你如何在 Python 中建模随时间变化的东西？

Question

I'm looking for a data type to help me model resource availability over fluid time.

• We're open from 9 til 6 and can handle 5 parallel jobs. In my imaginary programming land, I've just initialised an object with that range with a value of 3 across the board.
• We have appointments on the books, each with start and end times.
• I need to punch each of those out of the day
• That leaves me with a graph of sorts where the availability goes up and down, but ultimately allowing me to quickly find time ranges where there is remaining availability.

I've come at this problem from many directions but always come back to the fundamental problem of not knowing a data type to model something as simple as an integer over time.

I could convert my appointments into time series events (eg appointment arrives means -1 availability, appointment leaves means +1) but I still don't know how to manipulate that data so that I can distil out periods where the availability is greater than zero.

---

Somebody's left a close-vote citing lack of focus, but my goal here seems pretty singular so I'll try to explain the problem graphically. I'm trying to infer the periods of time where the number of active jobs falls below a given capacity.

Turning a range of known parallel capacity (eg 3 between 9-6) and a list of jobs with variable start/ends, into a list of time ranges of available time.

Answer 1

My approach would be to build the time series, but include the availability object with a value set to the availability in that period.

availability: 
[
  {
    "start": 09:00,
    "end": 12:00,
    "value": 4
  },
  {
     "start": 12:00,
     "end": 13:00,
     "value": 3
  }
]

data: [
  {
    "start": 10:00,
    "end": 10:30,
  }
]

Build the time series indexing on start/end times, with the value as the value. A start time for availability is +value, end time -value. While for an event, it'd be -1 or +1 as you said.

"09:00" 4
"10:00" -1
"10:30" 1
"12:00" -4
"12:00" 3
"13:00" -3

Then group by index, sum and cumulative sum.

getting:

"09:00" 4
"10:00" 3
"10:30" 4
"12:00" 3
"13:00" 0

Example code in pandas:

import numpy as np
import pandas as pd


data = [
  {
    "start": "10:00",
    "end": "10:30",
  }
]

breakpoints = [
  {
    "start": "00:00",
    "end": "09:00",
    "value": 0
  },
  {
    "start": "09:00",
    "end": "12:00",
    "value": 4
  },
  {
    "start": "12:00",
    "end": "12:30",
    "value": 4
  },
  {
    "start": "12:30",
    "end": "13:00",
    "value": 3
  },
  {
    "start": "13:00",
    "end": "00:00",
    "value": 0
  }
]

df = pd.DataFrame(data, columns=['start', 'end'])

print(df.head(5))

starts = pd.DataFrame(data, columns=['start'])
starts["value"] = -1
starts = starts.set_index("start")

ends = pd.DataFrame(data, columns=['end'])
ends["value"] = 1
ends = ends.set_index("end")

breakpointsStarts = pd.DataFrame(breakpoints, columns=['start', 'value']).set_index("start")

breakpointsEnds = pd.DataFrame(breakpoints, columns=['end', 'value'])
breakpointsEnds["value"] = breakpointsEnds["value"].transform(lambda x: -x)
breakpointsEnds = breakpointsEnds.set_index("end")

countsDf = pd.concat([starts, ends, breakpointsEnds, breakpointsStarts]).sort_index()
countsDf = countsDf.groupby(countsDf.index).sum().cumsum()

print(countsDf)

# Periods that are available

df = countsDf
df["available"] = df["value"] > 0

# Indexes where the value of available changes
# Alternatively swap out available for the value.
time_changes = df["available"].diff()[df["available"].diff() != 0].index.values
newDf = pd.DataFrame(time_changes, columns= ["start"])

# Setting the end column to the value of the next start
newDf['end'] = newDf.transform(np.roll, shift=-1)
print(newDf)

# Join this back in to get the actual value of available
mergedDf = newDf.merge(df, left_on="start", right_index=True)

print(mergedDf)

returning at the end:

   start    end  value  available
0  00:00  09:00      0      False
1  09:00  13:00      4       True
2  13:00  00:00      0      False

Answer 2

I'd approach it the same way you did with the appointments. Model the free time as appointments on its own. For each ending appointment check if theres another on ongoing, if so, skip here. If not, find the next starting appointment (one with a start date greater than this ones enddate.)

