I have a program with some user inputs and I need to check if what the user entered was a string or an integer value between 1 and 10 million.
My code looks like this (simplified):
while True:
inp = raw_input("Enter a value between 1 and 10 million: ")
if inp < 1:
print "Must be higher than 1"
continue
elif inp > 10.000.000:
print "Must be less than 10.000.000"
continue
elif 'inp is a string': #here's my problem
print "Must be an integer value!"
continue
else:
'execute the rest of the code'
I don't know how to solve this. My program always terminates when I enter a string by mistake.
Thank you!
First, you're using Python 2, which will happily compare strings to integers. You don't want to do that. Secondly, raw_input() will always return a string. What you're hoping to do is check if that string could possibly represent a number. Third, 10.000.000 is improper syntax. Don't use separators. Fourth, you only need to continue if you want to go to the top of the loop early. If everything's in an if..elif..else block at the end of the loop, only one of those will be executed, so you don't need to put a continue at the end of each branch. You can use continue statements or restructure your branch. Finally, don't use in as a variable name, because that's a Python keyword.
while True:
inp = raw_input("Enter a value between 1 and 10 million: ")
if not inp.isdigit():
print "Must be an integer value!"
continue # each of these continue statements acts like a "failed, try again"
inp = int(inp)
if inp < 1:
print "Must be higher than 1"
continue # same for this one
if inp > 10000000:
print "Must be less than 10.000.000"
continue # and this one
# execute the rest of the code
You can use .isdigit() to check if string consists of numbers to make sure it can be convertible to integer:
while True:
in = raw_input("Enter a value between 1 and 10 million: ")
if in.isdigit():
number = int(in)
if number < 1:
print "Must be higher than 1"
continue
elif number > 10**6:
print "Must be less than 10.000.000"
continue
else:
'execute the rest of the code'
else:
print "Must be an integer value!"
continue
I have no idea how you came up with your code but here are different suggestions:
Don't use "in" as variable name because it's a python operator.
After the input you can check whether it is an int or a string.
Your program can look like this:
while True:
try:
input_int = int(raw_input("Enter a value between 1 and 10 million: "))
if input_int < 1 :
print "Must be higher than 1"
elif input_int > 10**7:
print "Must be less than 10.000.000"
except:
print "Must be an integer value!"
else: #You can use else with try/except block
#'execute the rest of the code'
Voilà
Related
I'm building my first calculator. Trying to implement While loop to get a number through user input. I want the While to break once user put a number.
num1 = raw_input("Add mumber one: " )
try:
input = int(num1)
except ValueError:
print "This is not a number"
attempt = 0
while type(num1) != int and attempt < 5:
num1 = raw_input("Add Number one again: " )
attempt += 1
break
print "You are not putting number. So, goodbuy"
operation = raw_input("Add Operator: ")
Try this:
for _ in range(5):
num1 = unicode(raw_input("Add number one: "))
if num1.isnumeric():
break
Another method you can try is to have num1 and do the following in the while loop:
if type(num1) is int:
break
What this does is check the type of the variable num1. If it is an integer, int, then it breaks out of the while loop.
I want the While to break once user put a number.
You already are doing that in the condition for your while loop by having type(num) != int so your while loop should stop after the user enters an integer.
You have several errors here:
while type(num1) != int and attempt < 5:
num1 = raw_input("Add Number one again: " )
attempt += 1
break
The break statement shoud be inside the loop, but the indentation is wrong. As you write, it is after the loop (hence useless). Also, when you read from standard input, you always get a string even if it is a number. So when you check type(num1) != int this is always false. You must convert num1 with int() each time you read from standard input, not only the first time:
while True:
num1 = raw_input("Add Number one again: " )
try:
input = int(num1)
break
except ValueError:
print "This is not a number"
if attempt == 5:
break
attempt += 1
Here I try to convert to an integer the string read from stdin. If it works, I break the loop immediately. If it does not work (the except clause) I check how many attempts have been done, and break the loop if the attempts are equal to 5.
