我有一個這樣的清單：
dates = [
datetime.date(2014, 11, 24),
datetime.date(2014, 11, 25),
datetime.date(2014, 11, 26),
# datetime.date(2014, 11, 27), # This one is missing
datetime.date(2014, 11, 28),
datetime.date(2014, 11, 29),
datetime.date(2014, 11, 30),
datetime.date(2014, 12, 1)]
我正在嘗試使用此 expr 查找開始日期和結束日期之間缺少的日期：
date_set = {dates[0] timedelta(x) for x in range((dates[-1] - dates[0]).days)}
奇怪的是，它拋出了一個錯誤——它無法訪問dates變數。但是這個運算式運行良好：
date_set = {date(2015,2,11) timedelta(x) for x in range((dates[-1] - dates[0]).days)}
我寫了一個滿足我想要的運算式：
def find_missing_dates(dates: list[date]) -> list[date]:
"""Find the missing dates in a list of dates (that should already be sorted)."""
date_set = {(first_date timedelta(x)) for first_date, x in zip([dates[0]] * len(dates), range((dates[-1] - dates[0]).days))}
missing = sorted(date_set - set(dates))
return missing
這是一個丑陋的表達方式，迫使我用相同的變數填充第二個串列。有人有更干凈的表達嗎？
uj5u.com熱心網友回復：
如果您dates已排序，則只需對其進行迭代并將其間的日期添加到新串列中。我已經在此評論中提供了可能的單行解決方案。
from datetime import date, timedelta
dates = [
date(2014, 11, 24), date(2014, 11, 25), date(2014, 11, 26),
date(2014, 11, 28), date(2014, 11, 29), date(2014, 11, 30),
date(2014, 12, 1)
]
missing = [d timedelta(days=j) for i, d in enumerate(dates[:-1], 1) for j in range(1, (dates[i] - d).days)]
您可以使用常規 for 回圈來做到這一點：
from datetime import date, timedelta
dates = [
date(2014, 11, 24), date(2014, 11, 25), date(2014, 11, 26),
date(2014, 11, 28), date(2014, 11, 29), date(2014, 11, 30),
date(2014, 12, 1)
]
missing = []
for next_index, current_date in enumerate(dates[:-1], 1):
for days_diff in range(1, (dates[next_index] - current_date).days):
missing.append(current_date timedelta(days=days_diff))
uj5u.com熱心網友回復：
像下面這樣的東西。找到最小值和最大值。從 min 到 max 回圈并查看缺少哪個日期。
from datetime import timedelta, date
dates = [
date(2014, 11, 21),
date(2014, 11, 24),
date(2014, 11, 25),
date(2014, 11, 26),
date(2014, 11, 27),
date(2014, 11, 28),
date(2014, 11, 29),
date(2014, 11, 30),
date(2014, 12, 1)
]
_min = min(dates)
_max = max(dates)
missing = []
while _min < _max:
if _min not in dates:
missing.append(_min)
_min = timedelta(days=1)
print(missing)
輸出
[datetime.date(2014, 11, 22), datetime.date(2014, 11, 23)]
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