Finding duplicates in an array

[Thekid]

I'm trying to figure out a way to find if there are duplicates in an array. My idea was to take the array as 'a' and make a second array as 'b' and remove the duplicates from 'b' using 'set' and then compare a to b. If they're different then it will print out 'duplicates found'. The problem is that even after trying different arrays, some with duplicates some without, that 'b' rearranges the numbers. Here's an example:

 
a='1934, 2311, 1001, 4056, 1001, 3459, 9078'
b=list(set(a))
if a != b:
    print "duplicates found"
else:
   print "nothing found"
 
 

Is there a simpler way to find if there are duplicates?
Thanks

[bvdet]

In your code, you have assigned variable 'a' to a string.

>>> list(set(a))
[' ', ',', '1', '0', '3', '2', '5', '4', '7', '6', '9', '8']
>>> 

To see if there are any duplicates, let's start a list. Sets are unordered, but you can compare the length of the list to the length of the set.

>>> a=[1934, 2311, 1001, 4056, 1001, 3459, 9078]
>>> b = set(a)
>>> len(b)
6
>>> len(a)
7
>>> 

[bvdet]

To find the items that have duplicates:

>>> for item in a:
...     dd[item] = dd.get(item, 0) + 1
...     
>>> dd
{3459: 1, 2311: 1, 1001: 2, 1934: 1, 9078: 1, 4056: 1}
>>> 

OR (less efficient)

>>> for item in set(a):
...     if a.count(item) > 1:
...         print "Duplicate found: %s" % (item)
...         
Duplicate found: 1001
>>> 

[Thekid]

Thanks! I went with your first suggestion, which was along the lines of what I was thinking but I didn't consider comparing the lengths since set() is unordered.