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}
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}
To break it down, set(mylist) uniquifies the list and makes it more compact:
>>> set(mylist)
set([1, 2, 3, 4, 5])
Then the dictionary comprehension steps through the unique values and sets the count from the list.
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))
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
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))
Prints:
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
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.
尝试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替换Counter并编写
mydict[index] += 1
代替
mydict[index] = +1
纠正代码：
mydict[index] = +1
应该：
mydict[index] = mydict.setdefault(index, 0) + 1
setdefault该方法的一个变体是collections.defaultdict. 这有点快。
def foo(mylist):
d=defaultdict(int)
for i in mylist:
d[i] += 1
return d
itertools.groupBy提供了另一种选择。Counter它的速度与（至少在 2.7 上）大致相同
{x[0]:len(list(x[1])) for x in itertools.groupby(sorted(mylist))}
但是，在处理 OP 在评论中提到的 32Gb 数据时，这个小测试列表上的时间测试可能不一样。
我在python top N word count的 word count 案例中运行了其中几个选项，为什么多进程比单进程慢
在那里，OP 使用了 Counter，并试图通过使用多处理来加快速度。使用 1.2Mb 的文本文件，计数器使用defaultdict速度很快，只需 0.2 秒。对输出进行排序以获得前 40 个单词所花费的时间与计数本身一样长。
Counter有点慢3.2，而且慢得多2.7。那是因为3.2编译版本（.so文件）。
mylist.count但是在处理大量列表时，使用地面的计数器会停止；将近200秒。它必须多次搜索该大列表，一次收集密钥，然后在计算每个密钥时搜索一次。
您的代码将 1 分配为每个键的值。替换mydict[index] = +1为mylist.count(index)
这应该有效：
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] = mylist.count(index)
mydict