# How to loop through a list with a set amount of positions and start at the beginning when you reach the end

## Question:

as the title says i need to loop through a list that contains 7 values but i need to start at the beginning each cycle in order to get the right value based off of another value that my code gets, also needs to be done with pure python. I know that there are libraries to do it but i want to not use them. so the idea is that i have a list of the days of the week and whatever value i get for how many days later the time is i need to get the specific day of the week

## Answers:

As suggested by one of the comments, you could use the modulo (`%`

) operator, used to get the remainder of a number divided by another.

It is especially useful to continuously "roll over" an increased index value back from the beginning (or a decreased index back towards the end) when it goes outside the range.

```
# List of the days of the week
days = ['Monday', 'Tuesday', 'Wednesday', 'Thursday', 'Friday', 'Saturday', 'Sunday']
# The total number of days
n_days = len(days)
# Start at a certain day (in this case, Sunday)
start_day = "Sunday"
# Get the corresponding index
start_index = days.index(start_day)
# The number of days later
n = 20
# Get the index of the day of the week n days later:
d_index = (start_index + 20) % n_days
print("Start day:", start_day)
print(n, "days later:", days[d_index])
```

Start day: Sunday 20 days later: Saturday

But if every step matters, then a loop would be necessary:

```
days = ['Monday', 'Tuesday', 'Wednesday', 'Thursday', 'Friday', 'Saturday', 'Sunday']
n_days = len(days)
start_day = "Sunday"
start_index = days.index(start_day)
n = 20
print("Start day:", start_day)
print("nNext", n, "days:")
# Iterate over the next n days
for i in range(1, n + 1):
# Get the index corresponding to the current day.
d_index = (start_index + i) % n_days
print(i, days[d_index])
```

Start day: Sunday Next 20 days: 1 Monday 2 Tuesday 3 Wednesday 4 Thursday 5 Friday 6 Saturday 7 Sunday 8 Monday 9 Tuesday 10 Wednesday 11 Thursday 12 Friday 13 Saturday 14 Sunday 15 Monday 16 Tuesday 17 Wednesday 18 Thursday 19 Friday 20 Saturday

### Written as functions

```
days_of_week = ['Monday', 'Tuesday', 'Wednesday', 'Thursday', 'Friday', 'Saturday', 'Sunday']
n_days = len(days_of_week)
# Outputs the day of the week n days after one given.
def advance_day(start_day, n=1):
start_index = days_of_week.index(start_day)
d_index = (start_index + n) % n_days
return days_of_week[d_index]
# Outputs a list of the next n days after the one given.
def next_n_days(start_day, n):
start_index = days_of_week.index(start_day)
next_days = []
for i in range(1, n + 1):
d_index = (start_index + i) % n_days
next_days.append(days_of_week[d_index])
return next_days
```

Usage:

```
# Get the day of the week 20 days after Sunday.
advance_day("Sunday", 20)
# Get the next 20 days after Sunday (as a list).
next_n_days("Sunday", 20)
```