(python) sorting date without using build in function

Question

python sorting date without using build in function sort() , but the expect i want is

['2010-01-12', '2010-01-14', '2010-02-07', '2010-02-11', '2010-11-16',
'2010-11-22', '2010-11-23', '2010-11-26', '2010-12-02', '2010-12-13',
'2011-02-04', '2011-06-02', '2011-08-05', '2011-11-30']

I did read the similar problem and try to sloved it but still stock
How to sort things, without using the sorted build-in function?
first version code:

import datetime

def date_sorting_operation(input_list):
  
  input_list = [datetime.datetime.strptime(ts, "%Y-%m-%d") for ts in input_list] 

  for i in range(len(input_list)):
    for j in range(i + 1, len(input_list)):

        if l[i] > l[j]:
           l[i], l[j] = l[j], l[i]  
  return input_list

customer_date_list = ['2011-06-02', '2011-08-05', '2011-02-04', '2010-1-14', '2010-12-13', '2010-1-12', '2010-2-11', '2010-2-07', '2010-12-02', '2011-11-30']

print (date_sorting_operation(customer_date_list))

second version code:

import datetime

def date_sorting_operation(input_list):
  
  input_list = [datetime.datetime.strptime(ts, "%Y-%m-%d") for ts in input_list] 

  for i in range(len(input_list) - 1):
     for j in range(0, len(input_list) - i - 1):
         if input_list[j].repeats > input_list[j + 1].repeats:
             input_list[j], input_list[j + 1] = input_list[j + 1], input_list[j] 
  return input_list

customer_date_list = ['2011-06-02', '2011-08-05', '2011-02-04', '2010-1-14', '2010-12-13', '2010-1-12', '2010-2-11', '2010-2-07', '2010-12-02', '2011-11-30']

print (date_sorting_operation(customer_date_list))

(pic) for code and output First version:

(pic) for code and output Second version:

(Third version) Try just to use "input_list", remove "new_list" to save memory if that, how can I do so?

   

import datetime
 
def date_sorting_operation(input_list):
  
  new_list = []
  dates = [datetime.datetime.strptime(ts, "%Y-%m-%d") for ts in input_list] 
 
  while input_list:
      min = input_list[0]  
      for x in input_list: 
          if x < min:
              min = x
      new_list.append(min)
      input_list.remove(min)    
  return new_list
 
customer_date_list = ['2011-06-2', '2011-08-05', '2011-02-04', '2010-1-14', '2010-12-13', '2010-1-12', '2010-2-11', '2010-2-07', '2010-12-02', '2011-11-30']
 
print (date_sorting_operation(customer_date_list))

(pic) try just to use "input_list", remove "new_list" to save memory:

Tim Biegeleisen · Accepted Answer · 2022-09-29 02:39:09Z

3

Actually your list of text dates are almost in an ISO sortable format, except that occasionally the month may be one digit instead of two. If we left pad the month with zero, sorting should work:

customer_date_list = ['2011-06-02', '2011-08-05', '2011-02-04', '2010-1-14', '2010-12-13', '2010-1-12', '2010-2-11', '2010-2-07', '2010-12-02', '2011-11-30']
output = [re.sub(r'-(\d)-', r'-0\1-', x) for x in customer_date_list]
output.sort()
print(output)

['2010-01-12', '2010-01-14', '2010-02-07', '2010-02-11', '2010-12-02',
 '2010-12-13', '2011-02-04', '2011-06-02', '2011-08-05', '2011-11-30']

answered Sep 29, 2022 at 2:39

Tim Biegeleisen

526k32 gold badges323 silver badges399 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

j ton Over a year ago

hi @Tim Biegeleisen That I wa trying to not use .sort() function to sort the date myself but seems the date format date.time stop me for doing this. Therefore, how can I correct it though.

Tim Biegeleisen Over a year ago

See the other answer. You can also cast/parse each datetime string to a bona fide datetime, and then sort that.

Sharim09 · Accepted Answer · 2022-09-29 04:02:20Z

2

Try this

import datetime


customer_date_list = ['2011-06-02', '2011-08-05', '2011-02-04', '2010-1-14', '2010-12-13', '2010-1-12', '2010-2-11', '2010-2-07', '2010-12-02', '2011-11-30']

lst = [datetime.datetime.strptime(ts, "%Y-%m-%d") for ts in customer_date_list]

outLst = []
for _ in customer_date_list:
    m = min(lst)
    lst.remove(m)
    outLst.append(m.strftime('%Y-%m-%d'))

print(outLst)

Output

['2010-01-12', '2010-01-14', '2010-02-07', '2010-02-11', '2010-12-02',
 '2010-12-13', '2011-02-04', '2011-06-02', '2011-08-05', '2011-11-30']

AS OP says "that can I just use one list lst, which without outLst, to save on the same list. (In saving memory purpose)"


import datetime


customer_date_list = ['2011-06-02', '2011-08-05', '2011-02-04', '2010-1-14', '2010-12-13', '2010-1-12', '2010-2-11', '2010-2-07', '2010-12-02', '2011-11-30']

lst = [datetime.datetime.strptime(ts, "%Y-%m-%d") for ts in customer_date_list]


for i in range(len(customer_date_list)):
    m = max(lst[i:])
    lst.remove(m)
    lst.insert(0,m.strftime('%Y-%m-%d'))

print(lst)

output

['2010-01-12', '2010-01-14', '2010-02-07', '2010-02-11', '2010-12-02',
 '2010-12-13', '2011-02-04', '2011-06-02', '2011-08-05', '2011-11-30']

edited Sep 29, 2022 at 4:02

answered Sep 29, 2022 at 2:45

Sharim09

6,2162 gold badges10 silver badges31 bronze badges

1 Comment

j ton Over a year ago

fantastic @codester_09 one more question that can I just use one list lst , which without outLst , to save one the same list. (In saving memory purpose)

Collectives™ on Stack Overflow

(python) sorting date without using build in function

2 Answers 2

2 Comments

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

2 Comments

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related