在Python中使用循环在字典中计数元素的高效方法 [英] efficient way to count the element in a dictionary in Python using a loop

问题描述

我有一个值列表。我希望在循环期间计算每个类的元素数（即1,2,3,4,5）

I have a list of values. I wish to count during a loop the number of element for each class (i.e. 1,2,3,4,5)

mylist = [1,1,1,1,1,1,2,3,2,2,2,2,3,3,4,5,5,5,5]
mydict = dict()
for index in mylist:
    mydict[index] = +1
mydict
Out[344]: {1: 1, 2: 1, 3: 1, 4: 1, 5: 1}

我希望得到这个结果

Out[344]: {1: 6, 2: 5, 3: 3, 4: 1, 5: 4}

推荐答案

对于你的小例子，使用有限的元素，你可以使用set和dict comprehension ：

For your smaller example, with a limited diversity of elements, you can use a set and a dict comprehension:

>>> mylist = [1,1,1,1,1,1,2,3,2,2,2,2,3,3,4,5,5,5,5]
>>> {k:mylist.count(k) for k in set(mylist)}
{1: 6, 2: 5, 3: 3, 4: 1, 5: 4}

要将其分解， set（mylist） compact：

To break it down, set(mylist) uniquifies the list and makes it more compact:

>>> set(mylist)
set([1, 2, 3, 4, 5])

然后字典解析逐步通过唯一值并设置列表中的计数。

Then the dictionary comprehension steps through the unique values and sets the count from the list.

这也比使用计数器显着更快，而且比使用setdefault更快。

This also is significantly faster than using Counter and faster than using setdefault:

from __future__ import print_function
from collections import Counter
from collections import defaultdict
import random

mylist=[1,1,1,1,1,1,2,3,2,2,2,2,3,3,4,5,5,5,5]*10

def s1(mylist):
    return {k:mylist.count(k) for k in set(mylist)}

def s2(mlist):
    return Counter(mylist)

def s3(mylist):
    mydict=dict()
    for index in mylist:
        mydict[index] = mydict.setdefault(index, 0) + 1
    return mydict   

def s4(mylist):
    mydict={}.fromkeys(mylist,0)
    for k in mydict:
        mydict[k]=mylist.count(k)    
    return mydict    

def s5(mylist):
    mydict={}
    for k in mylist:
        mydict[k]=mydict.get(k,0)+1
    return mydict     

def s6(mylist):
    mydict=defaultdict(int)
    for i in mylist:
        mydict[i] += 1
    return mydict       

def s7(mylist):
    mydict={}.fromkeys(mylist,0)
    for e in mylist:
        mydict[e]+=1    
    return mydict    

if __name__ == '__main__':   
    import timeit 
    n=1000000
    print(timeit.timeit("s1(mylist)", setup="from __main__ import s1, mylist",number=n))
    print(timeit.timeit("s2(mylist)", setup="from __main__ import s2, mylist, Counter",number=n))
    print(timeit.timeit("s3(mylist)", setup="from __main__ import s3, mylist",number=n))
    print(timeit.timeit("s4(mylist)", setup="from __main__ import s4, mylist",number=n))
    print(timeit.timeit("s5(mylist)", setup="from __main__ import s5, mylist",number=n))
    print(timeit.timeit("s6(mylist)", setup="from __main__ import s6, mylist, defaultdict",number=n))
    print(timeit.timeit("s7(mylist)", setup="from __main__ import s7, mylist",number=n))

在打印（Python 3）的机器上：

On my machine that prints (Python 3):

18.123854104997008          # set and dict comprehension 
78.54796334600542           # Counter 
33.98185228800867           # setdefault 
19.0563529439969            # fromkeys / count 
34.54294775899325           # dict.get 
21.134678319009254          # defaultdict 
22.760544238000875          # fromkeys / loop

对于更大的列表，如10万个整数， ，在循环中使用defaultdict或fromkeys：

For Larger lists, like 10 million integers, with more diverse elements (1,500 random ints), use defaultdict or fromkeys in a loop:

from __future__ import print_function
from collections import Counter
from collections import defaultdict
import random

mylist = [random.randint(0,1500) for _ in range(10000000)]

def s1(mylist):
    return {k:mylist.count(k) for k in set(mylist)}

def s2(mlist):
    return Counter(mylist)

def s3(mylist):
    mydict=dict()
    for index in mylist:
        mydict[index] = mydict.setdefault(index, 0) + 1
    return mydict   

def s4(mylist):
    mydict={}.fromkeys(mylist,0)
    for k in mydict:
        mydict[k]=mylist.count(k)    
    return mydict    

def s5(mylist):
    mydict={}
    for k in mylist:
        mydict[k]=mydict.get(k,0)+1
    return mydict     

def s6(mylist):
    mydict=defaultdict(int)
    for i in mylist:
        mydict[i] += 1
    return mydict       

def s7(mylist):
    mydict={}.fromkeys(mylist,0)
    for e in mylist:
        mydict[e]+=1    
    return mydict    

if __name__ == '__main__':   
    import timeit 
    n=1
    print(timeit.timeit("s1(mylist)", setup="from __main__ import s1, mylist",number=n))
    print(timeit.timeit("s2(mylist)", setup="from __main__ import s2, mylist, Counter",number=n))
    print(timeit.timeit("s3(mylist)", setup="from __main__ import s3, mylist",number=n))
    print(timeit.timeit("s4(mylist)", setup="from __main__ import s4, mylist",number=n))
    print(timeit.timeit("s5(mylist)", setup="from __main__ import s5, mylist",number=n))
    print(timeit.timeit("s6(mylist)", setup="from __main__ import s6, mylist, defaultdict",number=n))
    print(timeit.timeit("s7(mylist)", setup="from __main__ import s7, mylist",number=n))

列印：

2825.2697427899984              # set and dict comprehension 
42.607481333994656              # Counter 
22.77713537499949               # setdefault 
2853.11187016801                # fromkeys / count 
23.241977066005347              # dict.get 
15.023175164998975              # defaultdict 
18.28165417900891               # fromkeys / loop

您可以看到，在计数时间通过大列表将遭受严重/灾难性与其他解决方案相比。

You can see that solutions that relay on count with a moderate number of times through the large list will suffer badly/catastrophically in comparison to other solutions.

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