def when(predicate, when_true):
  return lambda x: when_true(x) if predicate(x) else x

double_even_numbers = when(lambda x: x % 2 == 0, lambda x : x * 2)
print(double_even_numbers(2)) # 4
print(double_even_numbers(1)) # 1

from statistics import median, mean, mode

def print_stats(array):
  print(array)
  print("median =", median(array))
  print("mean   =", mean(array))
  print("mode   =", mode(array))
  print()

print_stats([1, 2, 3, 3, 4])
print_stats([1, 2, 3, 3])

# Deleting all even numbers from a list
a = [1,2,3,4,5]
del a[1::2]
print(a)

# function which return reverse of a string
def isPalindrome(s):
    return s == s[::-1]
 
# Driver code
s = "malayalam"
ans = isPalindrome(s)
 
if ans:
    print("Yes")
else:
    print("No")

# 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))