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:

Expand|Select|Wrap|Line Numbers

  
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.

Expand|Select|Wrap|Line Numbers

 >>> 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.

Expand|Select|Wrap|Line Numbers

 >>> 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:

Expand|Select|Wrap|Line Numbers

 >>> 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)

Expand|Select|Wrap|Line Numbers

 >>> 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.

Finding duplicates in an array

re: Finding duplicates in an array

re: Finding duplicates in an array

re: Finding duplicates in an array

Similar Python bytes

/bytes/development

/bytes/it

/bytes/about