How to divide time interval into parts of varying length?

问题

I have a time interval from 0 to t. I want to divide this interval into a cumulative sequence in a cycle of 2.25, 2.25 and 1.5, in the following manner:

input:

start = 0
stop = 19

output:

sequence = [0, 2.25, 4.5, 6, 8.25, 10.5, 12, 14.25, 16.5, 18, 19] 

How can I do this in Python?

---

The idea is to divide a time period into cycles of 6 hours, each cycle consisting of three sequential operations that last 2.25 h, 2.25 h and 1.5 h respectively. Or is there an alternative to using 'milestones' for this purpose?

回答1:

You could use a generator:

def interval(start, stop):
    cur = start
    yield cur                # return the start value
    while cur < stop:
        for increment in (2.25, 2.25, 1.5):
            cur += increment
            if cur >= stop:  # stop as soon as the value is above the stop (or equal)
                break
            yield cur
    yield stop               # also return the stop value

It works for the start and stop you proposed:

>>> list(interval(0, 19))
[0, 2.25, 4.5, 6.0, 8.25, 10.5, 12.0, 14.25, 16.5, 18.0, 19]

---

You could also use itertools.cycle to avoid the outer loop:

import itertools

def interval(start, stop):
    cur = start
    yield start
    for increment in itertools.cycle((2.25, 2.25, 1.5)):
        cur += increment
        if cur >= stop:
            break
        yield cur
    yield stop

回答2:

Not the cleanest. But it works.

>>> start = 0
>>> stop = 19
>>> step = [2.25, 2.25, 1.5]
>>> L = [start]
>>> while L[-1] <= stop:
...    L.append(L[-1] + step[i % 3])
...    i += 1
... 
>>> L[-1] = stop
>>> L
[0, 2.25, 4.5, 6.0, 8.25, 10.5, 12.0, 14.25, 16.5, 18.0, 19]

Keep your step values in a list. Just iterate over and keep adding them in rotation till you hit the cap.

来源：https://stackoverflow.com/questions/44419487/how-to-divide-time-interval-into-parts-of-varying-length