After you iterated all off your appointments, you should have an inverted mask of it.

Answer 3

To me, this problem would be well-represented by a list of boolean values. For ease of explanation, let's assume the length of every potential job is a multiple of 15 minutes. So, from 9 to 6, we have 135 "time slots" that we want to track availability for. We represent a queue's availability in a time slot with boolean variables: False if the queue is processing a job, True if the queue is available.

First, we create a list of time slots for every queue as well as the output. So, every queue and the output has time slots t _k , 1 <= k <= 135.

Then, given five job queues, q _j , 1 <= j <= 5, we say that t _k is "open" at time k if there exists at least one q _j where the time slot list at index k is True .

We can implement this in standalone Python as follows:

slots = [ True ] * 135
queues = [ slots ] * 5
output = [ False ] * 135

def available (k):

 for q in queues:
  if q[k]:
   return True

 return False

We can then assume there exists some function dispatch (length) that assigns a job to an available queue, setting the appropriate slots in queue[q] to False .

Finally, to update the output, we simply call:

def update():

 for k in range(0, 135):
  output[k] = available[k]

Or, for increased efficiency:

def update(i, j):
 for k in range(i, j):
  output[k] = available[k]

Then, you could simply call update(i, j) whenever dispatch() updates time slots i thru j for a new job. In this way, dispatching and updating is an O(n) operation, where n is how many time slots are being changed, regardless of how many time slots there are.

This would allow you to make a simple function that maps human-readable time onto the range of time slot values, which would allow for making time slots larger or smaller as you wish.

You could also easily extend this idea to use a pandas data frame where each column is one queue, allowing you to use Series.any() on every row at once to quickly update the output column.

Would love to hear suggestions regarding this approach! Perhaps there's a complexity of the problem I've missed, but I think this is a nice solution.

Answer 4

You can use (datetime, increment) tuples to keep track of the changes in availability. A job-start event has increment = 1 and a job-end event has increment = -1 . Then itertools.accumulate allows for computing the cumulative availability as jobs start and end over time. Here's an example implementation:

from datetime import time
import itertools as it

def compute_availability(jobs, opening_hours, capacity):
    jobs = [((x, -1), (y, +1)) for x, y in jobs]
    opens, closes = opening_hours
    events = [[opens, capacity]] + sorted(t for job in jobs for t in job) + [(closes, 0)]
    availability = list(it.accumulate(events,
                                      lambda x, y: [y[0], x[1] + y[1]]))
    for x, y in zip(availability, availability[1:]):
        # If multiple events happen at the same time, only yield the last one.
        if y[0] > x[0]:
            yield x

This adds artificial (opens, capacity) and (closes, 0) events to initialize the computation. The above example considers a single day but it is easy to extend it to multiple days by creating opens and closes datetime objects that share the day of the first and last job respectively.

Example

Here is the output for the OP's example schedule:

from pprint import pprint

jobs = [(time(10), time(15)),
        (time(9), time(11)),
        (time(12, 30), time(16)),
        (time(10), time(18))]

availability = list(compute_availability(
    jobs, opening_hours=(time(9), time(18)), capacity=3
))
pprint(availability)

which prints:

[[datetime.time(9, 0), 2],
 [datetime.time(10, 0), 0],
 [datetime.time(11, 0), 1],
 [datetime.time(12, 30), 0],
 [datetime.time(15, 0), 1],
 [datetime.time(16, 0), 2]]

The first element indicates when the availability changes and the second element denotes the availability that results from that change. For example at 9am one job is submitted causing the availability to drop from 3 to 2 and then at 10am two more jobs are submitted while the first one is still running (hence availability drops to 0).

Adding new jobs

Now that we have the initial availability computed an important aspect is to update it as new jobs are added. Here it is desirable not to recompute the availability from the full job list since that might be costly if many jobs are being tracked. Because the availability is already sorted we can use the bisect module to determine the relevant update range in O(log(N)). Then a number of steps need to be performed. Let's say the job is scheduled as [x, y] where x , y are two datetime objects.

1. Check if the availability in the [x, y] interval is greater than zero (including the event to the left of x (ie the previous event)).
2. Decrease the availability of all events in [x, y] by 1.
3. If x is not in the list of events we need to add it, otherwise we need to check whether we can merge the x event with the one left to it.
4. If y is not in the list of events we need to add it.

