Styjun

I am currently doing Udemy course for Python by Jose Portilla. I am still a complete beginner. The excercise is the following:

"Given a list of ints, return True if the array contains a 3 next to a 3 somewhere."

has_33([1, 3, 3]) → True
has_33([1, 3, 1, 3]) → False
has_33([3, 1, 3]) → False

My code is the following. For my logic it should give me the above results, but it gives False, False, True, which is completely off:

def has_33(nums):
 for n in nums:
 a = nums.index(n)
 x = nums[a + 1]
 y = nums[a - 1]
 if n == 3 and (x == n or y == n):
 return True
 else:
 return False

I try to explain the working of your function for has_33([1, 3, 3]). This way it gets obvious why your function is not doing what you expect it to do.

def has_33(nums):
 # Call function and nums = [1, 3, 3]

 for n in nums:
 # Iterate over nums. n = 1

 a = nums.index(n)
 # Get the index of the list a where element = 1; a = 0

 x = nums[a + 1]
 # 0 + 1 = 1 -> x = nums[1] = 3

 y = nums[a - 1]
 # 0 - 1 = -1 -> y = nums[-1] = 3 (index -1 points to the last element of the list)

 if n == 3 and (x == n or y == n):
 # False, since n != 3 (even though x = 3 and y = 3)

 return True
 else:
 return False
 # When returning, the functin stopts executing, hence n never reaches another element.

As pointed out in the comments, the main reason why this doesn't work is that you return false on the first iteration. Concerning the implementation, you can simplify things a bit to understand a bit more clearly what's going on:

def has_33(nums):
 for a in range(len(nums)-1):
 x = nums[a]
 y = nums[a + 1]
 if x == 3 and y == 3:
 return True
 return False

In your original solution, you were iterating over items by group of 3, adding print(y, n, x) right before your test would have output the following (for a call to has_33([1, 2, 3, 3, 4])):

item at: i-1 i i+1
 ---------------
i = 0 4 1 2
i = 1 1 2 3
i = 2 2 3 3

In the first line, 4 1 2, the 4 is the item at position 0-1 = -1, in python, negative indexes correspond to positions relative to the end of the list. For example, nums[-1] is the last item from nums and nums[-2] is the one before the last, etc.

In this second code we just iterate through nums's indexes to get every x item together with the following item y, printing print(x, y) would give:

item at: i i+1
 ---------
i = 0 1 2
i = 1 2 3
i = 2 3 3

Remark that, in the end we just used the indexes to get items, when the index is not used for anything else, you can often use the zip function instead:

def has_33(nums):
 for x, y in zip(nums, nums[1:]):
 if x == 3 and y == x:
 return True
 return False

has_33([1, 2, 3, 3, 1])

This outputs True, and for the record:

>>> nums
[1, 2, 3, 3, 1]
>>> nums[1:]
[2, 3, 3, 1]
>>> list(zip(nums, nums[1:]))
[(1, 2), (2, 3), (3, 3), (3, 1)]

The zip function will pair item at index i from the first list with item at index i from the second list. Here we just removed the first item from nums to form our second list, using this strategy we managed to pair item i with item i+1.

As mentioned in the comments, you can even use the function any that does exactly what our loop does

def has_33(nums):
 return any(p == (3, 3) for p in zip(nums, nums[1:]))

There is a logic error in your code at line 3. You mustn't use:

a = nums.index(n)

To get the index but you need to use this way:

for a, n in enumerate(nums):

Finally there is your corrected code:

def has_33(nums):
 for a, n in enumerate(nums):
 x = nums[a + 1]
 y = nums[a - 1]
 if n == 3 and (x == n or y == n):
 return True
 else:
 return False

WHY? nums.index(n) return the first position of n not the current position.