I am trying to validate user input to check that, when they enter their name, it is more than 2 characters and is alphabetic. I am attempting to do this using try/except as I have been told that it is the best loop for user validation. Unfortunately if the user enters characters that are not alphabetic, nothing happens, and the program proceeds like normal. I also do not know how to check if the input is longer than 2 characters in the try/except loop as it is very new to me. Any help is much appreciated.
list = []
def users_name():
while True:
try:
name = str(input("Please enter your first name: "))
list.append(name)
break
except TypeError:
print("Letters only please.")
continue
except EOFError:
print("Please input something....")
continue
users_name()
You program will continue to run because of the continue clause in the catch block.
You can also check to see if the length of what they entered is longer than two characters.
list = []
def users_name():
while True: # Never ending loop
try:
name = str(input("Please enter your first name: "))
if (len(name) > 2)
list.append(name)
break
except TypeError:
print("Letters only please.")
continue # This causes it to continue
except EOFError:
print("Please input something....")
continue # This causes it to continue
users_name()
Also, you might want to stop the loop somehow. Maybe put in a break clause when you insert into the array?
if(len(name) > 2)
list.append(name)
break
...
To check if the input is a digit or an alphabet character use isdigit() or isalpha()
if a.isalpha():
#do something
elif a.isdigit():
#do something
Your code will then look like this:
list = []
def users_name():
while True: # Never ending loop
try:
name = str(input("Please enter your first name: "))
if (len(name) > 2 && name.isalpha()):
list.append(name)
break
else:
raise TypeError
except TypeError:
print("Letters only please.")
continue # This causes it to continue
except EOFError:
print("Please input something....")
continue # This causes it to continue
users_name()
Also if you are using Python < 3 consider using raw_input(). input() will actually evaluate the input as Python code.
To check the length of your String you can use the method len() to obtain its character length. You can then select those that are greater than 2 or your desired length.
To check if you string has only alphabetic characters you can use the str.isalpha() method to check so.
Therefore, if you check len(name) > 2 and name.isalpha() will help you filter those string that do not match what you want:
>>> name = "MyName"
>>> len(name) > 2 and name.isalpha()
True
>>> wrong_name = "My42Name"
>>> len(wrong_name) > 2 and wrong_name.isalpha()
False
>>> short_name = "A"
>>> len(short_name) > 2 and short_name.isalpha()
False
I need to write a prog in Python that accomplishes the following:
Prompt for and accept the input of a number, either positive or negative.
Using a single alternative "decision" structure print a message only if the number is positive.
It's extremely simply, but I'm new to Python so I have trouble with even the most simple things. The program asks for a user to input a number. If the number is positive it will display a message. If the number is negative it will display nothing.
num = raw_input ("Please enter a number.")
if num >= 0 print "The number you entered is " + num
else:
return num
I'm using Wing IDE
I get the error "if num >= 0 print "The number you entered is " + num"
How do I return to start if the number entered is negative?
What am I doing wrong?
Try this:
def getNumFromUser():
num = input("Please enter a number: ")
if num >= 0:
print "The number you entered is " + str(num)
else:
getNumFromUser()
getNumFromUser()
The reason you received an error is because you omitted a colon after the condition of your if-statement. To be able to return to the start of the process if the number if negative, I put the code inside a function which calls itself if the if condition is not satisfied. You could also easily use a while loop.
while True:
num = input("Please enter a number: ")
if num >= 0:
print "The number you entered is " + str(num)
break
Try this:
inputnum = raw_input ("Please enter a number.")
num = int(inputnum)
if num >= 0:
print("The number you entered is " + str(num))
you don't need the else part just because the code is not inside a method/function.
I agree with the other comment - as a beginner you may want to change your IDE to one that will be of more help to you (especially with such easy to fix syntax related errors)
(I was pretty sure, that print should be on a new line and intended, but... I was wrong.)