Here is the relevant code:

import bisect

def add_job(availability, job, *, weight=1):
    """weight: how many lanes the job requires"""
    job = list(job)
    start = bisect.bisect(availability, job[:1])
    # Emulate a `bisect_right` which doens't work directly since
    # we're comparing lists of different length.
    if start < len(availability):
        start += (job[0] == availability[start][0])
    stop = bisect.bisect(availability, job[1:])

    if any(slot[1] < weight for slot in availability[start-1:stop]):
        raise ValueError('The requested time slot is not available')

    for slot in availability[start:stop]:
        slot[1] -= weight

    if job[0] > availability[start-1][0]:
        previous_availability = availability[start-1][1]
        availability.insert(start, [job[0], previous_availability - weight])
        stop += 1
    else:
        availability[start-1][1] -= weight
        if start >= 2 and availability[start-1][1] == availability[start-2][1]:
            del availability[start-1]
            stop -= 1

    if stop == len(availability) or job[1] < availability[stop][0]:
        previous_availability = availability[stop-1][1]
        availability.insert(stop, [job[1], previous_availability + weight])

Example schedule

We can test it by adding some jobs to the OP's example schedule:

for job in [[time(15), time(17)],
            [time(11, 30), time(12)],
            [time(13), time(14)]]:  # this one should raise since availability is zero
    print(f'\nAdding {job = }')
    add_job(availability, job)
    pprint(availability)

which outputs:

Adding job = [datetime.time(15, 0), datetime.time(17, 0)]
[[datetime.time(9, 0), 2],
 [datetime.time(10, 0), 0],
 [datetime.time(11, 0), 1],
 [datetime.time(12, 30), 0],
 [datetime.time(16, 0), 1],
 [datetime.time(17, 0), 2]]

Adding job = [datetime.time(11, 30), datetime.time(12, 0)]
[[datetime.time(9, 0), 2],
 [datetime.time(10, 0), 0],
 [datetime.time(11, 0), 1],
 [datetime.time(11, 30), 0],
 [datetime.time(12, 0), 1],
 [datetime.time(12, 30), 0],
 [datetime.time(16, 0), 1],
 [datetime.time(17, 0), 2]]

Adding job = [datetime.time(13, 0), datetime.time(14, 0)]
Traceback (most recent call last):
  [...]
ValueError: The requested time slot is not available

Blocking night hours

We can also use this interface to block all lanes during hours when the service is unavailable (eg from 6pm to 9am on the next day). Just submit a job with weight=capacity for that time span:

add_job(availability,
        [datetime(2020, 3, 14, 18), datetime(2020, 3, 15, 9)]
        weight=3)

Build full schedule from scratch

We can also use add_job to build the full schedule from scratch:

availability = availability = list(compute_availability(
    [], opening_hours=(time(9), time(18)), capacity=3
))
print('Initial availability')
pprint(availability)
for job in jobs:
    print(f'\nAdding {job = }')
    add_job(availability, job)
    pprint(availability)

which outputs:

Initial availability
[[datetime.time(9, 0), 3]]

Adding job = (datetime.time(10, 0), datetime.time(15, 0))
[[datetime.time(9, 0), 3],
 [datetime.time(10, 0), 2],
 [datetime.time(15, 0), 3]]

Adding job = (datetime.time(9, 0), datetime.time(11, 0))
[[datetime.time(9, 0), 2],
 [datetime.time(10, 0), 1],
 [datetime.time(11, 0), 2],
 [datetime.time(15, 0), 3]]

Adding job = (datetime.time(12, 30), datetime.time(16, 0))
[[datetime.time(9, 0), 2],
 [datetime.time(10, 0), 1],
 [datetime.time(11, 0), 2],
 [datetime.time(12, 30), 1],
 [datetime.time(15, 0), 2],
 [datetime.time(16, 0), 3]]

Adding job = (datetime.time(10, 0), datetime.time(18, 0))
[[datetime.time(9, 0), 2],
 [datetime.time(10, 0), 0],
 [datetime.time(11, 0), 1],
 [datetime.time(12, 30), 0],
 [datetime.time(15, 0), 1],
 [datetime.time(16, 0), 2],
 [datetime.time(18, 0), 3]]

