计算两个日期之间的星期几小时对
Counting day-of-week-hour pairs between two dates
考虑以下 星期几小时 对的列表，格式为 24H：
{
'Mon': [9,23],
'Thu': [12, 13, 14],
'Tue': [11, 12, 14],
'Wed': [11, 12, 13, 14]
'Fri': [13],
'Sat': [],
'Sun': [],
}
和两个时间点，例如：
开始：
datetime.datetime(2015, 7, 22, 17, 58, 54, 746784)
结束:
datetime.datetime(2015, 8, 30, 10, 22, 36, 363912)
假设我们需要知道上面指定的 day-of-week-hour 对中的每一个的这两个日期时间之间有多少小时（向上或向下舍入） .
如何解决 Python 中的这个问题？我在一般的详细级别上探索了 timedelta 和 relativedelta，但我没有找到任何提供接近此内容的内容。
为简单起见，我们可以假设所有内容都指代同一时区。
也许一个更简单的问题是专注于单个天-小时对，例如两个任意日期时间之间有多少个 Wednesdays: 14？
因此，如果我对您的问题的理解正确，我会先找到时间范围内 "hour" 的第一次出现，然后逐周查找下一次出现的情况。像这样：
#!/usr/bin/python
from __future__ import print_function
import datetime
import dateutil.relativedelta
def hours_between(start, end, weekday, hour):
first = start + dateutil.relativedelta.relativedelta(
weekday=weekday, hour=hour,
minute=0, second=0, microsecond=0)
week = dateutil.relativedelta.relativedelta(weeks=1)
all_dates = []
d = first
while d < end:
all_dates.append(d)
d += week
return all_dates
def main():
start = datetime.datetime(2015, 7, 22, 17, 58, 54, 746784)
end = datetime.datetime(2015, 8, 30, 10, 22, 36, 363912)
all_dates = hours_between(start, end, dateutil.relativedelta.WE, 14)
print(all_dates)
print(len(all_dates))
main()
也许我没有完全理解你的问题，但你可以得到两个日期之间的所有时间，然后计算两个日期之间每小时和每天出现的次数：
from datetime import datetime
from dateutil import rrule,parser
d={
'Mon': [9, 23],
'Thu': [12, 13, 14],
'Tue': [11, 12, 14],
'Wed': [11, 12, 13, 14],
'Fri': [13],
'Sat': [],
'Sun': [],
}
st = datetime(2015, 7, 22, 17, 58, 54, 746784)
ed = datetime(2015, 8, 30, 10, 22, 36, 363912)
dates = list(rrule.rrule(rrule.HOURLY,
dtstart=parser.parse(st.strftime("%Y-%m-%d %H:%M:%S")),
until=parser.parse(ed.strftime("%Y-%m-%d %H:%M:%S"))))
days = {"Mon":0,"Tue": 1,"Wed":2,"Thu": 3,"Fri":4,"Sat":5,"Sun":6}
for k, val in d.items():
for v in val:
print("day: {} hour: {}".format(k,v))
day = days[k]
print(sum((v == dt.hour and dt.weekday() == day) for dt in dates))
输出：
day: Wed hour: 11
5
day: Wed hour: 12
5
day: Wed hour: 13
5
day: Wed hour: 14
5
day: Fri hour: 13
6
day: Tue hour: 11
5
day: Tue hour: 12
5
day: Tue hour: 14
5
day: Mon hour: 9
6
day: Mon hour: 23
5
day: Thu hour: 12
5
day: Thu hour: 13
5
day: Thu hour: 14
5
不确定您是想要每个列表中所有小时的总和还是每个单独小时的总和，但无论哪种方式，您都可以将输出存储在字典中。
counts = {'Thu':{}, 'Sun':{}, 'Fri':{}, 'Mon':{}, 'Tue':{}, 'Sat':{}, 'Wed':{}}
for k, val in d.items():
for v in val:
day = days[k]
sm = sum((v == dt.hour and dt.weekday() == day) for dt in dates)
counts[k][v] = sm
from pprint import pprint as pp
pp(counts)
输出：
{'Fri': {13: 6},
'Mon': {9: 5, 23: 5},
'Sat': {},
'Sun': {},
'Thu': {12: 6, 13: 6, 14: 6},
'Tue': {11: 5, 12: 5, 14: 5},
