确保要列出的多个用户输入是 while 循环 Python 3 中的整数

Question

the code below works as expected. I am trying to ensure that all data entered is integers. Non integers should get the error meessage. My while loop works for the first input where the user is asked what # of numbers to enter. However it does not catch errors for the subsequent inputs.

while True:
    try:
        userIn = int(input("Input no of numbers :"))     

        userNums = []
        uNums = []

        print("Type " + str(userIn) + " numbers in list :")          

        for index in range(int(userIn)):
            userNums.append(input("entry" + str(index+1) +" = "))

        for number in userNums:
            if number not in uNums:
                uNums.append(number)

        print("unique number counted")    
        for uNum in uNums:       
            print(str(uNum) + " count is " + str(userNums.count(uNum)) + "times")     
        break
    except:
        print("Error. only ints allowed")

Answer 1

You don't actually try and convert to integers at any point. Below just saves the strings.

for index in range(int(userIn)):
        userNums.append(input("entry" + str(index+1) +" = "))

And, by the way, you can get rid of the following:

for number in userNums:
    if number not in uNums:
        uNums.append(number)

If you made userNum a set - userNum = set() - which can only have unique values in it.

Answer 2

Your main mistake is a forgotten int():

for index in range(userIn):
    num = int(input("entry" + str(index+1) +" = "))
    userNums.append(num)

However, still the whole program has to start again in case of a wrong input, therefore I recommend a try&except whenever in need. You also tend to:

1.convert variables when not necessary
2.use + and str() , which might be not the best habit

This code should be straight forward and the customised INPUT-function might be useful in the future:)

def INPUT(InputText):
    """customized input handling"""
    while True:
        try:
            return int(input(InputText))
        except ValueError:
            print ("Error: Only integers accepted")


#note: you could wrap the code below in a function 
#      and call the function afterwards if you want:

userIn = INPUT("Input no of numbers :")
userNums = []
uNums = []

print("Type {} numbers in list :".format(userIn))

for index in range(userIn):
    num = INPUT("entry {} = ".format(index+1))
    userNums.append(num)

for number in userNums:
    if number not in uNums:
        uNums.append(number)

print("unique number counted")

for uNum in uNums:
    print("{} count is {} times".format(uNum,userNums.count(uNum)))

Answer 3

You're not casting the subsequent inputs to int().

You need something like

userNums.append(int(input("entry" + str(index+1) +" = ")))

Answer 4

The below solution should work. The main changes:

(1) Added explicit ValueError checks, as this is good practice.
(2) Added explicit continue , as this is also good practice.
(3) Incorporated else clauses to create try / except / else structure.
(4) Added try / except to loop of number input code.
(5) Use set to get unique items of list.

while True:

    try:
        userIn = int(input("Input no of numbers :"))

    except ValueError:
        print("Error: only ints allowed")
        continue

    else:

        userNums = []

        print("Type " + str(userIn) + " numbers in list :")          

        for index in range(int(userIn)):
            while True:
                try:
                    x = int(input("entry" + str(index+1) +" = "))
                    userNums.append(x)
                    break
                except ValueError:
                    print("Error: only ints allowed")
                    continue

        uNums = set(userNums)

        print("Unique numbers counted:")

        for uNum in uNums:       
            print(str(uNum) + " count is " + str(userNums.count(uNum)) + " times")

        break