检查列表中的所有元素是否唯一

What is the best way (best as in the conventional way) of checking whether all elements in a list are unique?

My current approach using a Counter is:

>>> x = [1, 1, 1, 2, 3, 4, 5, 6, 2]
>>> counter = Counter(x)
>>> for values in counter.itervalues():
        if values > 1: 
            # do something

Can I do better?

Not the most efficient, but straight forward and concise:

if len(x) > len(set(x)):
   pass # do something

Probably won’t make much of a difference for short lists.

Here is a two-liner that will also do early exit:

>>> def allUnique(x):
...     seen = set()
...     return not any(i in seen or seen.add(i) for i in x)
...
>>> allUnique("ABCDEF")
True
>>> allUnique("ABACDEF")
False

If the elements of x aren’t hashable, then you’ll have to resort to using a list for seen:

>>> def allUnique(x):
...     seen = list()
...     return not any(i in seen or seen.append(i) for i in x)
...
>>> allUnique([list("ABC"), list("DEF")])
True
>>> allUnique([list("ABC"), list("DEF"), list("ABC")])
False

An early-exit solution could be

def unique_values(g):
    s = set()
    for x in g:
        if x in s: return False
        s.add(x)
    return True

however for small cases or if early-exiting is not the common case then I would expect len(x) != len(set(x)) being the fastest method.

for speed:

import numpy as np
x = [1, 1, 1, 2, 3, 4, 5, 6, 2]
np.unique(x).size == len(x)

How about adding all the entries to a set and checking its length?

len(set(x)) == len(x)

Alternative to a set, you can use a dict.

len({}.fromkeys(x)) == len(x)

Another approach entirely, using sorted and groupby:

from itertools import groupby
is_unique = lambda seq: all(sum(1 for _ in x[1])==1 for x in groupby(sorted(seq)))

It requires a sort, but exits on the first repeated value.

Here is a recursive O(N²) version for fun:

def is_unique(lst):
    if len(lst) > 1:
        return is_unique(s[1:]) and (s[0] not in s[1:])
    return True

Here is a recursive early-exit function:

def distinct(L):
    if len(L) == 2:
        return L[0] != L[1]
    H = L[0]
    T = L[1:]
    if (H in T):
            return False
    else:
            return distinct(T)    

It’s fast enough for me without using weird(slow) conversions while having a functional-style approach.

How about this

def is_unique(lst):
    if not lst:
        return True
    else:
        return Counter(lst).most_common(1)[0][1]==1

You can use Yan’s syntax (len(x) > len(set(x))), but instead of set(x), define a function:

 def f5(seq, idfun=None): 
    # order preserving
    if idfun is None:
        def idfun(x): return x
    seen = {}
    result = []
    for item in seq:
        marker = idfun(item)
        # in old Python versions:
        # if seen.has_key(marker)
        # but in new ones:
        if marker in seen: continue
        seen[marker] = 1
        result.append(item)
    return result

and do len(x) > len(f5(x)). This will be fast and is also order preserving.

Code there is taken from: http://www.peterbe.com/plog/uniqifiers-benchmark

Using a similar approach in a Pandas dataframe to test if the contents of a column contains unique values:

if tempDF['var1'].size == tempDF['var1'].unique().size:
    print("Unique")
else:
    print("Not unique")

For me, this is instantaneous on an int variable in a dateframe containing over a million rows.

all answer above are good but i prefer to use all_unique example from 30 seconds of python

you need to use set() on the given list to remove duplicates, compare its length with the length of the list.

def all_unique(lst):
  return len(lst) == len(set(lst))

it returns True if all the values in a flat list are unique, False otherwise

x = [1,2,3,4,5,6]
y = [1,2,2,3,4,5]
all_unique(x) # True
all_unique(y) # False

For begginers:

def AllDifferent(s):
    for i in range(len(s)):
        for i2 in range(len(s)):
            if i != i2:
                if s[i] == s[i2]:
                    return False
    return True