如何正确处理 Python 时钟中的累加器？

Question

我正在用 Python 编写一个可以启动、停止和重置的小时钟对象。我很难知道如何正确跟踪时间的累积。现在，更新过程的编写方式，时钟累积了太多时间（因为调用了更新方法）。我不确定如何编写它以累积正确的时间。

import time
import datetime

class Clock(object):

    def __init__(
        self,
        name  = None,
        start = True
        ):
        self._name  = name
        self._start = start # Boolean start clock on instantiation
        # If a global clock list is detected, add a clock instance to it.
        if "clocks" in globals():
            clocks.add(self)
        self.reset()
        if self._start:
            self.start()

    def start(self):
        self._startTime = datetime.datetime.utcnow()

    def stop(self):
        self._startTime = None

    # Update the clock accumulator.
    def update(self):
        self.accumulator += (
            datetime.datetime.utcnow() - self._startTime
        )

    def reset(self):
        self.accumulator = datetime.timedelta(0)
        self._startTime  = None

    # If the clock has a start time, add the difference between now and the
    # start time to the accumulator and return the accumulation. If the clock
    # does not have a start time, return the accumulation.
    def elapsed(self):
        if self._startTime:
            self.update()
        return(self.accumulator)

    def time(self):
        return(self.elapsed().total_seconds())

clock = Clock()
print clock.time()
print "starting"; clock.start()
for i in range(4):
    time.sleep(3)
    print clock.time()
print "stopping"; clock.stop()
for i in range(4):
    time.sleep(3)
    print clock.time()
print "starting"; clock.start()
for i in range(4):
    time.sleep(3)
    print clock.time()
print "resetting"; clock.reset()
print clock.time()

Answer 1

如果您只是在update 末尾添加对self.start() 的调用，那么我认为它会按照您的预期方式工作。

您看到当前代码问题的原因是您每次更新时都将总偏移量从 _startTime 添加到 now()，但您只想要 now() 和计时器启动时间之间的差异第一个电话。在随后的调用中，您需要 now() 和上一次调用 update 之间的差异。