我有这样的二维列表
s = [
[1, '2017-08-15 13:30:00', 'pickup'],
[1, '2017-08-15 14:00:00', 'dropoff'],
[2, '2017-08-15 14:30:00', 'pickup'],
[3, '2017-08-15 14:35:00', 'pickup'],
[2, '2017-08-15 15:00:00', 'dropoff'],
[4, '2017-08-15 15:05:00', 'pickup'],
[3, '2017-08-15 15:10:00', 'dropoff'],
[4, '2017-08-15 15:40:00', 'dropoff'],
]
我正在尝试创建一个程序，该程序将计算该人活动的时间段，即他实际持有至少一个包裹的时间。我无法弄清楚应该对重叠包应用什么逻辑，因为同一时间段将被多次添加。任何帮助表示赞赏。
编辑： 这是我目前拥有的代码
for i in range(len(s)-1):
if s[i+1][0] == s[i][0]:
print(s[i])
diff = datetime.datetime.strptime(s[i+1][1], datetimeFormat) - datetime.datetime.strptime(s[i][1], datetimeFormat)
print(diff)
我可以得到两次之间的差，但我需要处理重叠的时间。
答案 0 :(得分：3)
我刚刚快速输入了一些内容，也许您从这里得到了一个主意
from datetime import datetime, date
# stores each time fram between pickup and dropoff
time = []
# inits a global counter, that is updated dependend on you array
counter = 0
# inits global time constants that are updated, based on logic
starttime = datetime(2009, 10, 5, 18, 0)
endtime = datetime(2009, 10, 5, 18, 0)
for i in range(len(s)):
# check if initial pickup, if 0 that means inital pickup, therefore update time
if(counter == 0):
starttime = datetime.strptime(s[i][1],'%Y-%m-%d %H:%M:%S')
# in case pickup add 1 to counter to indicate the number of packages on hand
if(s[i][2] == 'pickup'):
counter = counter + 1
# in case dropoff subtract 1 from counter to update packages on hand
else:
counter = counter - 1
# in case counter reaches 0 meaning no more packages, therefore measure time differences
if(counter == 0):
endtime = datetime.strptime(s[i][1],'%Y-%m-%d %H:%M:%S')
print(endtime - starttime)
time.append(endtime - starttime)
答案 1 :(得分：1)
from datetime import datetime
from datetime import time
from datetime import timedelta
s = [
# package_name time event
[1, '2017-08-15 13:30:00', 'pickup'], #begin
[1, '2017-08-15 14:00:00', 'dropoff'], # end (00:30:00)
[2, '2017-08-15 14:30:00', 'pickup'], #begin
[3, '2017-08-15 14:35:00', 'pickup'],
[2, '2017-08-15 15:00:00', 'dropoff'],
[4, '2017-08-15 15:05:00', 'pickup'],
[3, '2017-08-15 15:10:00', 'dropoff'],
[4, '2017-08-15 15:40:00', 'dropoff'], #end (01:10:00)
] #total_active_time 1:40:00
datetime_format = '%Y-%m-%d %H:%M:%S'
pckg_inHand = 0
st_time = ''
total_active_time = timedelta(days=0, seconds=0, microseconds=0, milliseconds=0, minutes=0, hours=0, weeks=0)
for log in s:
if log[2] == 'pickup':
if pckg_inHand == 0: #person state is not_active
st_time = log[1] #set active state begin time
pckg_inHand += 1
else: #it's a dropoff
pckg_inHand -= 1
if pckg_inHand == 0:
start = datetime.strptime(st_time, datetime_format)
end = datetime.strptime(log[1], datetime_format)
total_active_time = total_active_time + (end - start)
print(total_active_time)
答案 2 :(得分：1)
尝试以下代码：
from datetime import datetime
def calculateTime(frm, to):
p = '%Y-%m-%d %H:%M:%S'
epoch = datetime(1970, 1, 1)
frmTime = (datetime.strptime(frm, p) - epoch).total_seconds()
toTime = (datetime.strptime(to, p) - epoch).total_seconds()
return toTime - frmTime
def printTime(t, form='H:M'):
h = int(t/(60*60))
m = int(60 * (t/(60*60) - h))
s = int(t - ((h * 60 * 60) + (m * 60)))
if form == 'H:M':
print h,'hrs ',m,'mins'
if form == 'H:M:S':
print h,'hrs ',m,'mins',s,' secs'
#changed
def totalActiveTime(lst):
orders = []
activeTime = 0
prevTime = None
for i in lst:
if i[2] == 'pickup':
orders.append(i[0])
if prevTime == None:
prevTime = i[1]
elif i[2] == 'dropoff':
orders.remove(i[0])
if len(orders) == 0:
activeTime += calculateTime(prevTime, i[1])
prevTime = None
return activeTime
s = [
[1, '2017-08-15 13:30:00', 'pickup'],
[1, '2017-08-15 14:00:00', 'dropoff'],
[2, '2017-08-15 14:30:00', 'pickup'],
[3, '2017-08-15 14:35:00', 'pickup'],
[2, '2017-08-15 15:00:00', 'dropoff'],
[4, '2017-08-15 15:05:00', 'pickup'],
[3, '2017-08-15 15:10:00', 'dropoff'],
[4, '2017-08-15 15:40:00', 'dropoff'],
]
printTime(totalActiveTime(s))
1 hrs 40 mins