在不手动检查的情况下格式化时差时省略零部分的好方法是什么？

Question

我想像这样向用户显示一个完整的句子：

The difference is 5 days, 3 hours, 16 minutes, 10 seconds, and 150 microseconds.

但排除零项。

例如：91天，23:00:00打印为：

The difference is 91 days and 23 hours.

我现在能想到的唯一方法是创建一堆 if 语句，正如我在下面发布的那样。我认为这可能过于复杂，我希望有更简单的方法。

# prompts the user to enter the date following the order
d1 = input('Please enter the date time (YYYY-mm-dd hour:minute:second:microsecond): ')
m_d1 = datetime.strptime(d1, '%Y-%m-%d %H:%M:%S:%f')  # a datetime variable with the entered info
# prompts the user to enter another date following the order
d2 = input('Please enter the date time (YYYY-mm-dd hour:minute:second:microsecond): ')
m_d2 = datetime.strptime(d2, '%Y-%m-%d %H:%M:%S:%f')  # another datetime variable with the entered info
if m_d1 > m_d2:
    result_date = m_d1 - m_d2
else:
    result_date = m_d2 - m_d1

# total difference in number of days with fraction
difference_days = result_date.total_seconds() / timedelta(days=1).total_seconds()
# total difference in number of hours with fraction
difference_hours = result_date.total_seconds() / timedelta(hours=1).total_seconds()
# total difference in number of minutes with fraction
difference_minutes = result_date.total_seconds() / timedelta(minutes=1).total_seconds()
# total difference in number of seconds with fraction
difference_seconds = result_date.total_seconds()
# total difference in number of microseconds with fraction
difference_microseconds = result_date.total_seconds() / timedelta(microseconds=1).total_seconds()

# now, A complete sentence that breaks everything down to the
# correct units and excluding any 0s.
if difference_days.is_integer():
    print(f'The difference is {int(difference_days)} days.')
else:
    hours = (Decimal(difference_days) - int(difference_days)) * 24
    if difference_hours.is_integer():
        if difference_days > 1:
            print(f'The difference is {int(difference_days)} days and {int(hours)} hours.')
        else:
            print(f'The difference is {int(hours)} hours.')
    else:
        minutes = (hours - int(hours)) * 60
        if difference_minutes.is_integer():
            if difference_days < 1 and hours < 1:
                print(f'The difference is {int(minutes)} minutes.')
            elif difference_days > 1 and hours < 1:
                print(f'The difference is {int(difference_days)} days and {int(minutes)} minutes.')
            else:
                print(
                    f'The difference is {int(difference_days)} days, {int(hours)} hours, and {int(minutes)} minutes.')
        else:
            seconds = (minutes - int(minutes)) * 60
            if difference_seconds.is_integer():
                if difference_days < 1 and hours < 1 and minutes < 1:
                    print(f'The difference is {int(seconds)} seconds.')
                elif difference_days > 1 and hours < 1 and minutes < 1:
                    print(f'The difference is {int(difference_days)} days and {int(seconds)} seconds.')
                elif difference_days < 1 and hours > 1 and minutes < 1:
                    print(f'The difference is {int(hours)} hours and {int(seconds)} seconds.')
                else:
                    # ... other code

Answer 1

使用 DRY（不要重复自己）原则。不用每单位时间都重写代码，因为都是一样的：

non_zero_counts = {}
rest = result_date
for duration in ["days", "hours", "minutes", "seconds", "microseconds"]:
    td = timedelta(**{duration: 1})
    count = int(rest / td)
    rest = rest - count * td
    if count:
        non_zero_counts[duration] = count

结果是字典：

non_zero_counts
Out[27]: {'days': 31, 'hours': 3, 'seconds': 34, 'microseconds': 50000}

我将处理所有特殊情况，让您掌握正确的语言。请随意编辑此答案以添加您的解决方案以形成句子本身。

编辑

造句：

if non_zero_counts:
    result = ', '.join(f'{value} {key}' for key, value in 
    non_zero_counts.items())
    print(f'The difference is {result}')
else:
    print('The two dates are the same.')

输入：

Please enter the date time (YYYY-mm-dd hour:minute:second:microsecond): 2022-08-04 15:30:20:12470
Please enter the date time (YYYY-mm-dd hour:minute:second:microsecond): 2022-09-08 12:44:33:12348

输出：

The difference is 34 days, 21 hours, 14 minutes, 12 seconds, 998780 microseconds

在不手动检查的情况下格式化时差时省略零部分的好方法是什么？

What are good ways to omit the zero parts when formatting a time difference without manual checking?

python

datetime

timedelta