1

我有一个看起来像这样的代码：

import random
import numpy as np
from operator import itemgetter
import sys

equal_to = {"a":"u_plus", "b":"u_minus", "c":"v_plus", "d":"v_minus"} #bell state

p = 8 #length of generated binary

key1 = [] #list of generated binary 
for i in range(p): 
    temp = random.randint(0,1)
    key1.append(temp) 

tmplist2 = [] #list of random sample_letters
for i in range(p):
    while True:
        attempt = str(random.choice(list(equal_to)))
        tmplist2.append(attempt)

        if attempt == 'b':
            break

#evaluate result of alice binary and bell state
def eva(alice, bell):
    if alice == 1:
        if bell == 'a' or bell == 'b':
            return 1
        elif bell == 'c' or bell == 'd':
            return 0
    elif alice == 0:
        if bell == 'c' or bell == 'd':
            return 1
        elif bell == 'a' or bell == 'b':
            return 0


for_bob = [] #list of generated binary and bell state through logic gate

for k in key1:
    for t in tmplist2:
        e = eva(k, t)
        for_bob.append(e)

#tr = [[eva(k,t) for t in tmplist2] for k in key1] #list comprehension split the key properly
print("generated random binary strings:", key1)
print("generated bell states:", tmplist2)
print("encrypted strings:", for_bob)

它打印出如下内容：

generated random binary strings:
  [0, 0, 0, 0, 0, 1, 1, 0]
generated bell states:
  ['b', 'c', 'b', 'a', 'a', 'c', 'b', 'b', 'd', 'd', 'd', 'c', 'c', 'b', 'a', 'd', 'd', 'b', 'c', 'c', 'd', 'c', 'd', 'b', 'd', 'c', 'a', 'd', 'c', 'd', 'c', 'c', 'b']
encrypted strings:
  [0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0]

如果字符串“b”出现，我正在寻找一种tmplist2无需从头开始的循环方式。

循环应该按如下方式运行： 应使用'b'key1[0]进行评估，返回 0。由于'b'出来了，外循环应该继续到下一个元素，它应该使用'c'和'b'进行评估，返回 1 0 . 再次因为'b'出来了，外循环将继续到下一个元素并用'a'、'a'、'c'和'b'评估自身，返回 0 0 1 0。这应该一直持续到完成。key1[1]key[2]tmplist2

有人有解决这个问题的方法吗？我正在寻找如何在循环中循环而不是在给出某个条件时返回开头的解决方案。在这种情况下'b'。

注意：这个程序旨在模拟量子隐形传态。

4

1 回答 1

0

主要数据

for_bob = []
key1 = [0, 0, 0, 0, 0, 1, 1, 0]
tmplist2 = ['b', 'c', 'b', 'a', 'a', 'c', 'b', 'b', 'd', 'd', 'd', 'c', 'c', 'b', 'a', 'd', 'd', 'b', 'c', 'c', 'd', 'c', 'd', 'b', 'd', 'c', 'a', 'd', 'c', 'd', 'c', 'c', 'b']

方法一

_tmp = tmplist2[:]
for k in key1:
    while _tmp:
        if _tmp[:1] != ['b']:
            for_bob.append(eva(k, *_tmp[:1]))
            _tmp = _tmp[1:]
        else:
            for_bob.append(eva(k, *_tmp[:1]))
            _tmp = _tmp[1:]
            break

方法二：

_tmp = iter(tmplist2)

for i in key1:
    for j in iter(lambda: next(_tmp), None):
        if j != 'b':
            for_bob.append(eva(i, j))
        else:
            for_bob.append(eva(i, j))
            break

generated random binary strings: [0, 0, 0, 0, 0, 1, 1, 0]
generated bell states: ['b', 'c', 'b', 'a', 'a', 'c', 'b', 'b', 'd', 'd', 'd', 'c', 'c', 'b', 'a', 'd', 'd', 'b', 'c', 'c', 'd', 'c', 'd', 'b', 'd', 'c', 'a', 'd', 'c', 'd', 'c', 'c', 'b']
encrypted strings: [0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0]
于 2020-11-13T09:31:10.170 回答