I am working on small python payroll project where you enter employee name, wage, and hours worked. When I enter decimals for the wage input, I am getting "invalid entry" because of my exception handling. Why are decimals being returned as invalid? Also, how can I loop this program so that it keeps the same 3 questions until the user types "Done"?
Any help will be greatly appreciated!
Thanks!
import cPickle
def getName():
strName="dummy"
lstNames=[]
strName=raw_input("Enter employee's Name: ")
lstNames.append(strName.title() + " \n")
def getWage():
lstWage=[]
strNum="0"
blnDone=False
while blnDone==False: #loop to stay in program until valid data is entered
try:
intWage=int(raw_input("Enter employee's wage: "))
if intWage >= 6.0 and intWage <=20.0:
lstWage.append(float(strNum)) #convert to float
blnDone=True
else:
print "Wage must be between $6.00 and $20.00"
except(ValueError): #if you have Value Error exception. Explicit on error type
print "Invalid entry"
def getHours():
lstHours=[]
blnDone=False
while blnDone==False: #loop to stay in program until valid data is entered
try:
intHrs=int(raw_input("Enter number of hours worked: "))
if intHrs >= 1.0 and intHrs <=60.0:
blnDone=True
else:
print "Hours worked must be 1 through 60."
except(ValueError): #if you have Value Error exception. Explicit on error type
print "Invalid entry"
def getDone():
strDone=""
blnDone=False
while blnDone==False:
try:
srtDone=raw_input("Type \"DONE\" if you are finished entering names, otherwise press enter: ")
if strDone.lower()=="done":
blnDone=True
else:
print "Type another empolyee name"
except(ValueError): #if you have Value Error exception. Explicit on error type
print "Invalid entry"
##### Mainline ########
strUserName=getName()
strWage=getWage()
strHours=getHours()
srtDone1=getDone()
Here's the core of it:
>>> int("4.3")
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
ValueError: invalid literal for int() with base 10: '4.3'
You can't convert a string to an integer if it's not an integer. So when you do intWage=int(raw_input("Enter employee's wage: ")) it throws the ValueError. Perhaps you should convert it directly to float.
Because you're converting the input to int:
intWage=int(raw_input("Enter employee's wage: "))
You are assuming that the wage is an integer, which by definition does not have a decimal place. Try this:
intWage=float(raw_input("Enter employee's wage: "))
Try using
intWage=int(float(raw_input("Enter employee's wage: ")))
This will accept a decimal number as input.
Error w/ Floats
As others said before, you are assuming the input will be a float. Either use float() or eval()instead of int():
intWage = float(raw_input("Enter employee's wage: "))
Or use input() instead of int(raw_input()):
intWage = input("Enter employee's wage:")
Both will accomplish the same thing. Oh, and change intWage to floatWage atleast, or even better, don't use Hungarian Notation.
The Code
As for your code, I did a couple of things:
Used break and/or return to terminate loops instead of keeping track of booleans (that's the whole purpose of the break and continue statements)
Changed intWage to floatWage
Rewrote number comparisons in a more concise way (x <= y and x >= z can be written as z >= x >= y)
Added return statements. I don't get why you didn't put them yourself, unless you wanted to assign None to strUserName, strWage and strHours)
Added a loop as you requested when asking for an employee's details.
Modified getDone() to work w/ the loop.
import cPickle
def getName():
strName = "dummy"
lstNames = []
strName = raw_input("Enter employee's Name: ")
lstNames.append(strName.title() + " \n")
return strName
def getWage():
lstWage = []
strNum = "0"
while True: #Loop to stay in program until valid data is entered
try:
floatWage = float(raw_input("Enter employee's wage: "))
if 6.0 <= floatWage <= 20.0:
lstWage.append(floatWage)
return floatWage
else:
print "Wage must be between $6.00 and $20.00"
except ValueError: #Catches ValueErrors from conversion to float
print "Invalid entry"
def getHours():
lstHours = []
while True: #loop to stay in program until valid data is entered
try:
intHrs=int(raw_input("Enter number of hours worked: "))
if 1.0 <= intHrs <= 60.0:
return intHrs
else:
print "Hours worked must be 1 through 60."
