Python while loop in try except instruction - python

I have simple code:
def simpleMethod():
try:
list_to_app = []
number_of_a = input('\nHow many a you want to create? ')
number_of_a = int(number_of_a)
for i in range(number_of_a):
user_a = input('\nPlease type here some track: ')
list_to_app.append(user_a.lower())
except ValueError:
stderr.write('Type error - please provide an integer character')
simpleMethod()
I do not want to use something like while True:...
How to make some loop (I think while will be fine in this case) for this kind of flow:
user types non-integer
program shows Type error - please provide an integer character
program goes back to 1. step
This is simple method but I'm stuck.

Add a while True loop to repeat the try block until it can successfully break. You probably also want to return the list you're building so that the caller can access it:
def simpleMethod():
list_to_app = []
while True:
try:
number_of_a = input('\nHow many a you want to create? ')
number_of_a = int(number_of_a)
break
except ValueError:
stderr.write('Type error - please provide an integer character')
for i in range(number_of_a):
user_a = input('\nPlease type here some track: ')
list_to_app.append(user_a.lower())
return list_to_app # no point building a list you don't return!
tracks = simpleMethod()
A simpler way to build a list is a comprehension -- you can actually put almost the entire function in a single comprehension statement within that while, since the exception will raise to the outer loop. You might also want to make the function name more descriptive:
from typing import List
def get_user_tracks() -> List[str]:
"""Prompt the user for a list of tracks."""
while True:
try:
return [
input('\nPlease type here some track: ')
for _ in range(int(input('\nHow many a you want to create? ')))
]
except ValueError:
stderr.write('Type error - please provide an integer character')
tracks = get_user_tracks()

Related

How to detect if input is not given in a python program

I have a python program where it prompts a user input for position or index and deletes the element in the list based on the position or index. The python program works but I'm having issues with the condition where if no user input is given, it automatically deletes the whole line in the list.
Example:
lst = [1,2,3,4,5]
enter position: 2
output: [1,2,4,5]
enter position: #user just pressed enter without giving any input
output: []
I'm writing the function within a class whereby:
def delete(self,index):
"""
This function deletes an item based on the index
:param self: the array
:param index: the index of an item in the array
:return: the array is updated
:raises: IndexError if out of range
"""
if not index:
self.__init__()
if index<0:
index = index + self.count
for i in range(index, self.count -1):
self._array[i] = self._array[i+1]
self.count-=1
and prompting the user input is as such:
position = int(input("Enter position:"))
it's not possible to just press 'enter' without receiving an error due to the position only receiving integers hence I'm looking for a method where if the user doesn't give any position, it registers it and prints just an empty list instead of an error message.
What you're looking for is the try-except block. See the following for an example:
input_invalid = true
while input_invalid:
user_input = input("Enter position: ")
try:
user_input = int(user_input)
input_invalid = false
except ValueError:
print("Please enter a valid integer!")
Here, the try-except block catches any errors (of the type specified
in except) thrown within the code block. In this case, the error results from trying to call int() on a string that does not contain an integer (ValueError). You can use this to explicitly prevent the error and control the logic flow of your program like shown above.
An alternate solution without using try-except is to use the .isdigit() method to validate the data beforehand. If you were to use .isdigit() (which I personally think is better), your code would look something like this:
input_invalid = true
while input_invalid:
user_input = input("Enter position: ")
if user_input.isdigit():
input_invalid = false
else:
print("Please enter a valid integer!")
Hope this helped!

