that has the required list as its only element, due to the extra square brackets around your return item.

You can fill the list once per loop-5-times too:

def next_nontriples(N):
# return a list of 5 next numbers that are not divisible by 3
    ls = list()
    nx = N + 1
    for ix in range(5):
        if nx%3 == 0:
            nx += 1 # skip this one 
        ls.append(nx)
        nx += 1
    return ls

Here we just add an increment to our candidate value if it fails the test (by being divisible by three). We know that the next number will qualify.

Finally you could just handle the three cases separately. Not recommended particularly here but hard-coding is sometimes your friend:

def next_nontriples(N):
# return a list of 5 next numbers that are not divisible by 3
    if N%3 == 0:
        return [N+1, N+2, N+4, N+5, N+7]
    if N%3 == 1:
        return [N+1, N+3, N+4, N+6, N+7]
    if N%3 == 2:
        return [N+2, N+3, N+5, N+6, N+8]

Answer 3

def main(N):
    
#print a list of 5 next numbers that are not divided by 3
    l = list()
    i = N + 1
    while len(l)<5:
        if i%3!=0:
            l.append(i)
        i += 1
    return [l]

Answer 4

Another way - >

def main(N):
    gen = (i for i in range(int(N), 150) if i % 3 != 0)  # create generator
    return [next(gen) for j in range(5)]  # iterate 5 times


main(99) # [100, 101, 103, 104, 106]