Answer 5

Unless your time resolution is finer than a minute, I would suggest using a map of minutes in the day with a set of jobIds assigned over the time span of each job

For example:

# convert time to minute of the day (assumes24H time, but you can make this your own way)
def toMinute(time): 
    return sum(p*t for p,t in zip(map(int,time.split(":")),(60,1)))

def toTime(minute):
    return f"{minute//60}:{minute%60:02d}"

# booking a job adds it to all minutes covered by its duration
def book(timeMap,jobId,start,duration):
    startMin = toMinute(start)
    for m in range(startMin,startMin+duration):
        timeMap[m].add(jobId)

# unbooking a job removes it from all minutes where it was present
def unbook(timeMap,jobId):
    for s in timeMap:
        s.discard(jobId)

# return time ranges for minutes meeting a given condition
def minuteSpans(timeMap,condition,start="09:00",end="18:00"):
    start,end  = toMinute(start),toMinute(end)
    timeRange  = timeMap[start:end]
    match      = [condition(s) for s in timeRange]
    breaks     = [True] + [a!=b for a,b in zip(match,match[1:])]
    starts     = [i for (i,a),b in zip(enumerate(match),breaks) if b]
    return [(start+s,start+e) for s,e in zip(starts,starts[1:]+[len(match)]) if match[s]]

def timeSpans(timeMap,condition,start="09:00",end="18:00"):
    return [(toTime(s),toTime(e)) for s,e in minuteSpans(timeMap,condition,start,end)]

# availability is ranges of minutes where the number of jobs is less than your capacity
def available(timeMap,start="09:00",end="18:00",maxJobs=5):
    return timeSpans(timeMap,lambda s:len(s)<maxJobs,start,end)

sample usage:

timeMap = [set() for _ in range(1440)]

book(timeMap,"job1","9:45",25)
book(timeMap,"job2","9:30",45)
book(timeMap,"job3","9:00",90)

print(available(timeMap,maxJobs=3))
[('9:00', '9:45'), ('10:10', '18:00')]

print(timeSpans(timeMap,lambda s:"job3" in s))
[('9:00', '10:30')]

With a few adjustments you could even have discontinuous jobs that skip over some periods (eg lunch time). You can also block out some periods by placing fake jobs in them.

If you need to manage job queues individually, you can have separate time maps (one per queue) and combine them into one when you need to have a global picture:

 print(available(timeMap1,maxJobs=1))
 print(available(timeMap2,maxJobs=1))
 print(available(timeMap3,maxJobs=1))

 globalMap = list(set.union(*qs) for qs in zip(timeMap1,timeMap2,timeMap3))
 print(available(globalMap),maxJobs=3)

Put all this into a TimeMap class (instead of individual functions) and you should have a pretty good toolset to work with.

Answer 6

You can use a dedicated class representing a lane that can run jobs. These objects can keep track of jobs and correspondingly of their availability:

import bisect
from datetime import time
from functools import total_ordering
import math


@total_ordering
class TimeSlot:
    def __init__(self, start, stop, lane):
        self.start = start
        self.stop = stop
        self.lane = lane

    def __contains__(self,  other):
        return self.start <= other.start and self.stop >= other.stop

    def __lt__(self, other):
        return (self.start, -self.stop.second) < (other.start, -other.stop.second)

    def __eq__(self, other):
        return (self.start, -self.stop.second) == (other.start, -other.stop.second)

    def __str__(self):
        return f'({self.lane}) {[self.start, self.stop]}'

    __repr__ = __str__


class Lane:
    @total_ordering
    class TimeHorizon:
        def __repr__(self):
            return '...'
        def __lt__(self, other):
            return False
        def __eq__(self, other):
            return False
        @property
        def second(self):
            return math.inf
        @property
        def timestamp(self):
            return math.inf

    time_horizon = TimeHorizon()
    del TimeHorizon

    def __init__(self, start, nr):
        self.nr = nr
        self.availability = [TimeSlot(start, self.time_horizon, self)]