'Wed': {11: 5, 12: 5, 13: 5, 14: 5}}
也许是这样的：
from calendar import day_abbr
from datetime import datetime, timedelta
def solve(start, end, data):
days = list(day_abbr)
output = dict.fromkeys(days, 0)
while start <= end:
day = days[start.weekday()]
if start.hour in data[day]:
output[day] += 1
start = start + timedelta(minutes=60)
return output
data = {
'Mon': [9, 23],
'Thu': [12, 13, 14],
'Tue': [11, 12, 14],
'Wed': [11, 12, 13, 14],
'Fri': [13],
'Sat': [],
'Sun': [],
}
start = datetime(2015, 7, 22, 17, 58, 54, 746784)
end = datetime(2015, 8, 30, 10, 22, 36, 363912)
print solve(start, end, data)
# {'Wed': 20, 'Sun': 0, 'Fri': 6, 'Tue': 15, 'Mon': 10, 'Thu': 18, 'Sat': 0}
按小时获取每天的计数：
from calendar import day_abbr
from collections import defaultdict
from datetime import datetime, timedelta
from pprint import pprint
def solve(start, end, data):
days = list(day_abbr)
output = defaultdict(lambda: defaultdict(int))
while start <= end:
day = days[start.weekday()]
if start.hour in data[day]:
output[day][start.hour] += 1
start = start + timedelta(minutes=60)
return {k: dict(v) for k, v in output.items()}
data = {
'Mon': [9, 23],
'Thu': [12, 13, 14],
'Tue': [11, 12, 14],
'Wed': [11, 12, 13, 14],
'Fri': [13],
'Sat': [],
'Sun': [],
}
start = datetime(2015, 7, 22, 17, 58, 54, 746784)
end = datetime(2015, 8, 30, 10, 22, 36, 363912)
pprint(solve(start, end, data))
# output
{'Fri': {13: 6},
'Mon': {9: 5, 23: 5},
'Thu': {12: 6, 13: 6, 14: 6},
'Tue': {11: 5, 12: 5, 14: 5},
'Wed': {11: 5, 12: 5, 13: 5, 14: 5}}
这是一个带循环的解决方案 datetime:
import datetime
pairs = {1: [9,23],
2: [11, 12, 14],
3: [11, 12, 13, 14],
4: [12, 13, 14],
5: [13],
6: [],
7: []
}
start = datetime.datetime(2015, 7, 22, 17, 58, 54, 746784)
end = datetime.datetime(2015, 8, 30, 10, 22, 36, 363912)
result={}
for d,hl in pairs.items():
for h in hl:
result[(d,h)] = 0
for diff in range((end-start).days*24):
comp = start + datetime.timedelta(hours=diff)
if comp.isoweekday() == d and comp.hour == h:
result[(d,h)] += 1
>>> result
{(3, 12): 5, (5, 13): 6, (3, 13): 5, (1, 23): 5, (2, 11): 5, (3, 11): 5, (4, 14): 6, (4, 13): 6, (4, 12): 6, (2, 12): 5, (2, 14): 5, (3, 14): 5, (1, 9): 5}
我也会尝试 timestamp() 和 % 的解决方案。
这是另一种算法解决方案：
import datetime
pairs = {1: [9,23],
2: [11, 12, 14],
3: [11, 12, 13, 14],
4: [12, 13, 14],
5: [13],
6: [],
7: []
}
start = datetime.datetime(2015, 7, 22, 17, 58, 54, 746784)
end = datetime.datetime(2015, 8, 30, 10, 22, 36, 363912)
result={}
weeks = (end-start).days//7
for d,hl in pairs.items():
for h in hl:
initial = weeks
if d > start.isoweekday() or (
d == start.isoweekday() and h >= start.hour):
initial += 1
result[(d,h)] = initial
>>> for k in sorted(result):
... print(k, result[k])
...
