最近我一直在使用python中的文字游戏，但是我遇到了一些细节上的问题。游戏的想法是用随机给出的字母创建单词。协调游戏的功能要求用户输入一个字母，并根据它，让他/她玩新手，再次玩同一只手或退出游戏。问题是我不能让游戏返回到前一手牌并让用户再次播放。相反，代码会返回前一手的修改版本，这样如果您正确输入了一个单词，那么该单词中的字母就不会出现在其中。我认为这是不可能的，因为“更新手”的功能位于不同的范围。我在python 3.2.1
工作我发布了整个代码，因此您可以对其进行测试并查看其功能。然而，问题在于：play_game，play_hand和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)
答案 0 :(得分：1)
撰写y = x会使y将引用为与x相同的。它确实不复制x。对于“不可变”对象，这看起来像一个副本，因为你必须创建一个新对象才能使任何“变化”变得明显。对于像列表这样的“可变”对象，因为x和y都引用同一个对象，所以对另一个名称的更改会出现在另一个名称中。
您可能希望复制列表：
new_hand = hand[:] # this funny slicing notation produces a shallow copy of the original list