我怎样才能缩短这段代码？ （Python 3.3） [英] How may I shorten this code? (Python 3.3)

问题描述

我为一个名为Codebreaker的项目完成了一小段代码，其中生成了一个数字并且必须猜测。您的猜测与名为Code的生成数组进行比较。目前，我编程的部分是程序在猜测中输出错误位置的正确数字的AMOUNT。有没有办法缩短或将其变成子程序？

如果guess1 == code [1]：
 wronPlace =（wronPlace + 1）
如果guess1 == code [2]：
 wronPlace =（wronPlace + 1）
如果guess1 == code [3]：
 wronPlace =（wronPlace + 1）
 
如果guess2 == code [0]：
 wronPlace =（wronPlace + 1）
如果guess2 == code [2]：
 wronPlace =（wronPlace + 1）
如果guess2 == code [3]：
 wronPlace =（wronPlace + 1）
 
如果guess3 == code [0]：
 wronPlace =（wronPlace + 1）
如果guess3 == code [1]：
 wronPlace =（wronPlace + 1）
 if guess3 == code [3]：
 wronPlace =（wronPlace + 1）
 
如果guess4 == code [0]：
 wronPlace =（wronPlace + 1）
如果guess4 == code [1]：
 wronPlace =（wronPlace + 1）
如果guess4 =代码[2]：
 wronPlace =（wronPlace + 1）

我尝试了什么：



我试图将其变成子程序或手动缩短它但没有运气。任何人都可以帮忙吗？

解决方案

不短，但可以调试：

  if  guess4 == code [ 0 ]：
 wronPlace =（wronPlace +  1 ）
 如果  guess3  ==代码[ 1 ]：
 wronPlace =（wronPlace +  1 ）
 如果  guess3  == code [ 2 ]：
 wronPlace =（wronPlace +  1 ） 

你确定这部分的 guess3 吗？



[更新]

Quote：

Guess1 Guess2 Guess3和Guess4是我分开的四位数猜测。他们需要与四个生成的数字进行比较

仔细阅读你的代码！

你检查 guess1 3次，

然后你检查 guess2 3次，

然后你检查 guess3 3次，

此时，人们可以指望你检查 guess4 3次但是，

你检查 guess4 1次， guess3 2次。

是吗你想要什么？



[更新]

引用：

Codebreaker：这是一个猜测代码的4位数的游戏。它输出正确位置的正确位数和错误位置的正确位数。代码存储为数组。您有12个猜测猜测代码

好​​的。所以你的代码是完全错误的，因为当代码或猜测中有2个相同的数字时你就不会处理。

示例：代码是1123，猜测是3211，你的程序会说有6个错位数字。

您必须重新考虑逻辑。

使用数组或列表： 5。数据结构 - Python 3.4.8文档 [ ^ ]。

I have finished a small piece of code for a project called Codebreaker where a number is generated and one has to guess. Your guess is compared to the generated array named "Code". Currently the part I have programmed is where the program outputs the AMOUNT of correct digits in the wrong place within your guess. Is there a way to shorten this or turn it into a sub-routine?

if guess1== code[1]:
    wronPlace= (wronPlace+1)
if guess1== code[2]:
    wronPlace= (wronPlace+1)
if guess1== code[3]:
    wronPlace= (wronPlace+1)

if guess2== code[0]:
    wronPlace= (wronPlace+1)
if guess2== code[2]:
    wronPlace= (wronPlace+1)
if guess2== code[3]:
    wronPlace= (wronPlace+1)

if guess3== code[0]:
    wronPlace= (wronPlace+1)
if guess3== code[1]:
    wronPlace= (wronPlace+1)
if guess3== code[3]:
    wronPlace= (wronPlace+1)

if guess4== code[0]:
    wronPlace= (wronPlace+1)
if guess4== code[1]:
    wronPlace= (wronPlace+1)
if guess4= code[2]:
    wronPlace= (wronPlace+1)

What I have tried:

I've tried to turn it into a sub-routine or shorten it manually but have had no luck. Could anyone help?

解决方案

No shorter, but may be debugged:

if guess4== code[0]:
    wronPlace= (wronPlace+1)
if guess3== code[1]:
    wronPlace= (wronPlace+1)
if guess3== code[2]:
    wronPlace= (wronPlace+1)

Are you sure about the guess3 in this part?

[Update]

Quote:

Guess1 Guess2 Guess3 and Guess4 are the four digits I split apart from guess. They need to be compared to the four generated digits

Read carefully your code!
you check guess1 3 times,
then you check guess2 3 times,
then you check guess3 3 times,
at this point, one can expect you to check guess4 3 times but,
you check guess4 1 time and guess3 2 times.
Is it what you want?

[Update]

Quote:

Codebreaker: It is a game where you guess 4 digits of a code. It outputs the amount of correct digits in the correct place and amount of correct digits in the WRONG place. The code is stored as a array. You have 12 guesses to guess the code

OK. So your code is plain wrong because you don't handle when there is 2 identical digits in code or guesses.
Example: code is 1123 and guess is 3211, your program will say there is 6 wrong placed digits.
You have to rethink the logic.

Use arrays or lists: 5. Data Structures — Python 3.4.8 documentation[^].

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