except ValueError: #Catches ValueErrors from conversion to int
print "Invalid entry"
def getDone():
strDone = ""
while True:
srtDone = raw_input('Type "DONE" if you are finished entering names, otherwise press enter: ')
if strDone.strip().lower() == "done":
return True
else:
print "Type another empolyee name"
while not getDone():
strUserName = getName()
strWage = getWage()
strHours = getHours()
An Example of break and continue
The break statements inside a loop (for and while) terminate the loop and skip all 'else' clauses (if there are any).
Thecontinue statements skips the rest of the code in the loop and the continues the loop as if nothing happened.
The else clause in a for...else construct, executes its code block when the loop exhausted all the items and exited normally, i.e., when it's not terminated by break or something.
for no in range(2, 10):
for factor in range(2, no):
if no % factor == 0:
if factor == 2:
print "%d is even" % no
continue
# we want to skip the rest of the code in this for loop for now
# as we've already done the printing
print "%d = %d * %d" % (no, factor, n/x)
break
# We've asserted that the no. isn't prime,
# we don't need to test the other factors
else:
# if 'break' wasn't called
# i.e., if the loop fell through w/o finding any factor
print no, 'is a prime number'
I'm trying to take a raw input and detect whether it is in a range.
Here's my code.
def gold_room():
print "This room is full of gold. How much do you take?"
next = raw_input("> ")
if next == int in range(50):
how_much = int(next)
else:
dead("Man, learn how to type a number.")
if how_much < 50:
print "Nice, you're not greedy, you win!"
exit(0)
else:
dead("You greedy bastard!")
When I enter a number it gives me the else: "Man, learn how to type a number."
I guess the line that isn't working is "if next == int in range(50):
Could anyone help me out?
Thanks in advance!
Edit:
I'm a noob so that line was just me ballparking.
I thought it would check next to see if it was an integer in the range of numbers 0-50.
If you want to "get an integer input in a range", you'll need two things:
Check if the input is an int
Check if it's in your range
Check if the input is an int
try/except will be perfect here:
n = raw_input('> ')
try:
n = int(n)
except ValueError:
dead() # or what you want
Why this is good?
Because if n is an int you'll have it convert it to an integer and if it's not an exception get raised and you can call your dead() funcion.
Check if it's in your range
If you get to this point it means that the exception before it was not raised and n was converted to an integer.
So you just need to do:
if 0 <= n <= 50:
print 'You win'
else:
print 'You lose'
Don't do:
if n in range(50):
# ...
Beacuse it will build a list of 50 numbers for nothing.
Note: don't use next as a variable, beacuse it'll shadow the built-in next()
Since raw_input returns a string, you need to convert to an int first. Try replacing the line with this:
if int(next) in range(50):
The result of raw_input() will be a string, so first you need to check to see if next is all digits. There are a few ways to do this, the easiest is to use the str.isdigit() function:
next = raw_input("> ")
if next.isdigit():
how_much = int(next)
else:
dead("Man, learn how to type a number.")
Note that you do not need to check to see if the value for next is in the range from 0 to 50, since your next if statement already checks to see if the value is less than 50 and negative numbers will be excluded by next.isdigit().
The test "next == int in range(50)" evaluates to "(next == int) and (int in range(50))" which is fairly meaningless and always equates to false.
Instead you could try
try:
how_much = int(next)
if not (0<=how_much<50):
print 'Too greedy'
except ValueError:
dead("Man, learn how to type a number.")
Using try .. except will allow you to make sure entered value is an int. # raise is a place holder for your handling a non-int contition:
try:
next = int(raw_input("> "))
except ValueError:
# raise
if not 0 <= next <= 50:
print 'Too greedy'