-2

最近、私はPythonで単語ゲームに取り組んでいますが、それにもかかわらず、少し詳細に問題があります.ゲームのアイデアは、ランダムに与えられた手で文字を使って単語を作成することです. ゲームをコーディネートする機能は、ユーザーに文字の入力を求め、それに従って新しいハンドをプレイしたり、同じハンドを再プレイしたり、ゲームを終了したりします。問題は、ゲームを前のハンドに戻して、ユーザーにもう一度プレイさせることができないことです。代わりに、コードは前の手を修正したバージョンを返すため、単語を正しく入力した場合、その単語の文字は含まれません。「手札を更新する」という機能は別のスコープにあるので、これは不可能だと思いました。私はpython 3.2.1で作業しています

コード全体を投稿して、テストしてその動作を確認できるようにしました。ただし、問題は次の関数にあります: play_game、play_hand、および update_hand。

import random
import string

VOWELS = 'aeiou'

CONSONANTS = 'bcdfghjklmnpqrstvwxyz'

HAND_SIZE = 7

SCRABBLE_LETTER_VALUES = {
    'a': 1, 'b': 3, 'c': 3, 'd': 2, 'e': 1, 'f': 4, 'g': 2, 'h': 4, 'i': 1, 'j': 8, 'k': 5, 'l': 1, 'm': 3, 'n': 1, 'o': 1, 'p': 3, 'q': 10, 'r': 1, 's': 1, 't': 1, 'u': 1, 'v': 4, 'w': 4, 'x': 8, 'y': 4, 'z': 10
}

WORDLIST_FILENAME = "words.txt"

def load_words():

    """
    Returns a list of valid words. Words are strings of lowercase letters.

    Depending on the size of the word list, this function may
    take a while to finish.
    """
    print ("Loading word list from file...")
    # inFile: file
    inFile = open(WORDLIST_FILENAME, 'r')
    # wordlist: list of strings
    wordlist = []
    for line in inFile:
        wordlist.append(line.strip().lower())
    print ("  ", len(wordlist), "words loaded.")
    return wordlist

def get_frequency_dict(sequence):

    """
    Returns a dictionary where the keys are elements of the sequence
    and the values are integer counts, for the number of times that
    an element is repeated in the sequence.

    sequence: string or list
    return: dictionary
    """
    # freqs: dictionary (element_type -> int)
    freq = {}
    for x in sequence:
        freq[x] = freq.get(x,0) + 1
    return freq


def get_word_score(word, n):

    """
    Returns the score for a word. Assumes the word is a
    valid word.

    The score for a word is the sum of the points for letters
    in the word multiplied by the length of the word, plus 50
    points if all n letters are used on the first go.

    Letters are scored as in Scrabble; A is worth 1, B is
    worth 3, C is worth 3, D is worth 2, E is worth 1, and so on.

    word: string (lowercase letters)
    returns: int >= 0
    """
    score = 0
    for character in word:
        score = score + (SCRABBLE_LETTER_VALUES[character])
    score = score * len(word)
    if len(word) == 7:
        score = score + 50
    return score


def display_hand(hand):

    """
    Displays the letters currently in the hand.

    For example:
       display_hand({'a':1, 'x':2, 'l':3, 'e':1})
    Should print out something like:
       a x x l l l e
    The order of the letters is unimportant.

    hand: dictionary (string -> int)
    """
    for letter in hand.keys():
        for j in range(hand[letter]):
             print (letter,)             


def deal_hand(n):

    """
    Returns a random hand containing n lowercase letters.
    At least n/3 the letters in the hand should be VOWELS.

    Hands are represented as dictionaries. The keys are
    letters and the values are the number of times the
    particular letter is repeated in that hand.

    n: int >= 0
    returns: dictionary (string -> int)
    """
    hand={}
    num_vowels = int(n / 3)

    for i in range(num_vowels):
        x = VOWELS[random.randrange(0,len(VOWELS))]
        hand[x] = hand.get(x, 0) + 1

    for i in range(num_vowels, n):    
        x = CONSONANTS[random.randrange(0,len(CONSONANTS))]
        hand[x] = hand.get(x, 0) + 1

    return hand


def update_hand(hand, word):

    """
    Assumes that 'hand' has all the letters in word.
    In other words, this assumes that however many times
    a letter appears in 'word', 'hand' has at least as
    many of that letter in it. 

