Question

我制作了一个赠品模块，现在我正在尝试创建一个函数，该函数使用两个这样的列表为所有获奖者返回一个祝贺字符串。

winners = ["john", "anna", "max", "george"]
prizes = [(1, "apple"), (1, "pear"), (2, "carrots")]
# (number of these prize avaible, the prize)

它们的长度不必与您看到的长度相同，但是奖品中的数字总和始终与获胜者中的长度相同。

我理想的字符串应该是这样的。

"Congratulations to john for winning 1 apple, anna for winning 1 pear, and max and george for winning 1 carrot each."

我知道我可能可以扩大奖项清单并在以后压缩两个奖项。但我实际上是在尝试将拥有多个获奖者的奖项分组归纳。 100个奖项和总和（x个奖项中x的x [0]）。

这是我到目前为止一直在尝试的方法，但是当每个奖项的获奖者超过两个时，它并没有真正的用处，而且我真的不知道一种方法来知道我何时在上次迭代中替换，与和。

winners = ["john", "anna", "max", "george"]
prizes = [(2, "peanuts"), (2, "carrot")]


def congratulations(winners, prizes):
    s = "Congratulations to "
    for w in winners:
        for p in prizes:
            s += "{} for winning {} {}, ".format(w, p[0], p[1])
            break
    return s


print(congratulations(winners, prizes))
#Congratulations to john for winning 2 peanuts, anna for winning 2 peanuts, max for winning 2 peanuts, george for winning 2 peanuts,

有些事情我不在乎，例如奖品写成复数形式，或者您需要在末尾添加。感谢您的帮助。

感谢亨利，我做了一些小的修改，但是代码是相同的。

winners = ["john", "anna", "max", "george", "carlos"]
prizes = [(1, "apple"), (3, "pears"), (1, "carrots")]


def congratulations(winners, prizes):
    s = "Congratulations to "
    i = 0
    for p in prizes:
        w = ", ".join(winners[i : i + p[0]])
        w = " and ".join(w.rsplit(", ", 1))
        s += "{} for winning {} {}{}; ".format(w, 1, p[1], "" if p[0] == 1 else " each")
        i += p[0]
    s = s[:-2] + "."
    k = s.rfind(";")
    s = s[:k] + "; and" + s[k + 1 :]
    return s


w = congratulations(winners, prizes)
print(w)
# Congratulations to john for winning 1 apple, anna, max and george for winning 1 pears each, and carlos for winning 1 carrots.

Answer 1

最好先遍历奖品列表，然后将优胜者放在字符串结构中，例如：

winners = ["john", "anna", "max", "george"]
prizes = [(2, "peanuts"), (2, "carrot")]

def congratulations(winners, prizes):
    s = "Congratulations to "
    i = 0
    for p in prizes:
        w = ', '.join(winners[i: i + p[0]])
        s += "{} for winning {} {}{}, ".format(w, 1, p[1], '' if p[0]==1 else " each")
        i += p[0]

    return s


print(congratulations(winners, prizes))
# Congratulations to john, anna for winning 1 peanuts each, max, george for winning 1 carrot each,

在首选使用w而不是and的情况下，您需要修改,的用语，例如：

w = ', '.join(winners[i: i + p[0]])
w = ' and '.join(w.rsplit(', ', 1))

它将用', '代替最后一次出现的' and '，最后一个字符串看起来像 Congratulations to john and anna for winning 1 peanuts each, max and george for winning 1 carrot each,

Answer 2

请尝试以下操作：

vec.reserve(2);

从两个列表创建自定义字符串

2 个答案: