使用循环在Python中计算字典中元素的有效方法

Question

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)

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}

Answer 1

对于较小的示例,使用有限的元素,您可以使用集合和字典理解:

>>> 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}

要打破它,set(mylist)将列表统一并使其更紧凑:

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

然后字典理解逐步执行唯一值并从列表中设置计数.

这也是显著比使用计数器比存在使用它的速度越来越快:

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))

在我打印的机器上(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

对于较大的列表,如1000万个整数,具有更多不同的元素(1,500个随机整数),在循环中使用defaultdict或fromkeys:

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))

打印:

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,count与其他解决方案相比,通过大型列表中等次数的中继解决方案将遭受严重/灾难性后果.

Answer 2

试试collections.Counter:

   >>> from collections import Counter
   >>> Counter([1,1,1,1,1,1,2,3,2,2,2,2,3,3,4,5,5,5,5])
   Counter({1: 6, 2: 5, 5: 4, 3: 3, 4: 1})

在您的代码中,您基本上可以mydict用a Counter和write 替换

mydict[index] += 1

代替

mydict[index] = +1