    def add_job(self, job):
        if not isinstance(job, TimeSlot):
            job = TimeSlot(*job, self)
        # We want to bisect_right but only on the start time,
        # so we need to do it manually if they are equal.
        index = bisect.bisect_left(self.availability, job)
        if index < len(self.availability):
            index += (job.start == self.availability[index].start)
        index -= 1  # select the corresponding free slot
        slot = self.availability[index]
        if slot.start > job.start or slot.stop is not self.time_horizon and job.stop > slot.stop:
            raise ValueError('Requested time slot not available')
        if job == slot:
            del self.availability[index]
        elif job.start == slot.start:
            slot.start = job.stop
        elif job.stop == slot.stop:
            slot.stop = job.start
        else:
            slot_end = slot.stop
            slot.stop = job.start
            self.availability.insert(index+1, TimeSlot(job.stop, slot_end, self))

A Lane object can be used as follows:

lane = Lane(start=time(9), nr=1)
print(lane.availability)
lane.add_job([time(11), time(14)])
print(lane.availability)

which outputs:

[(1) [datetime.time(9, 0), ...]]
[(1) [datetime.time(9, 0), datetime.time(11, 0)],
 (1) [datetime.time(14, 0), ...]]

After adding the job, the availability gets updated as well.

Now we could use multiple of these lane objects together to represent a full schedule. Jobs can be added as required and the availability will be updated automatically:

class Schedule:
    def __init__(self, n_lanes, start):
        self.lanes = [Lane(start, nr=i) for i in range(n_lanes)]

    def add_job(self, job):
        for lane in self.lanes:
            try:
                lane.add_job(job)
            except ValueError:
                pass
            else:
                break

Testing on example schedule

from pprint import pprint

# Example jobs from OP.
jobs = [(time(10), time(15)),
        (time(9), time(11)),
        (time(12, 30), time(16)),
        (time(10), time(18))]

schedule = Schedule(3, start=time(9))
for job in jobs:
    schedule.add_job(job)

for lane in schedule.lanes:
    pprint(lane.availability)

which outputs:

[(0) [datetime.time(9, 0), datetime.time(10, 0)],
 (0) [datetime.time(15, 0), ...]]
[(1) [datetime.time(11, 0), datetime.time(12, 30)],
 (1) [datetime.time(16, 0), ...]]
[(2) [datetime.time(9, 0), datetime.time(10, 0)],
 (2) [datetime.time(18, 0), ...]]

Automatic load balancing for jobs

We can create a dedicated tree-like structure which keeps track of time slots of all lanes for selecting the best suited slot when registering a new job. A node in the tree represents a single time slot and its children are all time slots that are contained within that slot. Then, when registering a new job, we can search the tree to find an optimal slot. The tree and the lanes share the same time slots so we only need to adjust slots manually when they are either deleted or new ones are inserted. Here is the relevant code, it is a bit lengthy (just a quick draft):

import itertools as it

class OneStepBuffered:
    """Can back up elements that are consumed by `it.takewhile`.
       From: https://stackoverflow.com/a/30615837/3767239
    """
    _sentinel = object()

    def __init__(self, it):
        self._it = iter(it)
        self._last = self._sentinel
        self._next = self._sentinel

    def __iter__(self):
        return self

    def __next__(self):
        sentinel = self._sentinel
        if self._next is not sentinel:
            next_val, self._next = self._next, sentinel
            return next_val
        try:
            self._last = next(self._it)
            return self._last
        except StopIteration:
            self._last = self._next = sentinel
            raise

    def step_back(self):
        if self._last is self._sentinel:
            raise ValueError("Can't back up a step")
        self._next, self._last = self._last, self._sentinel


class SlotTree:
    def __init__(self, slot, subslots, parent=None):
        self.parent = parent
        self.slot = slot
        self.subslots = []
        slots = OneStepBuffered(subslots)
        for slot in slots:
            subslots = it.takewhile(lambda x: x.stop <= slot.stop, slots)
            self.subslots.append(SlotTree(slot, subslots, self))
            try:
                slots.step_back()
            except ValueError:
                break

    def __str__(self):
        sub_repr = ['\n|   '.join(str(slot).split('\n'))
                    for slot in self.subslots]
        sub_repr = [f'|   {x}' for x in sub_repr]
        sub_repr = '\n'.join(sub_repr)
        sep = '\n' if sub_repr else ''
        return f'{self.slot}{sep}{sub_repr}'

    def find_minimal_containing_slot(self, slot):
        try:
            return min(self.find_containing_slots(slot),
                       key=lambda x: x.slot.stop.second - x.slot.start.second)
        except ValueError:
            raise ValueError('Requested time slot not available') from None

    def find_containing_slots(self, slot):
        for candidate in self.subslots:
            if slot in candidate.slot:
                yield from candidate.find_containing_slots(slot)
                yield candidate

