python列表去重

系统管理员 2021-12-14 10:53 269阅读 0赞

Suppose you have a list in python that looks like this:

['a','b','a']
    # or like this:
    [1,2,2,2,3,4,5,6,6,6,6]

and you want to remove all duplicates so you get this result:

['a','b']
    # or
    [1,2,3,4,5,6]

How do you do that? ...the fastest way? I wrote a couple of alternative implementations and did a quick benchmark loop on the various implementations to find out which way was the fastest. (I haven't looked at memory usage). The slowest function was 78 times slower than the fastest function.

However, there's one very important difference between the various functions. Some are order preserving and some are not. For example, in an order preserving function, apart from the duplicates, the order is guaranteed to be the same as it was inputted. Eg, `uniqify([1,2,2,3])==[1,2,3]`

Here are the functions:

def f1(seq):
       # not order preserving
       set = {}
       map(set.__setitem__, seq, [])
       return set.keys()
    
    def f2(seq): 
       # order preserving
       checked = []
       for e in seq:
           if e not in checked:
               checked.append(e)
       return checked
    
    def f3(seq):
       # Not order preserving
       keys = {}
       for e in seq:
           keys[e] = 1
       return keys.keys()
    
    def f4(seq): 
       # order preserving
       noDupes = []
       [noDupes.append(i) for i in seq if not noDupes.count(i)]
       return noDupes
    
    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
    
    def f6(seq):
       # Not order preserving    
       set = Set(seq)
       return list(set)

And what you've all been waiting for (if you're still reading). Here are the results:

* f2 13.24
    * f4 11.73
    * f5 0.37
    f1 0.18
    f3 0.17
    f6 0.19
    
    (* order preserving)

Clearly `f5` is the "best" solution. Not only is it really really fast; it's also order preserving and supports an optional transform function which makes it possible to do this:

>>> a=list('ABeeE')
    >>> f5(a)
    ['A','B','e','E']
    >>> f5(a, lambda x: x.lower())
    ['A','B','e']

[Download the benchmark script here][]

**UPDATE**

From the comments I've now added a couple of more functions to the benchmark. Some which don't support uniqify a list of objects that can't be hashed unless passed with a special hashing method. So see all the functions [download the file][Download the benchmark script here]

Here are the new results:

* f5 10.1
    * f5b 9.99
    * f8 6.49
    * f10 6.57
    * f11 6.6
    f1 4.28
    f3 3.55
    f6 4.03
    f7 2.59
    f9 2.58

(f2 and f4) were too slow for this testdata.

[Download the benchmark script here]: http://www.peterbe.com/plog/uniqifiers-benchmark/uniqifiers_benchmark.py