    Updates the hand: uses up the letters in the given word
    and returns the new hand, without those letters in it.

    Has no side effects: does not modify hand.

    word: string
    hand: dictionary (string -> int)    
    returns: dictionary (string -> int)
    """
    for i in word:
        new_hand = hand
        new_VOWELS = VOWELS
        new_CONSONANTS = CONSONANTS
        if i in VOWELS:
            new_VOWELS = new_VOWELS.replace("i", "")
            new_hand[i] = new_hand.get(i, 0) - 1
            if new_hand[i] <= 0:
                new_hand.pop(i, None)

        else:
            new_CONSONANTS = new_CONSONANTS.replace("i", "")
            new_hand[i] = new_hand.get(i, 0) - 1
            if new_hand[i] <= 0:
                new_hand.pop(i, None)
    return (new_hand)


def is_valid_word(word, hand, word_list):

    """
    Returns True if word is in the word_list and is entirely
    composed of letters in the hand. Otherwise, returns False.
    Does not mutate hand or word_list.

    word: string
    hand: dictionary (string -> int)
    word_list: list of lowercase strings
    """

    for character in word:
        x = 0
        if character in hand:
            x = x + 1
    if (x==(len(word)-1)) and (word in word_list):
        return (True)
    else:
        return (False)

def calculate_handlen(hand):
    handlen = 0
    for v in hand.values():
        handlen += v
    return handlen


def play_hand(hand, word_list):

    """
    Allows the user to play the given hand, as follows:

    * The hand is displayed.

    * The user may input a word.

    * An invalid word is rejected, and a message is displayed asking
      the user to choose another word.

    * When a valid word is entered, it uses up letters from the hand.

    * After every valid word: the score for that word is displayed,
      the remaining letters in the hand are displayed, and the user
      is asked to input another word.

    * The sum of the word scores is displayed when the hand finishes.

    * The hand finishes when there are no more unused letters.
      The user can also finish playing the hand by inputing a single
      period (the string '.') instead of a word.

      hand: dictionary (string -> int)
      word_list: list of lowercase strings

    """
    score = 0
    han = hand
    while True:
        print ("Score= ", score)
        word = str(input("Enter your word: "))
        value = is_valid_word(word, han, word_list)
        if value == True:
            print ("Congratulations, that word is worth", get_word_score(word, 7), "points")
            print ("Please input another word")
            han = update_hand(han, word)
            print ("Current hand=")
            display_hand(han)
            score = score + get_word_score(word, 7)
        elif word == "." or len(hand)==0:
            break
        else:
            print ("Current hand=")
            display_hand(han)
            print ("Sorry, that word is not valid")
            print ("Please choose another word")



def play_game(word_list):

    """
    Allow the user to play an arbitrary number of hands.

    * Asks the user to input 'n' or 'r' or 'e'.

    * If the user inputs 'n', let the user play a new (random) hand.
      When done playing the hand, ask the 'n' or 'e' question again.

    * If the user inputs 'r', let the user play the last hand again.

    * If the user inputs 'e', exit the game.

    * If the user inputs anything else, ask them again.
    """
    global handy 
    handy = deal_hand(7)
    while True:
        print ("Use 'n' to play a new hand, 'r' to play the last hand again or 'e' to exit game")
        inp = str(input("Enter a letter:'n' or 'r' or 'e': ="))
        if inp == 'n':
            hand = deal_hand(7)
            handy = hand
            print("Current hand =")
            display_hand(hand)
            play_hand(hand, word_list)
        elif inp == 'r':
            print("Current hand =")
            display_hand(handy)
            play_hand(handy, word_list)
        elif inp == 'e':
            exit(play_game)
        else:
            print ("Please input a valid letter")

if __name__ == '__main__':
    word_list = load_words()
    play_game(word_list)
4

1 に答える 1

1

と書くと と同じものy = xy指します。コピーしませ。「不変」オブジェクトでは、「変更」を明らかにするために新しいオブジェクトを作成する必要があるため、これはコピーのように見えます。リストのような「変更可能な」オブジェクトでは、と が同じオブジェクトを参照するため、一方の名前への変更がもう一方の名前に反映されます。xxxy

おそらくリストをコピーしたいと思うでしょう:

new_hand = hand[:] # this funny slicing notation produces a shallow copy of the original list
于 2013-08-13T04:17:45.207 に答える