Delete object in array within another array while checking for valid index

Question

The task is to delete an object from an array within another array (a touple if i'm correct). To bugproof i would need to check whether the index is actually valid. This means ranges from 0 to 4 (so input should be greater 0 smaller 4) and obviously not be a string or float of any kind.

I have tried to do that with my def function. i obviously want loopx to only be false if indeed both of the above mentioned criteria are met. otherwise i would want to jump to except. I felt like if that is what I've done but now been stuck for over an hour.

#Write a Python program to delete an existing item from the array

#function used to check for valid input
def valuecheck(checker):
  loopx = True
  while loopx:
    try:
      if checker == int(checker) and (checker>0 and checker<4):
      #first it checks if the input is actually an integer
      #checker = int(checker) 
      #checks if the input index number is within range of the array
        loopx = False 
      else:
        checker = input("Value isn't a valid input, try again: ")
      return(checker)
      #if input isn't an integer the below prompt is printed
    except:
      checker = input("Value isn't a valid input, try again: ")

#the example array is defined and printed
myarray = ['i', [1, 3, 5, 7, 9]]
print(myarray[1])

#input defined and checked by the loop
deletion = input("Please input the index of the element you want to remove (0 through 4). Indexes for the elements start at 0, increasing left to right: ")
deletion = valuecheck(deletion)

#pop is then used to remove the value with index "deletion" from the array
myarray[1].pop(deletion)
#finally the new array is printed
print ("This is the new array:",myarray[1])

return(checker) seems to be in the wrong place. It should be at the bottom of the function, outside of the while loop. Better to check the type with isinstance(checker, int) - but checker will never be an int, as input always returns a string. — CryptoFool
– CryptoFool, Commented Nov 9, 2022 at 17:16

CryptoFool · Accepted Answer · 2022-11-09 17:28:18Z

1

Here is a version of the valuecheck() function that should do what you want:

def valuecheck(checker):
    while True:
        try:
            checker = int(checker)
            if checker > 0 and checker < 4:
                return checker
            checker = input("Value must be between 1 and 3, try again: ")
        except:
            checker = input("Value must be an integer, try again: ")

The main difference to your code is that the value in checker is converted to an integer before it is tested to check if it's in the range 1-3. That operation will throw an exception if checker doesn't contain the string representation of a valid integer.

Rather than using the loop variable loopx, the function simply returns when it finds a valid value.

answered Nov 9, 2022 at 17:28

CryptoFool

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1 Comment

freen Over a year ago

thanks for much! this is truly a fantastic community

User051209 · Accepted Answer · 2022-11-13 14:37:44Z

Try this code (it contains only little differences respect yours):

#Write a Python program to delete an existing item from the array

#function used to check for valid input
def valuecheck(checker):
    loopx = True
    while loopx:
        try:
            if (int(checker)>=0 and int(checker)<=4):
                loopx = False
            else:
                checker = input("Value isn't a valid input, try again: ")
        except Exception as ex:
            checker = input("Value isn't a valid input, try again: ")
    return checker

#the example array is defined and printed
myarray = ['i', [1, 3, 5, 7, 9]]
print(myarray[1])

#input defined and checked by the loop
deletion = input("Please input the index of the element you want to remove (0 through 4). Indexes for the elements start at 0, increasing left to right: ")
deletion = valuecheck(deletion)

#pop is then used to remove the value with index "deletion" from the array
myarray[1].pop(int(deletion))
#finally the new array is printed
print ("This is the new array:",myarray[1])

Important: change the conditions in the if

In my opinion the condition is:

if (int(checker)>=0 and int(checker)<=4):

because your array contains 5 elements, so the indexes 0 and 4 are included.

Gábor Fekete · Accepted Answer · 2022-11-09 17:52:57Z

def valuecheck():
    prompt = "Please input the index of the element you want to remove (0 through 4). Indexes for the elements start at 0, increasing left to right: "
    while True:
        try:
            checker = int(input(prompt))
            if 0 <= checker <= 4:
                return checker
            prompt = "Value isn't in range [0,4], try again: "
        except ValueError:
            prompt = "Value isn't a valid integer, try again: "

#the example array is defined and printed
myarray = ['i', [1, 3, 5, 7, 9]]
print(myarray[1])

#input defined and checked by the loop
deletion = valuecheck()

#pop is then used to remove the value with index "deletion" from the array
myarray[1].pop(deletion)
#finally the new array is printed
print ("This is the new array:",myarray[1])

The function will only return with a value when it can convert the input string to an integer and it's between 0 and 4 (inclusive). In other cases it changes the prompt's text and the loop continues.

I also specified the type of exception in the except clause as your program won't exit if you press Ctrl+C (KeyboardInterrupt) if you don't specify.

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Important: change the conditions in the if

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