Question

Python 3.x.我试图创建一个工资单程序，接受每个员工工作小时数的输入，并将其乘以预定的工资。如果用户输入无效输入，则错误陷阱，例如＆＃34; abc＆＃34;用了几个小时。然而，它不会接受浮动值。我知道isdigit（）不适用于小数，因为＆＃34;。＆＃34;在isdigit（）中返回一个false值。如何修改此程序以接受int和float值作为有效输入？

#! python3
#This is a program to calculate payroll expenses.

employees = ['Oscar', 'Judy', 'Sandra', 'Tino', 'Andres', 'Rich', 'Matt', 'Daniel', 'Natalie', 'Elena']
employeeHourlyPay = [14.5, 14.5, 13.5, 13.0, 13.0, 11.0, 11.0, 10.0, 9.0, 10.0]
employeeHours = []
totalPay = []
#Iterate through employees and ask for hours worked. Program will check for
#valid digit inputs, and prompt you to only enter digits when anything else
#is entered as input.

#****Fix to accept decimals****#
for i in employees:
    while True:
        print('Enter hours for', i , ':')
        x = str(input())
        if x.isdigit():
            employeeHours.append(float(x))
            break
        else:
            print('Please use numbers or decimals only.')
            continue
#***End Fix***    
#Calculate pay per employee and add to list.
for i, j in zip(employeeHourlyPay, employeeHours):
    totalPay.append(i * float(j))

#Display pay per employee by iterating through employees and totalPay.
for i, j in zip(employees, totalPay):
    print(i + "'s pay is", str(j))


#Calculate and display total payroll by summing items in totalPay.
print('Total Payroll: ' + str(sum(totalPay)))

Answer 1

python中常见的习惯用法是EAFP（比请求更容易要求宽恕），这意味着基本上只是尝试转换并处理异常，例如：

x = str(input())
try:
    employeeHours.append(float(x))
    break
except ValueError:
    print('Please use numbers or decimals only.')
    continue

Answer 2

最好说你很抱歉而不是要求获得许可......只是尝试从输入中获得float并且很好......如果你得到例外，那是因为它不是浮动的：

for i in employees:
    while True:
        print('Enter hours for', i, ':')
        try:
            x = float(input())
            print("You entered x=%s" % x)
            employeeHours.append(x)
            break
        except ValueError:
            print('Please use numbers or decimals only.')
            continue

Python 3：接受工资核算程序的小数

2 个答案: