[FIXED] Python - How to replace just time numbers from date?

Issue

I am a beginner in python, and need some help from Pro!

I just want to delete only the time that appears on date output and not the date itself.

print(trh)

Output: September 16, 2022, 11:31 pm 20 Days

.

I tried several ways with .replace() but I didn't get what I want.

print(trh).replace('TimeString','')

For Example:

I just want to delete only , 11:31 pm in output with .replace()

To show on output like this: September 16, 2022 (20 Days)

.

Code:

def month_string_to_number(ay):
    m = {
        'jan': 1,
        'feb': 2,
        'mar': 3,
        'apr':4,
         'may':5,
         'jun':6,
         'jul':7,
         'aug':8,
         'sep':9,
         'oct':10,
         'nov':11,
         'dec':12
        }
    s = ay.strip()[:3].lower()

    try:
        out = m[s]
        return out
    except:
        raise ValueError('Not a month')
import time
from datetime import date

def tarih_clear(trh):
    ay=""
    gun=""
    yil=""
    trai=""
    my_date=""
    sontrh=""
    out=""
    ay=str(trh.split(' ')[0])
    gun=str(trh.split(', ')[0].split(' ')[1])
    yil=str(trh.split(', ')[1])
    ay=str(month_string_to_number(ay))
    trai=str(gun)+'/'+str(ay)+'/'+str(yil)
    my_date = str(trai)
    if 1==1:
        d = date(int(yil), int(ay), int(gun))
        sontrh = time.mktime(d.timetuple())
        out=(int((sontrh-time.time())/86400))
        return out

second Code

            if not data.count('phone')==0:
                hcr="\33[1;36m"
                hcc=hcc+1
                trh=""
                if 'end_date' in data:
                    trh=data.split('end_date":"')[1]
                    trh=trh.split('"')[0]
                else:
                      try:
                          trh=data.split('phone":"')[1]
                          trh=trh.split('"')[0]
                          if trh.lower()[:2] =='un':
                            DaysRemain=(" Days")
                          else:
                            DaysRemain=(str(tarih_clear(trh))+" Days")
                            trh=trh+' '+DaysRemain
                      except:pass

Thanks!

Solution

I would suggest you not to write all these logic for parsing and understanding the dates, if it is a standard format always - like the example you shared.

You can use the python's capability to parse and understand dates, also use regex to make string searches.

An example approach to reset the date-contained-string is below

import re
from datetime import datetime


def get_formatted_time(date_str):
    d_pattern = r"(\d+ Days)"
    parts = re.split(d_pattern, date_str) #Split the string based on the pattern of days

    date_string = parts[0] #The date time string part
    days_string = parts[1] #the days count

    date = datetime.strptime(date_string.strip(), "%B %d, %Y, %I:%M %p") #Parse to a valid date time object
    removed_time_str = date.strftime("%B %d, %Y") #format as per your need

    return f"{removed_time_str} ({days_string})" #concat and build your final representation. 


print(get_formatted_time("September 16, 2022, 11:31 am 20 Days"))

This gives you

September 16, 2022 (20 Days)

Answered By - Kris