(1, 9) 5
(1, 23) 5
(2, 11) 5
(2, 12) 5
(2, 14) 5
(3, 11) 5
(3, 12) 5
(3, 13) 5
(3, 14) 5
(4, 12) 6
(4, 13) 6
(4, 14) 6
(5, 13) 6
考虑以下 星期几小时 对的列表，格式为 24H：
{
'Mon': [9,23],
'Thu': [12, 13, 14],
'Tue': [11, 12, 14],
'Wed': [11, 12, 13, 14]
'Fri': [13],
'Sat': [],
'Sun': [],
}
和两个时间点，例如：
开始：
datetime.datetime(2015, 7, 22, 17, 58, 54, 746784)结束:
datetime.datetime(2015, 8, 30, 10, 22, 36, 363912)
假设我们需要知道上面指定的 day-of-week-hour 对中的每一个的这两个日期时间之间有多少小时（向上或向下舍入） .
如何解决 Python 中的这个问题？我在一般的详细级别上探索了 timedelta 和 relativedelta，但我没有找到任何提供接近此内容的内容。
为简单起见，我们可以假设所有内容都指代同一时区。
也许一个更简单的问题是专注于单个天-小时对，例如两个任意日期时间之间有多少个 Wednesdays: 14？
因此，如果我对您的问题的理解正确，我会先找到时间范围内 "hour" 的第一次出现，然后逐周查找下一次出现的情况。像这样：
#!/usr/bin/python
from __future__ import print_function
import datetime
import dateutil.relativedelta
def hours_between(start, end, weekday, hour):
first = start + dateutil.relativedelta.relativedelta(
weekday=weekday, hour=hour,
minute=0, second=0, microsecond=0)
week = dateutil.relativedelta.relativedelta(weeks=1)
all_dates = []
d = first
while d < end:
all_dates.append(d)
d += week
return all_dates
def main():
start = datetime.datetime(2015, 7, 22, 17, 58, 54, 746784)
end = datetime.datetime(2015, 8, 30, 10, 22, 36, 363912)
all_dates = hours_between(start, end, dateutil.relativedelta.WE, 14)
print(all_dates)
print(len(all_dates))
main()
也许我没有完全理解你的问题，但你可以得到两个日期之间的所有时间，然后计算两个日期之间每小时和每天出现的次数：
from datetime import datetime
from dateutil import rrule,parser
d={
'Mon': [9, 23],
'Thu': [12, 13, 14],
'Tue': [11, 12, 14],
'Wed': [11, 12, 13, 14],
'Fri': [13],
'Sat': [],
'Sun': [],
}
st = datetime(2015, 7, 22, 17, 58, 54, 746784)
ed = datetime(2015, 8, 30, 10, 22, 36, 363912)
dates = list(rrule.rrule(rrule.HOURLY,
dtstart=parser.parse(st.strftime("%Y-%m-%d %H:%M:%S")),
until=parser.parse(ed.strftime("%Y-%m-%d %H:%M:%S"))))
days = {"Mon":0,"Tue": 1,"Wed":2,"Thu": 3,"Fri":4,"Sat":5,"Sun":6}
for k, val in d.items():
for v in val:
print("day: {} hour: {}".format(k,v))
day = days[k]
print(sum((v == dt.hour and dt.weekday() == day) for dt in dates))
输出：
day: Wed hour: 11
5
day: Wed hour: 12
5
day: Wed hour: 13
5
day: Wed hour: 14
5
day: Fri hour: 13
6
day: Tue hour: 11
5
day: Tue hour: 12
5
day: Tue hour: 14
5
day: Mon hour: 9
6
day: Mon hour: 23
5
day: Thu hour: 12
5
day: Thu hour: 13
5
day: Thu hour: 14
5
不确定您是想要每个列表中所有小时的总和还是每个单独小时的总和，但无论哪种方式，您都可以将输出存储在字典中。
counts = {'Thu':{}, 'Sun':{}, 'Fri':{}, 'Mon':{}, 'Tue':{}, 'Sat':{}, 'Wed':{}}
for k, val in d.items():
for v in val:
day = days[k]
sm = sum((v == dt.hour and dt.weekday() == day) for dt in dates)
counts[k][v] = sm
from pprint import pprint as pp
pp(counts)
输出：
{'Fri': {13: 6},
'Mon': {9: 5, 23: 5},
'Sat': {},
'Sun': {},
'Thu': {12: 6, 13: 6, 14: 6},
'Tue': {11: 5, 12: 5, 14: 5},
'Wed': {11: 5, 12: 5, 13: 5, 14: 5}}
