用于添加的Python列表理解
我是Python新手,无法将函数转换为列表。理解涉及用于添加的Python列表理解,python,list-comprehension,Python,List Comprehension,我是Python新手,无法将函数转换为列表。理解涉及值函数,其包含类如下: class Card(object): # Lists containing valid candidates for a card's rank and suit. suits = [None, "spade", "club", "heart", "diamond"] ranks = [None, "ace", "two", "three", "four", "five", "six",
值
函数,其包含类如下:
class Card(object):
# Lists containing valid candidates for a card's rank and suit.
suits = [None, "spade", "club", "heart", "diamond"]
ranks = [None, "ace", "two", "three", "four", "five", "six",
"seven", "eight", "nine", "ten", "jack", "queen", "king"]
# Dictionary containing the ranks and their associative values.
values = {None:0, "ace":1, "two":2, "three":3, "four":4,
"five":5,"six":6,"seven":7, "eight":8,"nine":9,
"ten":10, "jack":10, "queen":10, "king":10}
def __init__(self, rank=None, suit=None):
"""Constructor."""
if rank not in self.ranks:
raise ValueError("Invalid rank.")
if suit not in self.suits:
raise ValueError("Invalid suit.")
self.rank = rank
self.suit = suit
def __str__(self):
"""A string representation of the Card."""
return "{0}:{1}".format(self.rank, self.suit)
另一个类创建卡对象列表,并定义以下函数:
def value(self):
"""Returns an int value containing the summed values of the hand's cards."""
result = 0
for card in self.cards:
result += Card.values[card.rank]
return result
据我所见,value
函数是列表理解的候选函数,但我无法让它工作。我相信下面的内容是正确的,但我仍然会遇到语法错误,我不知道我做错了什么。请注意,我不熟悉Python并列出理解:
def value(self):
result = [x += y for x = Card.values[y.rank] for y in self.cards]
您可以简单地使用如下函数和生成器表达式
def value(self):
return sum(Card.values[card.rank] for card in self.cards)
def value(self):
return sum([Card.values[card.rank] for card in self.cards])
如果您想使用列表理解,那么您可以简单地使用列表理解语法转换生成器表达式,如下所示
def value(self):
return sum(Card.values[card.rank] for card in self.cards)
def value(self):
return sum([Card.values[card.rank] for card in self.cards])
这太棒了,它工作得非常完美,似乎比在这种情况下使用列表理解更好。为了我自己,你能告诉我如何做列表理解吗?@Cristian list comprehension用于创建列表,但你只是尝试对值求和。只是想根据第一条评论增加清晰度。您使用的是列表理解:它是sum中的列表。Sum只添加列表的组件,列表理解返回列表。