变量+ =％可能或如何获得另一个数字之后的数字的下一个间隔

Question

I want to be able to do something like the following...

Say I want a list of every number from 100-500 that is a multiple of 33.

>>> a = 100
>>> b = 500
>>> range(a,b,33)
[100, 133, 166, 199, 232, 265, 298, 331, 364, 397, 430, 463, 496]

This is not what I want, this is because a is not a multiple of 33.

To get the next multiple of 33 from a I can do:

a = a - a % 33 + 33

I want to know if there is an easier way to do this so that if I want to create this range without knowing the actual values and without having to define them beforehand..

Such as:

>>> def multiple(a, b, c):
    return range(a+=%c, b, c) #if this was possible

And obviously it would return me a range which would be correct, for example:

>>> multiple(100, 500, 33)
[132, 165, 198, 231, 264, 297, 330, 363, 396, 429, 462, 495]

I am aware I can do something simple like:

range(a - a%c + c, b, c)

However, without getting into details, calling the value of a is expensive for me in my case, and I would like to be able to find a way to not have to call it a second time, and also the above method is really not nice looking at all.

I really was not sure what the title of my question should be, but I suppose what I am looking for is a way to find the next multiple of a number after another given number.

Thank you.

Answer 1

The following does what you want and only uses a once:

In [4]: range((a + c - 1) // c * c, b + 1, c)
Out[4]: [132, 165, 198, 231, 264, 297, 330, 363, 396, 429, 462, 495]

I would still place it into a helper function, rendering moot the question of how many times a is used.

Unlike the code in your question, this actually works correctly for cases when a is evenly divisible by c . Also, unlike the code in your question, it includes b in the range (as it should, according to your answer to my question in the comments):

In [15]: range((a + c - 1) // c * c, b + 1, c)
Out[15]: [66, 99, 132]

Answer 2

Well, to get the multiple of 33 everytime, you would need to start with a multiple of 33 only.

So, start with first multiple of 33 after 100 , because range function will just keep on adding 33 to get next multiples.

To get the first multiple of 33 after a number num , you can use: -

num / 33 * 33 + 33  <==>  (num / 33 + 1) * 33

So, for your range you can use: -

>>> a = 100
>>> b = 500
>>> a = (a / 33 + 1) * 33
>>> range(a, b, 33)
[132, 165, 198, 231, 264, 297, 330, 363, 396, 429, 462, 495]