• 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))
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from time import sleep def delay(fn, ms, *args): sleep(ms / 1000) return fn(*args) delay(lambda x: print(x), 1000, 'later') # prints 'later' after one second
from math import pi def rads_to_degrees(rad): return (rad * 180.0) / pi rads_to_degrees(pi / 2) # 90.0
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def print_x_pattern(size): i,j = 0,size - 1 while j >= 0 and i < size: initial_spaces = ' '*min(i,j) middle_spaces = ' '*(abs(i - j) - 1) final_spaces = ' '*(size - 1 - max(i,j)) if j == i: print(initial_spaces + '*' + final_spaces) else: print(initial_spaces + '*' + middle_spaces + '*' + final_spaces) i += 1 j -= 1 print_x_pattern(7)
• Mar 10, 2021 •Skrome
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color2 = (60, 74, 172) color1 = (19, 28, 87) percent = 1.0 for i in range(101): resultRed = round(color1[0] + percent * (color2[0] - color1[0])) resultGreen = round(color1[1] + percent * (color2[1] - color1[1])) resultBlue = round(color1[2] + percent * (color2[2] - color1[2])) print((resultRed, resultGreen, resultBlue)) percent -= 0.01
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def clamp_number(num, a, b): return max(min(num, max(a, b)), min(a, b)) clamp_number(2, 3, 5) # 3 clamp_number(1, -1, -5) # -1