計算兩個字符串中字符數的問題

此函數的目的是比較代碼中出現的顏色次數以及出現在猜測中的次數，以最終確定正確猜測了多少顏色。我現在的代碼遇到了一些錯誤，我相信線的數量可以大大簡化。更具體地說，我認爲我過於複雜了。計算兩個字符串中字符數的問題

有沒有人有更好的方法來解決這個問題的建議？我對編程非常陌生，不勝感激。

validColors = ("R", "G", "B", "Y", "O", "P") 

secretCode = "YBGG" 
guess = "BYYG" 

def correctColorCount(secretCode, guess, validColors): 

    count = 0 

    numRedGuess = guess.count("R") 
    numRedCode = secretCode.count("R") 
    sumR = numRedGuess + numRedCode 

    numGreenGuess = guess.count("G") 
    numGreenCode = secretCode.count("G") 
    sumG = numGreenGuess + numGreenCode 

    numBlueGuess = guess.count("B") 
    numBlueCode = secretCode.count("B") 
    sumB = numBlueGuess + numBlueCode 

    numYellowGuess = guess.count("Y") 
    numYellowCode = secretCode.count("Y") 
    sumY = numYellowGuess + numBlueCode 

    numOrangeGuess = guess.count("O") 
    numOrangeCode = secretCode.count("O") 
    sumO = numOrangeGuess + numOrangeCode 

    numPurpleGuess = guess.count("P") 
    numPurpleCode = secretCode.count("P") 
    sumP = numPurpleGuess + numPurpleCode 

    if numRedCode == numRedGuess and sumR != 0 and sumR != 1: 
     count += 1 
     if numRedGuess == 2: 
      count += 1 
     elif numRedGuess == 3: 
      count += 2 
     elif numRedGuess == 4: 
      count += 3 
    elif numRedGuess >= 1 and numRedCode >= 1 and sumR != 0 and sumR != 1: 
     count += 1 
    if numGreenCode == numGreenGuess and sumG != 0 and sumG != 1: 
     count += 1 
     if numGreenGuess == 2: 
      count += 1 
     elif numGreenGuess == 3: 
      count += 2 
     elif numGreenGuess == 4: 
      count += 3 
    elif numGreenGuess >= 1 and numGreenCode >= 1 and sumG != 0 and sumG != 1: 
     count += 1 
    if numBlueCode == numBlueGuess and sumB != 0 and sumB != 1: 
     count += 1 
     if numBlueGuess == 2: 
      count += 1 
     elif numBlueGuess == 3: 
      count += 2 
     elif numBlueGuess == 4: 
      count += 3 
    elif numBlueGuess >= 1 and numBlueCode >= 1 and sumB != 0 and sumB != 1: 
     count += 1 
    if numYellowCode == numYellowGuess and sumY != 0 and sumY != 1: 
     count += 1 
     if numYellowGuess == 2: 
      count += 1 
     elif numYellowGuess == 3: 
      count += 2 
     elif numYellowGuess == 4: 
      count += 3 
    elif numYellowGuess >= 1 and numYellowCode >= 1 and sumY != 0 and sumY != 1: 
     count += 1 
    if numOrangeCode == numOrangeGuess and sumO != 0 and sumO != 1: 
     count += 1 
     if numOrangeGuess == 2: 
      count += 1 
     elif numOrangeGuess == 3: 
      count += 2 
     elif numOrangeGuess == 4: 
      count += 3 
    elif numOrangeGuess >= 1 and numOrangeCode >= 1 and sumO != 0 and sumO != 1: 
     count += 1 
    if numPurpleCode == numPurpleGuess and sumP != 0 and sumP != 1: 
     count += 1 
     if numPurpleGuess == 2: 
      count += 1 
     elif numPurpleGuess == 3: 
      count += 2 
     elif numPurpleGuess == 4: 
      count += 3 
    elif numPurpleGuess >= 1 and numPurpleCode >= 1 and sumP != 0 and sumP != 1: 
     count += 1 

    return count

來源

2017-02-20 Marissa Hedlund

這是對[策劃]（https://en.wikipedia.org/wiki/Mastermind_（board_game））？ – TigerhawkT3

這是！如果我包括我的其他程序，會有幫助嗎？ –

因爲Mastermind是關於計數元素的，所以試一下['collections.Counter']（https://docs.python.org/3/library/collections.html#collections.Counter），我在[This Let's Code Python： Mastermind視頻]（https://www.youtube.com/watch?v=Hv47MO1vQAo）（25行）。 – TigerhawkT3

也許你可以從這樣的建立，使用collections.Counter：

from collections import Counter 


def correct_color_count(secret, guess): 
    correct = Counter(secret) & Counter(guess) 
    return sum(correct.values()) 

correct_color_count(secret="YBGG", guess="BYYG") 
>>> 3

這會給你猜對伯爵不管位置。

來源

2017-02-20 21:35:49 karlson

下面是做這件事：

def countColors(word, colors): 
    return (word.count(color) for color in colors) 

def correctColorCount(secretCode, guess, colors): 
    for index, (guessedColor, correctColor) in enumerate(zip(*map(lambda word: 
     countColors(word, colors), (secretCode, guess)))): 
     color = colors[index] 
     if guessedColor == correctColor: 
      print("You guessed", color, "correctly!") 
     else: 
      print("You guessed", color, "incorrectly. You guessed", guessedColor, "but it was", str(correctColor) + ".") 


>>> validColors = ("R", "G", "B", "Y", "O", "P") 
>>> secretCode = "YBGG" 
>>> guess = "BYYG" 
>>> correctColorCount(secretCode, guess, validColors) 
You guessed R correctly! 
You guessed G incorrectly. You guessed 2 but it was 1. 
You guessed B correctly! 
You guessed Y incorrectly. You guessed 1 but it was 2. 
You guessed O correctly! 
You guessed P correctly!

countColors將計算單詞的每種顏色的數量，給定的顏色來算的列表。這使用列表理解。

correctColorCount解釋說：

map(lambda word: countColors(word, colors), (secretCode, guess))

這將運行與SECRETCODE和猜測countColors。相當於寫：

[countColors(secretCode, colors), countColors(guess, colors)]

zip(*....)

這將需要他們對：每個第一位的，然後是第二等，所以，這基本上將創建一個列表，每種顏色： [actualNumberOfTimes, guessedNumberOfTimes]。

enumerate可能比較容易理解here。

來源

2017-02-20 21:32:55 rassar

哇，這是一個很大的代碼......可以簡化爲：

def correctColorCount(secretCode, guess): 
    guess_list = list(guess) 
    return len(guess) - sum(1 for color in secretCode 
          if color not in guess_list or guess_list.remove(color))

無需額外的模塊。此比較不需要validColors，用戶輸入組合時應檢查validColors。

來源

2017-02-20 21:45:28 zwer

假設我們忽略的顏色從兩個輸入字符串缺少，這裏是一個辦法：

def correctColorCount(secretCode, guess, validColors): 
    count = 0 
    # for all the color characters 
    for color in validColors: 
     # if color count is not zero and same for both strings 
     if secretCode.count(color) == guess.count(color) != 0: 
      count += 1 
    return count

來源

2017-02-20 21:47:11 thanasisp

計算兩個字符串中字符數的問題

回答

相關問題