既然你以前用过C++。而且写的方式也很像C。那就可以直接用numba优化一下。没有用多线程或者是多进程。  
我试了一下，iend=10是1.5s, iend=34, 大约是30分钟。

#双色球所有号码历次中奖查询代码
  

  
import time
import numba
import numpy as np

filename="i:/ssq.txt中存储的是双色球历次中奖号码，形式每行如01182630313303，六个红球一个蓝球号码，共1533行
  
start=time.time()

with open(filename) as file_object:
    numstrings = file_object.readlines()

#---把中奖号码存在这个数组里
nums = np.zeros((len(numstrings), 7), dtype=np.int32)
for i,s in enumerate(numstrings):
    nums[i,:] = [int(s[0:2]),int(s[2:4]),int(s[4:6]),int(s[6:8]),int(s[8:10]),int(s[10:12]),int(s[12:14])]

@numba.njit  #---用numba jit 优化，不用python
def run(nums):  
    iend = 10    # iend调试用，iend=34时为实际情况，共17721088种可能，调试时iend=10，共1344种可能
    length = len(nums)
    
    n = np.zeros(7, dtype=np.int32) #

    for i0 in range(1,17):
        for i1 in range(1,iend-5):
            for i2 in range(i1+1,iend-4):
                for i3 in range(i2+1,iend-3):
                    for i4 in range(i3+1,iend-2):
                        for i5 in range(i4+1,iend-1):
                            for i6 in range(i5+1,iend):
                                #构建循环框架，i0为蓝球，i1-i6为红球
                                #在我的破机子上，循环一遍17721088次，每次只计数加1，用时就到了4.698s
                                #然后是对每一个i0-i6，判断在1533期中，总共中了几个一等奖到六等奖，

                                n[0], n[1], n[2], n[3], n[4], n[5], n[6] = i1,i2,i3,i4,i5,i6,i0 #---快速赋值，避免用tuple
                                #n = i1,i2,i3,i4,i5,i6,i0

                                j1,j2,j3,j4,j5,j6 = 0,0,0,0,0,0 #---原程序忘记加了

                                for s in range(length):                                
                                    
                                    ired,iblue=0,0
                                    # s1,s2,s3,s4,s5,s6,s0=nums[s,:]
                                    
                                    for i in range(0,6):
                                        #if(n[i] in (s1,s2,s3,s4,s5,s6)):
                                        if(n[i] == nums[s,0]) or (n[i] == nums[s,1]) or (n[i] == nums[s,2]) or (n[i] == nums[s,3]) or (n[i] == nums[s,4]) or (n[i] == nums[s,5]):
                                            ired += 1
                                    
                                    # if( i0 == s0 ):
                                    if( i0 == nums[s,6]):
                                        iblue += 1
                                        
                                    if((ired,iblue)==(6,1)):
                                        j1 += 1
                                    if((ired,iblue)==(6,0)):
                                        j2 += 1
                                    if((ired,iblue)==(5,1)):
                                        j3 += 1
                                    if((ired,iblue)==(5,0))or((ired,iblue)==(4,1)):
                                        j4 += 1
                                    if((ired,iblue)==(4,0))or((ired,iblue)==(3,1)):
                                        j5 += 1
                                    if((ired,iblue)==(2,1))or((ired,iblue)==(1,1))or((ired,iblue)==(0,1)):
                                        j6 += 1
                                    

                                print(n,'   ',j1,j2,j3,j4,j5,j6)

result = run(nums)
end=time.time()
print(f"Running time %s seconds"%(end-start))
  


【 在 Eggcai 的大作中提到: 】
: 20年前用的是C++，现在学习python，写一个程序，算双色球中奖的。列举所有C（6，33）*16=17721088种可能的号码，从有据可查到现在的1500多期开奖，每个号码能够中多少个一等奖，多少个二等奖到六等奖。
: 程序写出来了，但是运行时间不能接受，求优化代码
: #双色球所有号码历次中奖查询代码
: ...................
--
FROM 123.118.98.*