Question

places = ["Jack", "John", "Sochi"]
count=0
multi_word=0
place  = places[count]
while place != "Sochi" and count < len(places):
    if ' ' in place:
        multi_word += 1

    count += 1
    place = places[count]

print ('Number of cities before Sochi:', count)

我的代码应该打印索契之前的城市数量，不包括索契。我不明白这一行（place = places [count]）是做什么的，我也不明白为什么我需要它两次。

Answer 1

foreach会把它搞砸了

places = ["Jack", "John", "Sochi"]
count = 0
for place in places:
    if ' ' in place:
        multi_word += 1
    if place == "Sochi":
        break
    count += 1

Answer 2

count=0
place = places[count]

现在place总是places[0]，即杰克。因此，while循环仅终止于第二个条件，为您提供列表长度为3。

place = places[count]应该进入循环。

Answer 3

您可以使用以下while循环来检查索契之前的地点数量：

places = ["Jack", "John", "Sochi"]
count = 0
multi_word = 0
while count < len(places):
    place = places[count]
    if ' ' in place:
        multi_word += 1
    if place == "Sochi":
        break
    count += 1

print('Number of cities before Sochi:', count)

break语句表示您将退出while循环。

Answer 4

为什么不尝试更多的pythonic解决方案？

places = ["Jack", "John", "Sochi"]

try:
    count = places.index("Sochi")
except ValueError:
    count = len(places)

multi_word = len([place for place in places[:count] if ' ' in place])

使用while循环计算列表中的元素

4 个答案: