Python 从两个列表创建自定义字符串

我制作了一个赠品模块，现在我正在尝试创建一个函数，使用两个类似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)

它们不需要和你看到的一样长，但是奖品中的数字总和总是和获奖者的长度一样

我理想的弦应该是这样的

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

我知道我可能会扩大奖品列表，稍后再压缩这两个奖项，但我真的在努力让那些有多个获奖者的奖项集中在一起。因为该列表最多可包含100个奖品，并且在奖品获奖者中，x的sumx[0]为x

这是我到目前为止尝试过的，但当每个奖项有两个以上的获奖者时，它并没有真正的用处，而且我真的不知道如何知道何时在最后一次迭代中用and替换a

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,

但有些事情我不在乎，比如奖品是用复数形式写的，或者你需要在最后加上一个。谢谢你的帮助

多亏了亨利，我做了一些小的修正，但代码是一样的

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.

---

最好先在奖品列表上循环，然后将获奖者放在字符串结构中，例如：

winners=[约翰、安娜、马克斯、乔治] 奖品=[2，花生，2，胡萝卜] def祝贺获奖者、奖品： s=祝贺您 i=0 奖品中的p： w='，'。[i:i+p[0]] s+={}表示赢得{}{}，.formatw，1，p[1]，如果p[0]==1 i+=p[0] 返回s 祝贺获奖者、奖品 祝贺约翰、安娜每人赢得1枚花生，祝贺马克斯、乔治每人赢得1枚胡萝卜， 您需要修改w Phase，以备不时之需，而不是，例如：

w='，'。[i:i+p[0]] w='and'.joinw.rsplit'，'1 它将替换最后出现的“，”by“and”，最后一个字符串如下所示

---

祝贺john和anna各赢了一个花生，max和george各赢了一个胡萝卜，

最好先在奖品列表上循环，然后将获奖者放在字符串结构中，例如：

winners=[约翰、安娜、马克斯、乔治] 奖品=[2，花生，2，胡萝卜] def祝贺获奖者、奖品： s=祝贺您 i=0 奖品中的p： w='，'。[i:i+p[0]] s+={}表示赢得{}{}，.formatw，1，p[1]，如果p[0]==1 i+=p[0] 返回s 祝贺获奖者、奖品 祝贺约翰、安娜每人赢得1枚花生，祝贺马克斯、乔治每人赢得1枚胡萝卜， 您需要修改w Phase，以备不时之需，而不是，例如：

w='，'。[i:i+p[0]] w='and'.joinw.rsplit'，'1 它将替换最后出现的“，”by“and”，最后一个字符串如下所示 祝贺约翰和安娜每人赢得1枚花生，祝贺马克斯和乔治每人赢得1枚胡萝卜，

请尝试以下方法：

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

s = "Congratulations to "
winners_index = 0
st = ""
for p in prizes:
    i = 0
    for w_in in range(winners_index,winners_index+p[0]):
        if i == 0:
            st = st + s +winners[w_in]+','
        elif i == p[0]-1:
            st = st + winners[w_in]
        else:
            st = st + winners[w_in]+','
        i = i + 1
    st= st+ ' for winning '+p[1]+','
    winners_index = winners_index + p[0]
print(st)

请试试这个：

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

s = "Congratulations to "
winners_index = 0
st = ""
for p in prizes:
    i = 0
    for w_in in range(winners_index,winners_index+p[0]):
        if i == 0:
            st = st + s +winners[w_in]+','
        elif i == p[0]-1:
            st = st + winners[w_in]
        else:
            st = st + winners[w_in]+','
        i = i + 1
    st= st+ ' for winning '+p[1]+','
    winners_index = winners_index + p[0]
print(st)

---

问题陈述假设>奖品中的数字之和始终与获奖者的长度相同问题陈述假设>奖品中的数字之和始终与获奖者的长度相同