也许是这样的：
from calendar import day_abbr
from datetime import datetime, timedelta
def solve(start, end, data):
days = list(day_abbr)
output = dict.fromkeys(days, 0)
while start <= end:
day = days[start.weekday()]
if start.hour in data[day]:
output[day] += 1
start = start + timedelta(minutes=60)
return output
data = {
'Mon': [9, 23],
'Thu': [12, 13, 14],
'Tue': [11, 12, 14],
'Wed': [11, 12, 13, 14],
'Fri': [13],
'Sat': [],
'Sun': [],
}
start = datetime(2015, 7, 22, 17, 58, 54, 746784)
end = datetime(2015, 8, 30, 10, 22, 36, 363912)
print solve(start, end, data)
# {'Wed': 20, 'Sun': 0, 'Fri': 6, 'Tue': 15, 'Mon': 10, 'Thu': 18, 'Sat': 0}
按小时获取每天的计数：
from calendar import day_abbr
from collections import defaultdict
from datetime import datetime, timedelta
from pprint import pprint
def solve(start, end, data):
days = list(day_abbr)
output = defaultdict(lambda: defaultdict(int))
while start <= end:
day = days[start.weekday()]
if start.hour in data[day]:
output[day][start.hour] += 1
start = start + timedelta(minutes=60)
return {k: dict(v) for k, v in output.items()}
data = {
'Mon': [9, 23],
'Thu': [12, 13, 14],
'Tue': [11, 12, 14],
'Wed': [11, 12, 13, 14],
'Fri': [13],
'Sat': [],
'Sun': [],
}
start = datetime(2015, 7, 22, 17, 58, 54, 746784)
end = datetime(2015, 8, 30, 10, 22, 36, 363912)
pprint(solve(start, end, data))
# output
{'Fri': {13: 6},
'Mon': {9: 5, 23: 5},
'Thu': {12: 6, 13: 6, 14: 6},
'Tue': {11: 5, 12: 5, 14: 5},
'Wed': {11: 5, 12: 5, 13: 5, 14: 5}}
这是一个带循环的解决方案 datetime:
import datetime
pairs = {1: [9,23],
2: [11, 12, 14],
3: [11, 12, 13, 14],
4: [12, 13, 14],
5: [13],
6: [],
7: []
}
start = datetime.datetime(2015, 7, 22, 17, 58, 54, 746784)
end = datetime.datetime(2015, 8, 30, 10, 22, 36, 363912)
result={}
for d,hl in pairs.items():
for h in hl:
result[(d,h)] = 0
for diff in range((end-start).days*24):
comp = start + datetime.timedelta(hours=diff)
if comp.isoweekday() == d and comp.hour == h:
result[(d,h)] += 1
>>> result
{(3, 12): 5, (5, 13): 6, (3, 13): 5, (1, 23): 5, (2, 11): 5, (3, 11): 5, (4, 14): 6, (4, 13): 6, (4, 12): 6, (2, 12): 5, (2, 14): 5, (3, 14): 5, (1, 9): 5}
我也会尝试 timestamp() 和 % 的解决方案。
这是另一种算法解决方案：
import datetime
pairs = {1: [9,23],
2: [11, 12, 14],
3: [11, 12, 13, 14],
4: [12, 13, 14],
5: [13],
6: [],
7: []
}
start = datetime.datetime(2015, 7, 22, 17, 58, 54, 746784)
end = datetime.datetime(2015, 8, 30, 10, 22, 36, 363912)
result={}
weeks = (end-start).days//7
for d,hl in pairs.items():
for h in hl:
initial = weeks
if d > start.isoweekday() or (
d == start.isoweekday() and h >= start.hour):
initial += 1
result[(d,h)] = initial
>>> for k in sorted(result):
... print(k, result[k])
...
(1, 9) 5
(1, 23) 5
(2, 11) 5
(2, 12) 5
(2, 14) 5
(3, 11) 5
(3, 12) 5
(3, 13) 5
(3, 14) 5
(4, 12) 6
(4, 13) 6
(4, 14) 6
(5, 13) 6