Question

我对Python很陌生，只想提取这些客户地址的城市：

clients = ["Peter, Calle Fantasia 15, Madrid", "Robert, Plaza de Perdas 2, 
            Sevilla", "Paul, Calle Polo, Madrid", "Francesco, Plaza de Opo I, Segovia"]

有人可以帮忙吗？提前非常感谢您！

Answer 1

您可以使用列表推导，并保留从第一个,开始的每个字符串中的最后一个元素。

为此，请使用string.split设置,作为分隔符，它将在出现逗号时拆分每个字符串，将结果列表切成最后一个元素，然后使用string.strip删除前导空格：

clients = ["Peter, Calle Fantasia 15, Madrid", "Robert, Plaza de Perdas 2, 
            Sevilla", "Paul, Calle Polo, Madrid", "Francesco, Plaza de Opo I, Segovia"]

[i.split(',')[-1].strip() for i in clients]
# ['Madrid', 'Sevilla', 'Madrid', 'Segovia']

有关上述方法的更多详细信息，建议您看看：

Answer 2

如果clients的元素始终为"name, address, city"格式，则可以像这样split：

# List comprehension, splits each element of client on commas,
# then takes the final element (stripping any whitespace)
clients = [client.split(',')[-1].strip() for client in clients]

>>> print(clients)

['Madrid', 'Sevilla', 'Madrid', 'Segovia']

Answer 3

不使用列表理解

clients = ["Peter, Calle Fantasia 15, Madrid", "Robert, Plaza de Perdas 2, Sevilla", "Paul, Calle Polo, Madrid", "Francesco, Plaza de Opo I, Segovia"]
list_of_cities =[]
for i in clients:
    index_last_comma = 0
    for j in range(len(i)-1,0,-1):
        if i[j]==',':
            index_last_comma  = j
            break
    city = i[j+1:len(i)].strip()
    list_of_cities.append(city)

print(list_of_cities)
# output ['Madrid', 'Sevilla', 'Madrid', 'Segovia']

从列表中提取某些值

3 个答案: