• Nov 19, 2022 •CodeCatch
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# Python code to find the URL from an input string # Using the regular expression import re def Find(string): # findall() has been used # with valid conditions for urls in string regex = r"(?i)\b((?:https?://|www\d{0,3}[.]|[a-z0-9.\-]+[.][a-z]{2,4}/)(?:[^\s()<>]+|\(([^\s()<>]+|(\([^\s()<>]+\)))*\))+(?:\(([^\s()<>]+|(\([^\s()<>]+\)))*\)|[^\s`!()\[\]{};:'\".,<>?«»“”‘’]))" url = re.findall(regex,string) return [x[0] for x in url] # Driver Code string = 'My Profile: https://codecatch.net' print("Urls: ", Find(string))
# Python code to demonstrate # method to remove i'th character # Naive Method # Initializing String test_str = "CodeCatch" # Printing original string print ("The original string is : " + test_str) # Removing char at pos 3 # using loop new_str = "" for i in range(len(test_str)): if i != 2: new_str = new_str + test_str[i] # Printing string after removal print ("The string after removal of i'th character : " + new_str)
• May 5, 2026 •CodeCatch
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• Aug 1, 2025 •AustinLeath
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from typing import Optional from datetime import datetime def convert_timestamp_string_to_epoch(timestamp: str) -> Optional[int]: epoch_time = None time_obj = datetime.strptime(timestamp, "%Y-%m-%d %H:%M:%S.%f") epoch_time = int((time_obj - datetime(1970, 1, 1)).total_seconds() * 1000) return epoch_time print(int(convert_timestamp_string_to_epoch("2025-08-01 13:11:47.171"))) #above outputs 1754053907171.0 #how to I remove the .0 ?
• Nov 18, 2022 •AustinLeath
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mydict = {'carl':40, 'alan':2, 'bob':1, 'danny':0} # How to sort a dict by value Python 3> sort = {key:value for key, value in sorted(mydict.items(), key=lambda kv: (kv[1], kv[0]))} print(sort) # How to sort a dict by key Python 3> sort = {key:mydict[key] for key in sorted(mydict.keys())} print(sort)