Python：检查嵌套列表中的下一个列表

Question

month = [[0,1,0,1,0,1,1], [1,1,1,1,1,1,1], [0,0,0,0,0,0,1], [1,0,0,0,0,0,1]]

each nested list corresponds to a week and each 1 corresponds to an "event" and each event has a random length between 2-14
Example:
I want to enter the event at month[0][5] with a length of 6 days how do I make it so that the next 6 days all the "events" (including the events that cross the current week) turn 0?

Expected Output:

month = [[0,1,0,1,0,1,0], [0,0,0,0,0,1,1], [0,0,0,0,0,0,1], [1,0,0,0,0,0,1]]

Answer 1

You can run a nested for loop starting from i = 0 and j =5 and let it run until j = 7 at that point increment i by 1. and loop until you complete iterations equal to the length of the event while replacing the values with 0.

    month[0][5] = 6
    for i in range(0,len(month)):
       for j in range(6,8):
          month[i][j] = 0

Answer 2

here is one way:

week, day = 0, 5 #starting point eg: month[0][5]
event_days = 6
for _ in range(event_days):
    week, day = (week + 1, 0) if day == len(month[week])-1 else (week, day + 1)
    month[week][day] = 0

output:

>>
[
  [0, 1, 0, 1, 0, 1, 0]
, [0, 0, 0, 0, 0, 1, 1]
, [0, 0, 0, 0, 0, 0, 1]
, [1, 0, 0, 0, 0, 0, 1]
]

Answer 3

instead of representing a month as a nested list you could also flatten your month into days. With this list most operations should be much more simple. Whenever needed you can simple select a week by days[7*week:7*(week+1)]

days = [0,1,0,1,0,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,1,1,0,0,0,0,0,1]

# set an event
event_week = 0
event_day_of_week = 5
event_days = 6

event_begin = 7 * event_week + event_day_of_week + 1
event_end = event_begin + event_days
days[event_begin:event_end] = [0] * event_days

# get a week
week_begin = 1*7
week_end = 2*7
second_week = days[week_begin:week_end]

print(second_week)
>>>[0, 0, 0, 0, 0, 1, 1]