Finding the closest and furthest value to a list of strings

Question

I've got a list of times and a current time. I'm trying to find the closest and furthest time to the current time. I find it difficult to solve such tasks in strings.

However, I can print the minimum and maximum time in the list of strings using min and max functions.

I saw @kennytm posting a similar solution to the problem using: min(myList, key=lambda x:abs(x-myNumber)), but I'm not entirely sure if that works on strings.

Can anyone provide a solution to my question?

current_time = '09:00'
time = ['09:30','11:50','11:55','11:55','12:00','12:10','12:15','12:25','12:35','12:50','12:55', '13:00']

print(min(time))
# 09:30
print(max(time))
# 13:00

Checkout [datetime](https://docs.python.org/3/library/datetime.html). You could convert your strings to datetime objects using `strftime` — Jacques, Jul 03 '21 at 22:01
@daniboy000 Well, I tried something similar to here: https://stackoverflow.com/questions/12141150/from-list-of-integers-get-number-closest-to-a-given-value and I was trying to convert a string to an integer but that wouldn't because of the time format of elements in the list. — B A C H A S H, Jul 03 '21 at 22:03

score 1 · Answer 1 · answered Jul 03 '21 at 22:18

When you convert the dates to a datetime object you can get the closest time with min(time,key=lambda x : abs(current_time-x)):

from datetime import datetime

current_time = '09:00'
time = ['09:30','11:50','11:55','11:55','12:00','12:10','12:15','12:25','12:35','12:50','12:55', '13:00']

def get_time(time):
  return datetime.strptime(time, '%H:%M')

current_time = get_time(current_time)
time = [get_time(i) for i in time]

print(min(time,key=lambda x : abs(current_time-x)).strftime('%H:%M'))

score 1 · Answer 2 · answered Jul 03 '21 at 22:21

Convert the hh:mm format to a single integer, for the current time and the list of times. Then iterate through the integers, and track the times using the original list of time.

def hhmm2int(t):
    hour_min = t.split(':')
    hours_as_minutes = int(hour_min[0]) * 60
    minutes = int(hour_min[1])
    return hours_as_minutes + minutes


current_time = '09:00'
time = ['09:30', '11:50', '11:55', '11:55', '12:00', '12:10', '12:15', '12:25', '12:35', '12:50', '12:55', '13:00']

current_time = hhmm2int(current_time)
time_as_int = [hhmm2int(t) for t in time]

min_delta = float('inf')
max_delta = 0
closest_time = ''
furthest_time = ''
for i in range(len(time_as_int)):
    delta = abs(time_as_int[i] - current_time)
    if delta < min_delta:
        min_delta = delta
        closest_time = time[i]
    if delta > max_delta:
        max_delta = delta
        furthest_time = time[i]

print(closest_time, furthest_time)

Corralien · Answer 3 · 2021-07-03T22:37:17.323

0

A simple solution:

diff_time = lambda t: abs(int(current_time.replace(':', '')) \
                        - int(t.replace(':', '')))

From 00:00 (int: 0) to 23:59 (int: 2359), time is already lexicographical ordered and the order is maintained while converting to integer.

>>> min(time, key=diff_time)
'09:30'

>>> max(time, key=diff_time)
'13:00'

edited Jul 03 '21 at 22:37

answered Jul 03 '21 at 22:31

Corralien

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Finding the closest and furthest value to a list of strings

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