Looping the function

I have this function below, which I have done something wrong in somewhere.
def quantityFunction(product):
valid = False
while True:
if product is not None:
quantity = input("Please enter the amount of this item you would like to purchase: ")
for i in quantity:
try:
int(i)
return int(quantity)
valid = True
except ValueError:
print("We didn't recognise that number. Please try again.")
#If I get here, I want to loop back to the start of this function
return True
return False
To run through, the function is called from the main part of the program like so: quantity = quantityFunction(product)
The return False at the bottom of the code is to do with if product is None, which is needed after a bit of code in another function but has had to go in this function.
If the user input for quantity is a number, all works fine. If it is anything else, the Value Error is printed and you can enter another input. If you put another letter etc in, it repeats again, if you put a number in, it accepts it.
However, it does not return the number you inputted after the letters. It just returns 0.
I suspect this is something to do with how I am repeating the code, i.e. the code should loop back to the start of the function if it hits the Value Error.
Any Ideas?
You said:
the code should loop back to the start of the function if it hits the Value Error.
Then you should not use return statements, otherwise the function will terminate, returning True or False.
Few issue:
1) return statement returns control to the calling function.
2) You are looping over the input, which is wrong.
3) valid=True isn't executed at all.
def quantityFunction(product):
valid = False
while True:
if product is not None:
quantity = raw_input("Please enter the amount of this item you would like to purchase: ")
try:
return int(quantity)
#valid = True (since it is never run)
except ValueError:
print("We didn't recognise that number. Please try again.")
#If I get here, I want to loop back to the start of this function
#return True
return False
quantityFunction("val")
Note : Use raw_input() in case of Python 2.7 and input() in case of 3.x
Try this (some formatting included too, but the functionality should be the same):
def determine_quantity(product): # descriptive function name
if not product: # avoiding nesting
return False
while True:
quantity = input("Please enter the amount of this item you would like to purchase: ")
try:
return int(quantity) # try to convert quantity straight away
except ValueError:
print("We didn't recognise that number. Please try again.")
# nothing here means we simply continue in the while loop
Ideally, you'd take product out. A function should do as little as possible, and this check is better off somewhere else.
def determine_quantity():
while True:
quantity = input("Please enter the amount of this item you would like to purchase: ")
try:
return int(quantity)
except ValueError:
print("We didn't recognise that number. Please try again.")
First, let's address the code. Simply stated, you want a function that will loop until the user enters a legal quantity.
product doesn't do much for the function; check it in the calling program, not here. Let the function have a single purpose: fetch a valid quantity.
Let's work from there in the standard recipe for "loop until good input". Very simply, it looks like:
Get first input
Until input is valid
... print warning message and get a new value.
In code, it looks like this.
def get_quantity():
quantity_str = input("Please enter the amount of this item you would like to purchase: ")
while not quantity_str.isdigit():
print("We didn't recognise that number. Please try again.")
quantity_str = input("Please enter the amount of this item you would like to purchase: ")
return quantity
As for coding practice ...
Develop incrementally: write a few lines of code to add one feature to what you have. Debug that. Get it working before you add more.
Learn your language features. In the code you've posted, you misuse for, in, return, and a function call.
Look up how to solve simple problems. try/except is a more difficult concept to handle than the simple isdigit.
You should try this..
def quantityFunction(product):
valid = False
while True:
if product is not None:
quantity = raw_input("Please enter the amount of this item you would like to purchase: ")
if quantity.isdigit():
return int(quantity)
valid = True
else:
print("We didn't recognise that number. Please try again.")
continue
return False
quantity = quantityFunction("myproduct")

NameError when programming in python

I made a program in python that is supposed to accept a name as user input. It will then check if the name given is contained inside a string that is already given and if it is then the program will print out the telephone next to that name. My code is as follows:
tilefwnikos_katalogos = "Christoforos 99111111: Eirini 99556677: Costas 99222222: George 99333333: Panayiotis 99444444: Katerina 96543217"
check=str(input("Give a name: "))
for check in tilefwnikos_katalogos:
if check=="Christoforos":
arxi=check.find("Christoforos")
elif check=="Eirini":
arxi=check.find("Eirini")
elif check=="Costas":
arxi=check.find("Costas")
elif check=="George":
arxi=check.find("George")
elif check=="Panayiotis":
arxi=check.find("Panayiotis")
elif check=="Katerina":
arxi=check.find("Katerina")
s=check.find(" ",arxi)
arxi=s
y=check.find(":",arxi)
telos=y
apotelesma=tilefwnikos_katalogos[arxi+1:telos]
print(apotelesma)
But when I try to run it, I input the name and then the following message pops up:
Traceback (most recent call last):
File "C:\Users\Sotiris\Desktop\test.py", line 16, in <module> s=check.find(" ",arxi)
NameError: name 'arxi' is not defined
What am I doing wrong?
You're getting your error because arxi isn't getting defined in the first place when then name the user gave is not present on your list.You can fix that by simply adding an unconditional else case to your if/else if bundle as pointed in the comments. But the very way you tackled this problem is faulty, storing data like this in a string is a bad idea, you want to use a dictionary:
phone_catalog = {'Christoforos': 99111111, 'Eirini': 99556677, 'Costas': 99222222, 'George':99333333, 'Panayiotis':99444444, 'Katerina': 96543217}
Also check isn't a very clear variable name, maybe you should try using something better like:
user_name = str(input("Give a name: "))
And now you can do your if/elif condition but replacing it for using dictionary logic and making sure you have a final else, like such:
if user_name in phone_catalog:
print(phone_catalog[user_name])
else:
print("Unknown user")
See how the dictionary made your life much easier and your code cleaner here? Read more on Python Data Structures.
so there are a few things you have overlooked / not going as expected, the first of which is how iterating over strings in python works:
tilefwnikos_katalogos = "Christoforos 99111111: Eirini 99556677: Costas 99222222: George 99333333: Panayiotis 99444444: Katerina 96543217"
for check in tilefwnikos_katalogos:
print(check)
#print(repr(check)) #this shows it as you would write it in code ('HI' instead of just HI)
so check can never be equal to any of the things you are checking it against, and without an else statement the variable arxi is never defined. I'm assuming you meant to use the check from the user input instead of the one in the loop but I'm not sure you need the loop at all:
tilefwnikos_katalogos = "Christoforos 99111111: Eirini 99556677: Costas 99222222: George 99333333: Panayiotis 99444444: Katerina 96543217"
check=str(input("Give a name: ")) #the str() isn't really necessary, it is already a str.
if check=="Christoforos":
arxi=check.find("Christoforos")
elif check=="Eirini":
arxi=check.find("Eirini")
elif check=="Costas":
arxi=check.find("Costas")
elif check=="George":
arxi=check.find("George")
elif check=="Panayiotis":
arxi=check.find("Panayiotis")
elif check=="Katerina":
arxi=check.find("Katerina")
else: raise NotImplementedError("need a case where input is invalid")
s=check.find(" ",arxi)
arxi=s
y=check.find(":",arxi)
telos=y
apotelesma=tilefwnikos_katalogos[arxi+1:telos]
print(apotelesma)
but you could also just see if check is a substring of tilefwnikos_katalogos and deal with other conditions:
if check.isalpha() and check in tilefwnikos_katalogos:
# ^ ^ see if check is within the string
# ^ make sure the input is all letters, don't want to accept number as input
arxi=check.find(check)
else:
raise NotImplementedError("need a case where input is invalid")
although this would make an input of C and t give Cristoforos' number since it retrieves the first occurrence of the letter. An alternative approach which includes the loop (but not calling the variable check!) would be to split up the string into a list:
tilefwnikos_katalogos = "..."
check = input(...)
for entry in tilefwnikos_katalogos.split(":"):
name, number = entry.strip().split(" ")
if check == name:
apotelesma=number
break
else:
raise NotImplementedError("need a case where input is invalid")
although if you are going to parse the string anyway and you may use the data more then once it would be even better to pack the data into a dict like #BernardMeurer suggested:
data = {}
for entry in tilefwnikos_katalogos.split(":"):
name, number = entry.strip().split(" ")
data[name] = number #maybe use int(number)?
if check in data:
apotelesma = data[check]
else:
raise NotImplementedError("need a case where input is invalid")

While loop to identify integer in Python

I'm trying to write a program to calculate densities, and I have tried to create a while loop the prevents the user from entering nothing or a non-number for the volume.
But when I run the program the it just loops "You have to type a value" forever. I've tried the same code in a for loop and it does work after inputing 2 numbers.
def GetVolume():
print("How many cublic cm of water does the item displace")
Volume = input()
while Volume == ("") or type(Volume) != int:
print("You have to type a value")
Volume = input()
return float(Volume)
This solution is written assuming you are using Python 3. The problem in your code is that you are assuming that if you type in a number, the input method will return a type int. This is incorrect. You will always get a string back from your input.
Furthermore, if you try to cast int around your input to force an int, your code will raise if you enter a string, with:
ValueError: invalid literal for int() with base 10:
So, what I suggest you do to make your implementation easier is to make use of try/exceptinstead to attempt to convert your value to a float. If it does not work, you prompt the user to keep entering a value until they do. Then you simply break your loop and return your number, type casted to a float.
Furthermore, because of the use of the try/except, you no longer need to put in a conditional check in your while loop. You simply can set your loop to while True and then break once you have satisfied your condition in your code.
Observe the code below re-written with what I mentioned above:
def GetVolume():
print("How many cublic cm of water does the item displace")
Volume = input()
while True:
try:
Volume = float(Volume)
break
except:
print("You have to type a value")
Volume = input()
return Volume
def GetVolume():
Volume = input("How many cublic cm of water does the item displace")
while not Volume.replace('.', '').replace(',', '').isdigit():
Volume = input("You have to type a value")
return float(Volume)
x = GetVolume()
print(x)
You have to modify your while because is validating that is str or different than int. An input will always be an str by default unless you modified the type with int() or float() in your case.
You can use 'try' instead to check for this:
while True:
x = input("How many cubic cm of water does the item displace")
try:
x = float(x)
break
except ValueError:
pass
print('out of loop')

Try statement - multiple conditions - Python 2

I have little problem with try statement along with multiple conditions. When there is error at 2nd condition, it asks for 1st condition. What I want from it to do is to repeat the same condition, not the whole cycle. I hope you understand me, since my English isn't very good and also I'm newbie to Python so I also don't know how to describe it in my native language.
I hope the following example will help you to better understand my thought.
while True:
try:
zacatek = float(raw_input("Zacatek: "))
konec = float(raw_input("Konec: "))
except Exception:
pass
else:
break
it does following:
Zacatek: 1
Konec: a
Zacatek:
but I want it to do this:
Zacatek: 1
Konec: a
Konec:
Thanks in advance for any help.
Write a function to query for a single float, and call it twice:
def input_float(msg):
while True:
try:
return float(raw_input(msg))
except ValueError:
pass
zacatek = input_float("Zacatek: ")
konec = input_float("Konec: ")
What's happening is that your except clause is catching a ValueError exception on your answer to Konec and returning to the top of the loop.
Your float function is trying to cast a non-numeric response "a" to a float and it throwing the exception.
Alternatively, you could write a different loop for each input:
zacatek = None
while not zacatek:
try:
zacatek = float(raw_input("Zacatek: "))
except Exception:
continue
konec = None
while not konec:
try:
konec = float(raw_input("Konec: "))
except Exception:
continue

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