    @classmethod
    def from_slots(cls, slots):
        # Ascending in start time, descending in stop time (secondary).
        return cls(cls.__name__, sorted(slots))


class Schedule:
    def __init__(self, n_lanes, start):
        self.lanes = [Lane(start, i+1) for i in range(n_lanes)]
        self.slots = SlotTree.from_slots(
            s for lane in self.lanes for s in lane.availability)

    def add_job(self, job):
        if not isinstance(job, TimeSlot):
            job = TimeSlot(*job, lane=None)
        # Minimal containing slot is one possible strategy,
        # others can be implemented as well.
        slot = self.slots.find_minimal_containing_slot(job)
        lane = slot.slot.lane
        if job == slot.slot:
            slot.parent.subslots.remove(slot)
        elif job.start > slot.slot.start and job.stop < slot.slot.stop:
            slot.parent.subslots.insert(
                slot.parent.subslots.index(slot) + 1,
                SlotTree(TimeSlot(job.stop, slot.slot.stop, lane), [], slot.parent))
        lane.add_job(job)

Now we can use the Schedule class to automatically assign jobs to lanes and update their availability:

if __name__ == '__main__':
    jobs = [(time(10), time(15)),  # example from OP
            (time(9), time(11)),
            (time(12, 30), time(16)),
            (time(10), time(18))]

    schedule = Schedule(3, start=time(9))
    print(schedule.slots, end='\n\n')
    for job in jobs:
        print(f'Adding {TimeSlot(*job, "new slot")}')
        schedule.add_job(job)
        print(schedule.slots, end='\n\n')

which outputs:

SlotTree
|   (1) [datetime.time(9, 0), ...]
|   (2) [datetime.time(9, 0), ...]
|   (3) [datetime.time(9, 0), ...]

Adding (new slot) [datetime.time(10, 0), datetime.time(15, 0)]
SlotTree
|   (1) [datetime.time(9, 0), datetime.time(10, 0)]
|   (1) [datetime.time(15, 0), ...]
|   (2) [datetime.time(9, 0), ...]
|   (3) [datetime.time(9, 0), ...]

Adding (new slot) [datetime.time(9, 0), datetime.time(11, 0)]
SlotTree
|   (1) [datetime.time(9, 0), datetime.time(10, 0)]
|   (1) [datetime.time(15, 0), ...]
|   (2) [datetime.time(11, 0), ...]
|   (3) [datetime.time(9, 0), ...]

Adding (new slot) [datetime.time(12, 30), datetime.time(16, 0)]
SlotTree
|   (1) [datetime.time(9, 0), datetime.time(10, 0)]
|   (1) [datetime.time(15, 0), ...]
|   (2) [datetime.time(11, 0), datetime.time(12, 30)]
|   (2) [datetime.time(16, 0), ...]
|   (3) [datetime.time(9, 0), ...]

Adding (new slot) [datetime.time(10, 0), datetime.time(18, 0)]
SlotTree
|   (1) [datetime.time(9, 0), datetime.time(10, 0)]
|   (1) [datetime.time(15, 0), ...]
|   (2) [datetime.time(11, 0), datetime.time(12, 30)]
|   (2) [datetime.time(16, 0), ...]
|   (3) [datetime.time(9, 0), datetime.time(10, 0)]
|   (3) [datetime.time(18, 0), ...]

The numbers (i) indicate the lane number and the [] indicate the available time slots on that lane. A ... indicates "open end" (time horizon). As we can see the tree doesn't restructure itself when time slots are adjusted; this would be a possible improvement. Ideally for each new job, the corresponding best fitting time slot would be popped from the tree and then, depending on how the job fits in the slot, an adjusted version and possibly new slots are pushed back to the tree (or none at all if the job fits the slot exactly).

The above examples consider just a single day and time objects, but the code is easy to extend for usage with datetime objects as well.