我正在积极学习Python（3.5）并且非常享受它。
我想对字符串操作感到满意，因此决定制作一个基本的加密器/解密器，在字母表中向前移动一个字符串。
以下是代码：
def encrypt(string):
string = string.replace('z', 'a')
string = string.replace('y', 'z')
string = string.replace('x', 'y')
string = string.replace('w', 'x')
string = string.replace('v', 'w')
string = string.replace('u', 'v')
string = string.replace('t', 'u')
string = string.replace('s', 't')
string = string.replace('r', 's')
string = string.replace('q', 'r')
string = string.replace('p', 'q')
string = string.replace('o', 'p')
string = string.replace('n', 'o')
string = string.replace('m', 'n')
string = string.replace('l', 'm')
string = string.replace('k', 'l')
string = string.replace('j', 'k')
string = string.replace('i', 'j')
string = string.replace('h', 'i')
string = string.replace('g', 'h')
string = string.replace('f', 'g')
string = string.replace('e', 'f')
string = string.replace('d', 'e')
string = string.replace('c', 'd')
string = string.replace('b', 'c')
string = string.replace('a', 'b')
return string
def decrypt(string):
string = string.replace('b', 'a')
string = string.replace('c', 'b')
string = string.replace('d', 'c')
string = string.replace('e', 'd')
string = string.replace('f', 'e')
string = string.replace('g', 'f')
string = string.replace('h', 'g')
string = string.replace('i', 'h')
string = string.replace('j', 'i')
string = string.replace('k', 'j')
string = string.replace('l', 'k')
string = string.replace('m', 'l')
string = string.replace('n', 'm')
string = string.replace('o', 'n')
string = string.replace('p', 'o')
string = string.replace('q', 'p')
string = string.replace('r', 'q')
string = string.replace('s', 'r')
string = string.replace('t', 's')
string = string.replace('u', 't')
string = string.replace('v', 'u')
string = string.replace('w', 'v')
string = string.replace('x', 'w')
string = string.replace('y', 'x')
string = string.replace('z', 'y')
string = string.replace('a', 'z')
return string
choice = input('Do you want to decrypt or encrypt a sentence? (d / e)')
question = 'Give me a sentence to %s\n'
if choice == 'd':
encrypted_str = input(question % 'decrypt')
decrypted_str = decrypt(encrypted_str)
print(decrypted_str)
elif choice == 'e':
plaintext = input(question % 'encrypt')
encrypted_str = encrypt(plaintext)
print(encrypted_str)
else:
print('That is not a valid option')
我知道这是不你应该怎么做但我不知道怎么回事。
我遇到的问题是，如果我尝试加密＆＃39; zaaz＆＃39;它给了我＆bbbb＆＃39;而不是＆＃39; abba＆＃39;。我知道我的错在哪里（＆＃39; z＆＃39;＆＃39; a＆＃39;＆＃39; a＆＃39;＆＃39;＆＃39;＆＃39;＆＃39;＆＃39;＆＃39;＆＃39;＆＃39;＆＃39;不是如何解决它。有人可能会建议一个更好的方法来做到这一点。
P.S。我看到有人使用modulo运算符来使用字母表作为字符串进行文本换行或字符串索引，但我不知道如何在此处实现它。
有什么建议吗？
答案 0 :(得分：0)
作为建议， 看一下python（https://docs.python.org/2/library/functions.html#chr，https://docs.python.org/2/library/functions.html#ord）的chr / ord函数，它基本上只需要一个char并将其转换为一个数字，你可以在其中添加一个常量并将其转换回一个使用ord的字符串。
然后你可以使用for循环或列表理解来遍历你的代码。
我没有显示代码，因为你想学习，但我可以很容易地提供一些帮助：）
答案 1 :(得分：0)
strssmall = 'abcdefghijklmnopqrstuvwxyz'
strscaps = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
shift = 1 # How many characters need to shift
def encrypt(inp):
data = []
for i in inp:
if i.strip() and i in strssmall:
data.append(strssmall[(strssmall.index(i) + shift) % 26])
elif i.strip() and i in strscaps:
data.append(strscaps[(strscaps.index(i) + shift) % 26])
else:
data.append(i)
output = ''.join(data)
return output
def decrypt(inp):
data = []
for i in inp:
if i.strip() and i in strssmall:
data.append(strssmall[(strssmall.index(i) - shift) % 26])
elif i.strip() and i in strscaps:
data.append(strscaps[(strscaps.index(i) - shift) % 26])
else:
data.append(i)
output = ''.join(data)
return output
choice = input('Do you want to decrypt or encrypt a sentence? (d / e) :')
question = 'Give me a sentence to %s\n'
if choice == 'd':
encrypted_str = input(question % 'decrypt')
decrypted_str = decrypt(encrypted_str)
print(decrypted_str)
elif choice == 'e':
plaintext = input(question % 'encrypt')
encrypted_str = encrypt(plaintext)
print(encrypted_str)
else:
print('That is not a valid option')
另外，您可以访问link
答案 2 :(得分：0)
一种简单易懂的方法是使用dictionaries将旧字符映射到新字符，然后在joining将它们放在一起之前单独转换每个字符，例如这样：
>>> encrypt("zaaz")
'abba'
>>>
使用密码函数就像上面显示的一样简单，使用字典的get方法转换字符，如果不是转换表，则保持不变
现在你得到了正确的结果
#property copyright "Copyright 2017, MetaQuotes Software Corp."
#property link "https://www.mql5.com"
#property version "1.00"
#property strict
extern double ema_red = iMA( NULL, 0, 6, 0, MODE_EMA, PRICE_CLOSE, 0 );
extern double ema_purple = iMA( NULL, 0, 30, 0, MODE_EMA, PRICE_CLOSE, 0 );
extern double ema_blue = iMA( NULL, 0, 8, 1, MODE_EMA, PRICE_CLOSE, 0 );
// int point;
/*
void start()
{
double ema_red = iMA( NULL, 0, 6, 0, MODE_EMA, PRICE_CLOSE, 0 );
double ema_purple = iMA( NULL, 0, 30, 0, MODE_EMA, PRICE_CLOSE, 0 );
double ema_blue = iMA( NULL, 0, 8, 1, MODE_EMA, PRICE_CLOSE, 0 );
return;
}
*/
void OnTick()
{
// double ema_red = iMA( NULL, 0, 6, 0, MODE_EMA, PRICE_CLOSE, 0 );
// double ema_purple = iMA( NULL, 0, 30, 0, MODE_EMA, PRICE_CLOSE, 0 );
// double ema_blue = iMA( NULL, 0, 8, 1, MODE_EMA, PRICE_CLOSE, 0 );
if (OrdersTotal()<=21)
{
if (ema_red == ema_purple)&&(ema_red > ema_blue) // check when the ea's cross/meet
Buy_Order();
if (ema_red == ema_purple)&&(ema_red < ema_blue) // check when the ea's cross/meet
Sell_Order();
}
}
void Buy_Order()
{
double TP = Ask +(PipsToPointFactor()*15);
double SL = (MarketInfo(Symbol(),MODE_TICKVALUE)) - (PipsToPointFactor()*5);
OrderSend(Symbol(),OP_BUY,0.6,Ask,(3*PipsToPointFactor()),SL,TP,"",0,0,Green);
}
void Sell_Order()
{
double TP = (Bid - (15*PipsToPointFactor()));
double SL = ((MarketInfo(Symbol(),MODE_TICKVALUE)) + (PipsToPointFactor()*5 );
OrderSend(Symbol(),OP_SELL,0.6,Bid,(3*PipsToPointFactor()),SL,TP,"",0,0,Red);
}
int OnInit()
{
return(INIT_SUCCEEDED);
}
void OnDeinit(const int reason)
{
}
int PipsToPointFactor() // NOT MY OWN FUNCTION, takes care of pips to points issue
{
int point;
if(Digits==5 || Digits==3) // Check whether it's a 5 digit broker ( 3 digits for Yen )
point=10; // 1 pip to 10 point if 5 digit
else if(Digits==4 || Digits==2)
point=1; // 1 pip to 1 point if 4 digit
return(point);
}