Python 如何检查一个阵列是否有按升序排列的卡，或两张相同的卡或三张相同的卡等

我正在做一个扑克模拟，当时我被困在如何让python检查一个数组中的对、直、三种等等。我已经编写了我的代码，以便每个卡数组生成一个第二个数组，其中包含每张卡的存储值。 比如说

a=[1,3,4,2,5,8,1]  b=[2,2,4,7,10] c=[5,6,5,5,5]

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如何检查a是否至少有5个连续的数字（a直线），b是否至少有2个相互相等的数字（一对），c是否有4个。对手进行排序。这使得检查直线很容易，因为您需要按顺序排列五个数字

对于N类，遍历列表，查看每个项目的数量。例如：

for pos in range(len(hand)):
    card_count = hand.count(hand[pos])
    if card_count >= 2:
        print "Hand has", card_count, hand[pos], "'s"

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我并没有在这里透露所有的细节——这将为每一双打印两次，3种打印三次，等等。我假设您要求的是您需要的基本列表方法。

对手进行排序。这使得检查直线很容易，因为您需要按顺序排列五个数字

对于N类，遍历列表，查看每个项目的数量。例如：

for pos in range(len(hand)):
    card_count = hand.count(hand[pos])
    if card_count >= 2:
        print "Hand has", card_count, hand[pos], "'s"

---

我并没有在这里透露所有的细节——这将为每一双打印两次，3种打印三次，等等。我假设您要求的是您需要的基本列表方法。

对手进行排序。这使得检查直线很容易，因为您需要按顺序排列五个数字

对于N类，遍历列表，查看每个项目的数量。例如：

for pos in range(len(hand)):
    card_count = hand.count(hand[pos])
    if card_count >= 2:
        print "Hand has", card_count, hand[pos], "'s"

---

我并没有在这里透露所有的细节——这将为每一双打印两次，3种打印三次，等等。我假设您要求的是您需要的基本列表方法。

对手进行排序。这使得检查直线很容易，因为您需要按顺序排列五个数字

对于N类，遍历列表，查看每个项目的数量。例如：

for pos in range(len(hand)):
    card_count = hand.count(hand[pos])
    if card_count >= 2:
        print "Hand has", card_count, hand[pos], "'s"

---

我并没有在这里透露所有的细节——每双打印两次，每种打印三次，等等。我想您要求的是最基本的列表方法。

这应该足以让您开始

def check_hand(_hand_):
    last_c = ''
    straight = 0
    # This is illustrative, you can use this to return
    # the greatest hand one could have
    for card_value, number_of_occurences in _hand_.iteritems():
        if number_of_occurences == 2:
            print("We have a 2 of a kind")
        elif number_of_occurences == 3:
            print("We have a 3 of a kind")
        elif number_of_occurences == 4:
            print("We have a 4 of a kind")

        if last_c == '':
            last_c = card_value
        else:
            if card_value - last_c == 1:
                straight += 1
            last_c = card_value

    if straight >= 4:
        print("we have a straight")

a = [1, 3, 4, 2, 5, 8, 1]
b = [2, 2, 4, 7, 10]
c = [5, 6, 5, 5, 5]

# nifty way of assigning a dictionary with how many
# occurrences of a number in a list, and sorting it
check = {
    'a': dict((i, a.count(i)) for i in sorted(a)),
    'b': dict((i, b.count(i)) for i in sorted(b)),
    'c': dict((i, c.count(i)) for i in sorted(c)),
}

for which, hand in check.iteritems():
    print "Hand: " + which
    check_hand(hand)

---

这应该足以让你开始

def check_hand(_hand_):
    last_c = ''
    straight = 0
    # This is illustrative, you can use this to return
    # the greatest hand one could have
    for card_value, number_of_occurences in _hand_.iteritems():
        if number_of_occurences == 2:
            print("We have a 2 of a kind")
        elif number_of_occurences == 3:
            print("We have a 3 of a kind")
        elif number_of_occurences == 4:
            print("We have a 4 of a kind")

        if last_c == '':
            last_c = card_value
        else:
            if card_value - last_c == 1:
                straight += 1
            last_c = card_value

    if straight >= 4:
        print("we have a straight")

a = [1, 3, 4, 2, 5, 8, 1]
b = [2, 2, 4, 7, 10]
c = [5, 6, 5, 5, 5]

# nifty way of assigning a dictionary with how many
# occurrences of a number in a list, and sorting it
check = {
    'a': dict((i, a.count(i)) for i in sorted(a)),
    'b': dict((i, b.count(i)) for i in sorted(b)),
    'c': dict((i, c.count(i)) for i in sorted(c)),
}

for which, hand in check.iteritems():
    print "Hand: " + which
    check_hand(hand)

---

这应该足以让你开始

def check_hand(_hand_):
    last_c = ''
    straight = 0
    # This is illustrative, you can use this to return
    # the greatest hand one could have
    for card_value, number_of_occurences in _hand_.iteritems():
        if number_of_occurences == 2:
            print("We have a 2 of a kind")
        elif number_of_occurences == 3:
            print("We have a 3 of a kind")
        elif number_of_occurences == 4:
            print("We have a 4 of a kind")

        if last_c == '':
            last_c = card_value
        else:
            if card_value - last_c == 1:
                straight += 1
            last_c = card_value

    if straight >= 4:
        print("we have a straight")

a = [1, 3, 4, 2, 5, 8, 1]
b = [2, 2, 4, 7, 10]
c = [5, 6, 5, 5, 5]

# nifty way of assigning a dictionary with how many
# occurrences of a number in a list, and sorting it
check = {
    'a': dict((i, a.count(i)) for i in sorted(a)),
    'b': dict((i, b.count(i)) for i in sorted(b)),
    'c': dict((i, c.count(i)) for i in sorted(c)),
}

for which, hand in check.iteritems():
    print "Hand: " + which
    check_hand(hand)

---

这应该足以让你开始

def check_hand(_hand_):
    last_c = ''
    straight = 0
    # This is illustrative, you can use this to return
    # the greatest hand one could have
    for card_value, number_of_occurences in _hand_.iteritems():
        if number_of_occurences == 2:
            print("We have a 2 of a kind")
        elif number_of_occurences == 3:
            print("We have a 3 of a kind")
        elif number_of_occurences == 4:
            print("We have a 4 of a kind")

        if last_c == '':
            last_c = card_value
        else:
            if card_value - last_c == 1:
                straight += 1
            last_c = card_value

    if straight >= 4:
        print("we have a straight")

a = [1, 3, 4, 2, 5, 8, 1]
b = [2, 2, 4, 7, 10]
c = [5, 6, 5, 5, 5]

# nifty way of assigning a dictionary with how many
# occurrences of a number in a list, and sorting it
check = {
    'a': dict((i, a.count(i)) for i in sorted(a)),
    'b': dict((i, b.count(i)) for i in sorted(b)),
    'c': dict((i, c.count(i)) for i in sorted(c)),
}

for which, hand in check.iteritems():
    print "Hand: " + which
    check_hand(hand)

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