There is a list of date strings, as follows:
lst = ['2017-06-01', '2017-06-08', '2017-06-15', '2017-06-22', '2017-06-29', ...]
Seeking s = ['2017-06-09']Which date string between and lst is most similar
Idea 1: Convert the values ????of s and lst into dates, traverse and compare the difference in seconds, and the smallest one is the date string you are looking for.
Is there a better way to achieve this? ?
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I will give you an idea for your reference
lst.append(s)
lst.sort()
num=lst.index(s)
Then compare lst[num-1] and lst[num+1] The smaller number of seconds difference is the result, so there is no need to traverse and calculate the timestamp.
If you think it’s good, please give it to Zanga and adopt it.
Convert the date into an integer by removing -, and then subtract the dates in s respectively. The number with the smallest absolute value is the closest date.
# Python code
lst = ['2017-06-01', '2017-06-08', '2017-06-15', '2017-06-22', '2017-06-29']
s = ['2017-06-09']
date = [''.join(x.split('-')) for x in lst]
datetoint = [int(x) for x in date]
gaps = [abs(x - int(''.join(s[0].split('-')))) for x in datetoint]
mostrecentdate = lst[gaps.index(min(gaps))]
print(mostrecentdate)
I feel that lz means not to traverse lst. Whether it is sort or subtraction, traversal actually occurs
I should use the dichotomy method. This is probably what it means
i = 0
j = len(list)
while True:
index = (i + j) / 2
if s > lst[index]:
i = index
else:
j = index
continueJust read it as pseudo code, that’s what it means anyway, this way the number of traversals is the least.
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