CODEWITHSUNDEEP
python
user1834857
user1834857

Reputation:

Python Lists: How can I search for another element or duplicate item in a list

I have a list say p = [1,2,3,4,2] Is there any way of returning bool value True if it contains a duplicate using only find, indexing, slicing, len() etc methods and not dict, tuple etc.

I used this code: 

for e in p: 
   duplicate = p.find(e, e+1) 
   if duplicate in p: 
       return True

Upvotes: 1

Views: 664

Answers (6)

NPE
NPE

Reputation: 500883

Here is the easy way:

return len(p) != len(set(p))

A less efficient way that doesn't use set:

for i in range(len(p)):
   if p[i] in p[i+1:]:
      return True
return False

This second approach is not very idiomatic, but avoids all but the most basic features of the language (including tuples).

Here is one more way:

while p:
   e = p.pop()
   if e in p:
      return True
return False

This is simple, but does modify the list.

One final way I am going to demonstrate is:

s = sorted(p)
for i in range(1, len(s)):
   if s[i] == s[i - 1]:
      return True
return False

This works by sorting p and then comparing every pair of consecutive elements.

Upvotes: 6

shantanoo
shantanoo

Reputation: 3704

Using collections.Counter

>>> import collections
>>> p
[1, 2, 3, 4, 2]
>>> if collections.Counter(p).most_common()[0][1] > 1: 
...     print('duplicate found')
... 
duplicate found
>>> if collections.Counter(set(p)).most_common()[0][1] > 1: 
...     print('duplicate found')
... 
>>>

Upvotes: 1

Brigand
Brigand

Reputation: 86260

Here's a very simple way to do it. It may be slow for very large lists.

def has_duplicates(lst):
    for e in lst:
        lst = lst[1:]
        if e in lst: return True
    return False

Upvotes: 0

Patricio Molina
Patricio Molina

Reputation: 390

>>> p = [1, 2, 3, 4, 2]
>>> len(set(p)) == len(p)
False

You can find more information about sets in the Python Documentation.

Upvotes: 2

Cairnarvon
Cairnarvon

Reputation: 27832

You could also use list.count:

def has_duplicates(p):
    for e in p:
        if p.count(e) > 1:
            return True
    return False

Upvotes: 5

David Robinson
David Robinson

Reputation: 78630

If you have to do it this way, you could do:

def has_duplicates(lst):
    for i, e in enumerate(lst[::-1]):
        if lst.index(e) != len(lst) - i - 1:
            return True
    return False

This iterates through the list in reverse order (since index searches from the start of the list). But it's better simply to do:

def has_duplicates(lst):
    return len(set(lst)) != len(lst)

Upvotes: 1

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