python - 计算两个字符串中的字符数的问题

Question

该函数的目的是比较某种颜色在代码中出现的次数和在猜测中出现的次数，最终确定有多少种颜色被猜对了。我当前的代码遇到一些错误，我相信行数可以大大简化。更具体地说，我认为我手头的任务过于复杂化了。

有人有更好的方法来解决这个问题吗？我对编程非常陌生，我将不胜感激。

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

Answer 1

假设我们忽略两个输入字符串中缺少的颜色，这是一种方法:

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