我正在为在线Python课程创建一个简单的拼字游戏风格游戏。我在这个问题集中很好地理解了这个概念，我的程序工作得很好。唯一的问题是，在用户玩游戏的控制台输出中，一个随机的“无”字样。正在两条线之间打印，似乎不是原因。问题似乎源于print Current hand: ', display_hand(hand)和inpt = str(raw_input('Enter a word or "." to finish hand: '))之间的play_hand函数
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
}
# -----------------------------------
# Helper code
# (you don't need to understand this helper code)
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', 0)
# wordlist: list of strings
wordlist = []
for line in inFile:
wordlist.append(line.strip().lower())
print " ", len(wordlist), "words loaded."
return wordlist
word_list = load_words()
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
# (end of helper code)
# -----------------------------------
#
# Problem #1: Scoring a word
#
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
word = word.lower()
for letter in word:
score += SCRABBLE_LETTER_VALUES[letter]
score *= len(word)
if len(word) == n:
score += 50
return score
#
# Make sure you understand how this function works and what it does!
#
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, # print all on the same line
print '\n'
#
# Make sure you understand how this function works and what it does!
#
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 = 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
#
# Problem #2: Update a hand by removing letters
#
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 letter in word:
if word.count(letter) <= hand[letter]:
hand[letter] -= word.count(letter)
for key in hand.keys():
if hand[key] == 0:
del hand[key]
return hand
#
# Problem #3: Test word validity
#
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
"""
if word in word_list:
for letter in word:
if not(word.count(letter) <= hand.get(letter, 0)):
return False
return True
def calculate_handlen(hand):
handlen = 0
for v in hand.values():
handlen += v
return handlen
#
# Problem #4: Playing a hand
#
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
"""
inpt = 0
score = 0
while len(hand) > 0:
print 'Current hand: ', display_hand(hand)
inpt = str(raw_input('Enter a word or "." to finish hand: '))
if inpt == '.':
break
if (is_valid_word(inpt, hand, word_list)):
print 'Word score: ', get_word_score(inpt, HAND_SIZE)
score += get_word_score(inpt, HAND_SIZE)
hand = update_hand(hand, inpt)
else:
print 'Not a valid word!'
print 'Hand score: ', score
play_hand(deal_hand(HAND_SIZE), word_list)
答案 0 :(得分：2)
没有return语句的函数实际返回None。
print 'Current hand: ', display_hand(hand)
上面的行打印“Current hand：”，然后执行display_hand(hand)打印作为副作用的东西，然后最后打印函数的返回值None
为了防止这种情况，请在print语句之外的display_hand(hand)中运行，在自己的行中，即
print 'Current hand: ',
display_hand(hand)
答案 1 :(得分：1)
正如@zehnpaard所说，你正在打印None，因为那是display_hand返回的内容。你可以简单地用：
display_hand(hand)
并将Current hand:打印移动到该方法：只需创建第一行：
print 'Current hand: ',
或者，您可以定义get_hand方法返回要打印的字符串：
...
print 'Current hand: ', get_hand(hand)
...
def get_hand(hand):
"""
Get the letters currently in the hand.
For example:
get_hand({'a':1, 'x':2, 'l':3, 'e':1})
Should return something like:
'a x x l l l e'
The order of the letters is unimportant.
hand: dictionary (string -> int)
"""
letters = []
for letter in hand.keys():
for j in range(hand[letter]):
letters.append(letter)
return ' '.join(letters)