letter = input("Enter a letter to show that particular country: ")
output_names = [name for name in Country if (name[0] in letter)]
print(output_names)
我有这个，我的.txt文件有点长，错误显示是 - ＆＃34; IndexError：字符串索引超出范围＆＃34;。 我怎样才能使它发挥作用？
答案 0 :(得分：0)
您可以使用多种因素组合对列表推导进行小编辑。
空的东西是假的，例如列表/字典/空字符串
interface ICustomer{
id: number;
name: string;
age: number
city: string
}
let customers: Array<ICustomer>;
function generateCustomers(): void {
let id: number = 0;
createCustomer("Drew", id++, 22, "Glassboro");
createCustomer("Mike", id++, 40, "Rineyville");
createCustomer("Justin", id++, 19, "Jonesboro");
createCustomer("Alex", id++, 15, "Paulsboro");
createCustomer("Phil", id++, 32, "Glassboro");
}
function getAllCustomers(): ICustomer[]{
generateCustomers();
return customers;
}
function createCustomer(name:string,id:number,age:number,city:string): void {
let newCustomer:ICustomer = {id:id,name:name,age:age,city:city};
customers.push(newCustomer);
}
const allCustomers = getAllCustomers;
function getCustomerInformation(id:number): ICustomer {
for (let customer of allCustomers()) {
if(customer.id === id){
return customer;
}
}
return null;
}
console.log(getCustomerInformation(1));
not exists
SELECT applicantWarehouse.first
, applicantWarehouse.last
, applicantWarehouse.title
, applicantWarehouse.ID
, jobTracking.applicantID
, jobTracking.jobID
FROM jobTracking INNER JOIN applicantWarehouse ON jobTracking.applicantID = applicantWarehouse.ID
WHERE Degree="MD"
AND NOT EXISTS (
SELECT 1
FROM jobTracking j
WHERE j.jobID = 56
AND j.applicantID = applicantWarehouse.ID )
bool('')
逻辑运算符Out[1]: False将短路。这意味着在这种情况下，如果它在第一次测试时看到bool('a')，它将不会尝试后续测试。
将它们放在一起，我们可以避免尝试索引空字符串。
Out[2]: True答案 1 :(得分：0)
只是这样做：
print [name for name in Country if name.lower().startswith(letter.lower())]