We had performed some logical operations before , but now let us do the same inside a function . Let us create a function that returns a True boolean value if it finds the number( given as argument) to be even else it returns False.
def even_check(n):
if n%2==0:
return True
else:
return False
print(even_check(4))
True
Let us do something more complex. In the example given below , we will be passing a list to our function and the function return a new list containing only the odd numbers.
def odd_numbers(my_list):
new_list = []
for x in my_list:
if x%2!=0:
new_list.append(x)
else:
continue
return new_list
print(odd_numbers([1,2,3,4,5,6,7,8,9,10]))
print(odd_numbers([x for x in range (500)]))
[1, 3, 5, 7, 9]
[1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99, 101, 103, 105, 107, 109, 111, 113, 115, 117, 119, 121, 123, 125, 127, 129, 131, 133, 135, 137, 139, 141, 143, 145, 147, 149, 151, 153, 155, 157, 159, 161, 163, 165, 167, 169, 171, 173, 175, 177, 179, 181, 183, 185, 187, 189, 191, 193, 195, 197, 199, 201, 203, 205, 207, 209, 211, 213, 215, 217, 219, 221, 223, 225, 227, 229, 231, 233, 235, 237, 239, 241, 243, 245, 247, 249, 251, 253, 255, 257, 259, 261, 263, 265, 267, 269, 271, 273, 275, 277, 279, 281, 283, 285, 287, 289, 291, 293, 295, 297, 299, 301, 303, 305, 307, 309, 311, 313, 315, 317, 319, 321, 323, 325, 327, 329, 331, 333, 335, 337, 339, 341, 343, 345, 347, 349, 351, 353, 355, 357, 359, 361, 363, 365, 367, 369, 371, 373, 375, 377, 379, 381, 383, 385, 387, 389, 391, 393, 395, 397, 399, 401, 403, 405, 407, 409, 411, 413, 415, 417, 419, 421, 423, 425, 427, 429, 431, 433, 435, 437, 439, 441, 443, 445, 447, 449, 451, 453, 455, 457, 459, 461, 463, 465, 467, 469, 471, 473, 475, 477, 479, 481, 483, 485, 487, 489, 491, 493, 495, 497, 499]
In the code given below , we pass a list to a function and we want our function to return a True value if there is even a single even number inside the whole list and if there isn't a single even number then return False.
First we will look at the incorrect way of doing this problem .
def even_numbers(my_list):
for x in my_list:
if x%2==0:
return True
else:
return False
print(even_numbers([1,2,3,4,5]))
False
Notice the output of the code . is it right? well , NO! because our list does contain even numbers so it should have returned True value instead. A number can either be even or odd. So out of the two conditions ,one of them will always be true. And because we have return statement inside each condition , our result will be solely based on the first element of the list. Python will never be able to iterate over to the next element because it exits the function ( because of the return statement ) . So that is why it returned False simply because the first element of the list was odd.
Let us look at the correct way of solving the above problem.
def even_numbers(my_list):
for x in my_list:
if x%2==0:
return True
else:
continue
return False
print(even_numbers([1,2,3,4,5]))
print(even_numbers([1,3,5,7,9]))
print(even_numbers([2,4,6,8]))
True
False
True
Well now it is working as expected . All we had to do was to make sure that the else statement had no return statement. Instead have a return statement out of the loop so that if there were no even numbers inside the list , we'll return False.