Parsing and rewriting a date-time string

Question

I have got a string of time data: "2019-08-02 18:18:06.02887" and I am trying to rewrite it as another string "EV190802_181802" in another file.

What I am trying now is splitting the string into lists and reconstructing another string by those lists:

hello=data.split(' ')
date=hello[0]
time=hello[1]
world=hello[0].split('-')
stack=time.split('.')
overflow=stack[0].split(':')
print('EV' + world[0] + world[1] + world[2] + '_' + overflow[0] + overflow[1] + overflow[2])

However, I have no idea how to remove 20 in 2019/world[0]. Is there any way I could remove '20'?

If there are alternative methods to rewrite the string, welcome to suggest as well.

Just to be sure: Is it "EV190802_181802" or "EV190802_181806" (with "06" at the end?) — akensert, Aug 08 '20 at 10:49
Sorry for duplication of the question. Anyway, thanks for the answers, guys. I am not sure should I delete the question as some people here post answers different from another one's: https://stackoverflow.com/questions/2265357/parse-date-string-and-change-format. — Simon219, Aug 08 '20 at 17:07

sushanth · Accepted Answer · 2020-08-08T11:18:52.970

2

Just another way to solve the problem,

>>> from datetime import datetime
>>> 
>>> format_ = datetime.strptime("2019-08-02 18:18:06.02887", 
...                             "%Y-%m-%d %H:%M:%S.%f")
>>> 
>>> print(
    format_.strftime('EV%y%m%d_%H%M') + format_.strftime('%f')[:2]
)

EV190802_181802

edited Aug 08 '20 at 11:18

answered Aug 08 '20 at 10:58

sushanth

6,960
3
13
23

1

Why not simply `dt.strftime('EV%y%m%d_%H%M%S')` => `'EV190802_181806'`? – ekhumoro Aug 08 '20 at 11:10
Okay - but also note that the OP doesn't want `EV190802_181802` as the output. You can see this by running the code they wrote, which prints `EV20190802_181806`. (The question has a typo at the beginning). – ekhumoro Aug 08 '20 at 11:32

score 0 · Answer 2 · answered Aug 08 '20 at 10:46

Using regex:

import re

data = "2019-08-02 18:18:06.02887"
res = re.match(r'(?P<year>\d{4})-(?P<month>\d{2})-(?P<day>\d{2})\s(?P<hours>\d{2}):(?P<minutes>\d{2}):(?P<seconds>\d{2}).(?P<miliseconds>\d+)',data)

out = f"EV{res.group('year')[2:]}{res.group('month')}{res.group('day')}_{res.group('hours')}{res.group('minutes')}{res.group('miliseconds')[:2]}"
print(out)

Output will be:

EV190802_181802

Tim Biegeleisen · Answer 3 · 2020-08-08T10:54:46.467

I would use re.sub here for a regex approach:

inp = "2019-08-02 18:18:06.02887"
output = re.sub(r'^\d{2}(\d{2})-(\d{2})-(\d{2}) (\d{2}):(\d{2}):(\d{2}).*$',
                'EV\\1\\2\\3_\\4\\5\\6',
                inp)
print(output)

This prints:

EV190802_181806

Note: Your expected output was actually given as EV190802_181802, but it appears to be a typo for my solution above, as I see no reason why you would not want to report seconds, but instead report hundreth fractions of a second.

score 0 · Answer 4 · answered Aug 08 '20 at 10:50

0

Remove all occurrences of -, . and :

 hello.replace("-","")
 hello.replace(".","")
 hello.replace(":","")

Get string in one line:

 print("EV" + hello[2:8] + "_" + hello[9:15])

answered Aug 08 '20 at 10:50

Adam Elm

46
6

2

Or just use `re.sub(r'[:.-]', '', hello)` for a one line way to strip off the punctuation. – Tim Biegeleisen Aug 08 '20 at 10:53

score 0 · Answer 5 · answered Aug 08 '20 at 10:53

Without regex, just string splits and joins:

string = "2019-08-02 18:18:06.02887"
target = "EV190802_181806" 

d, t = string.split("20")[-1].split(".")[0].split(" ")
print("date:", d, "\ntime:", t)

d = "".join(d.split('-'))
t = "".join(t.split(':'))
result = "EV" + d + "_" + t
print("\nresult: ", result)

assert result == target

# >> out:
# date: 19-08-02 
# time: 18:18:06
# 
# result:  EV190802_181806

I make the assumption that you'd like to have "06" at the end of the target string (sorry if I'm mistaken!).

Parsing and rewriting a date-time string

5 Answers5