How to divide a time interval into parts of different lengths?
I have a time interval from 0 to t
. I want to divide this interval into a cumulative sequence in loop 2.25, 2.25 and 1.5 like this:
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 the time period into cycles of 6 hours, each cycle consisting of three consecutive operations that last 2.25 hours, 2.25 hours and 1.5 hours, respectively. Or is there an alternative to using "stages" for this purpose?
+3
source to share
2 answers
You can 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 to start and stops:
>>> 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 can 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
+3
source to share
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]
Store the step values in a list. Just loop over and keep adding them to the spin until you hit the cap.
+1
source to share