what's the difference between iterator and array?

Question

I am trying to write a function that takes a sequence of numbers and determines if all the numbers are different from each other.

This was my first attempt

def differ(data):
    for i in data:
        print(i)
        for j in data:
            print(j)
            if i==j:
                return False
    return True

print(differ([1,23,4,1,2,3,1,2]))
print(differ([1,2,3,4]))

1
1
False
1
1
False

Apparently, the for loop didn't loop over all the numbers in the data. Why did this happen?

I wrote another function using range().

def differ(data):
    for i in range(1,len(data)):
        print("i:",i)
        for j in range(i):
            print("j:",j)
            if data[i]==data[j]:
                return False
    return True
print(differ([1,2,2,4,53]))

i: 1
j: 0
i: 2
j: 0
j: 1
False

It works, however, I don't understand why my first attempt didn't work.

You return False as soon as you have a match, and in your first snippet you have both loops start with the first element, Naturally, the first element is equal to itself. — Tim Roberts
– Tim Roberts, Commented Jul 6, 2021 at 3:07
in the first solution you re comparing the I and j. When the code runs I is 0(zero) and it prints I, and then inner loop has j which runs and prints j(zero). and then since I and j are equal it returns False and exits. — huzzzus
– huzzzus, Commented Jul 6, 2021 at 3:08
You could just sort the array, and then make sure no two consecutive elements are equal. That would only take one loop. — Tim Roberts
– Tim Roberts, Commented Jul 6, 2021 at 3:08

user12128336 · Accepted Answer · 2021-07-06 03:55:53Z

1

The problem with your first attempt is that it will eventually compare the same number with itself.

In the sequence [1,2,3,4] the first function will start off with i=1 and j=1. They are both looking at the first number in the list causing it to fail.

The second attempt avoids this by only looking at the numbers before it. range(i) doesn't actually include i, so j can only ever be less than i meaning they will never point to the same value in the list.

answered Jul 6, 2021 at 3:55

user12128336

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Jorge Alvarado · Accepted Answer · 2021-07-06 04:44:04Z

I think Tim Roberts answer explains perfectly why your code is not working as you expect. I would like to complement the answer:

Answering your question

Arrays are similar to lists, one of the differences is that the former consists only of elements with the same data type. In the other hand, iterators are objects which you can iterate through, these objects must implement at least the following 2 methods on their respective classes: __iter__() and __next__()

Cool way to do what you want in 1 line

You can achieve what you want with this:

def differ(data):
    return len(set(data)) == len(data)


print(differ([1, 2, 3, 4]))  # True
print(differ([1, 23, 4, 1, 2, 3, 1, 2])) # False

So basically, the magic here happens when you use set(). Sets are similar to lists, but one of the main differences is that they can't have repeated elements. By transforming the list to a set you are removing every duplicated element, so if there is a difference in the length after casting to a set, it means there was at least one duplicated element.

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