So I have a finished program that accepts an input file with bank account information and parses it up and allows for a few different utilities.
One such utility is adding a transaction to the "database" (just a log file).
The program prompts the user to enter 'w' or 'd' and then an amount (float). This represents a deposit or withdrawal of X amount of money.
I was wondering how to go about making sure that the user entered either 'w' or 'd' AND a correct amount (number).
So, I decided that a while loop with the above condition would work, however I am having trouble getting it work 100%
I initially had:
while input1 is not ("w" or "d")
where input1 would be the first input (w or d) the user enters
However, I also want to check that a number exists.
I had the idea of casting the string input to a float, then checking that but I wouldn't know how to checking if that is right since casting and checking the type wouldn't tell me much.
How would I also check that the user entered in some sort of number.
So to reiterate, I would like the program to re-prompt for input if the user did not enter either:
A) A w or d
B) A number (int/float)
Thanks
the expression ("w" or "d") will always evaluate to "w". Generally, here you want an in:
while input1 not in ("w", "d"):
...
As far as handling the case where the input is a number, this is a job for a helper function:
def string_is_number(s):
try:
float(s)
return True
except ValueError:
return False
Now you can use that with the above to put the whole thing together:
while (not string_is_number(input1)) and (input1 not in ("w", "d")): ...
Which can actually be simplified a little more:
while not (string_is_number(input1) or (input1 in ("w", "d"))): ...
And now a completely different approach, You can actually use a recursive function for this sort of thing. Combine that with python's exception handling and we could probably put together a pretty elegant solution in just a few lines:
def prog_loop():
# prompt for user's input here:
input1 = raw_input("Enter a number, or 'w' or 'd':")
# See if we got a number
try:
number = float(input1)
except ValueError:
# Nope, wasn't a number. Check to see if it was in our
# whitelisted strings. If so, break early.
if input1 in ('w', 'd'):
return function_handle_w_d(input1)
else:
# Yes, we got a number. Use the number and exit early
return function_handle_number(number)
# haven't exited yet, so we didn't get a whitelisted string or a number
# I guess we need to try again...
return prog_loop()
This will work as long as your user doesn't enter bad input 1000 times.
Try this:
while True:
if input1 == 'w':
withdraw()
elif input1 == 'd':
deposite()
else:
continue()
Related
Here is my code:
from random import randint
doorNum = randint(1, 3)
doorInp = input("Please Enter A Door Number Between 1 and 3: ")
x = 1
while (x == 1) :
if(doorNum == doorInp) :
print("You opened the wrong door and died.")
exit()
now, that works fine, if I happen to get the unlucky number.
else :
print("You entered a room.")
doorNum = randint(1, 3)
This is the part where it stops responding entirely. I am running it in a bash interactive shell (Terminal, on osx). It just ends up blank.
I am new to programming in Python, I spent most of my time as a web developer.
UPDATE:
Thanks #rawing, I can not yet upvote (newbie), so will put it here.
If you are using python3, then input returns a string and comparing a string to an int is always false, thus your exit() function can never run.
Your doorInp variable is a string type, which is causing the issue because you are comparing it to an integer in the if statement. You can easily check by adding something like print(type(doorInp)) after your input line.
To fix it, just enclose the input statement inside int() like:doorInp = int(input("...."))
In python3, the input function returns a string. You're comparing this string value to a random int value. This will always evaluate to False. Since you only ask for user input once, before the loop, the user never gets a chance to choose a new number and the loop keeps comparing a random number to a string forever.
I'm not sure what exactly your code is supposed to do, but you probably wanted to do something like this:
from random import randint
while True:
doorNum = randint(1, 3)
doorInp = int(input("Please Enter A Door Number Between 1 and 3: "))
if(doorNum == doorInp) :
print("You opened the wrong door and died.")
break
print("You entered a room.")
See also: Asking the user for input until they give a valid response
I am very new to Python (started 2 days ago). I was trying to validate positive integers. The code does validate the numbers but it asks twice after a wrong input is entered. For example if I enter the word Python, it says: This is not an integer! like is supposed to but if I enter 20 afterwards, it also says it is not an integer and if I enter 20 again it reads it.
def is_positive_integer(input):
#error: when a non-integer is input and then an integer is input it takes two tries to read the integer
flag = 0
while flag != 1:
try:
input = int(input)
if input <= 0:
print "This is not a positive integer!"
input = raw_input("Enter the number again:")
except ValueError:
print "This is not an integer!"
input = raw_input("Enter the number again: ")
if isinstance(input, int):
flag = 1
return input
number = raw_input("Enter the number to be expanded: ")
is_positive_integer(number)
number = int(is_positive_integer(number))
Any help is appreciated.
The main bug is that you call is_positive_integer(number) twice with the same input (the first thing you enter).
The first time you call is_positive_integer(number), you throw away the return value. Only the second time do you assign the result to number.
You can "fix" your program by removing the line with just is_positive_integer(number) on its own.
However, your code is a little messy, and the name is_positive_integer does not describe what the function actually does.
I would refactor a little like this:
def input_positive_integer(prompt):
input = raw_input(prompt)
while True:
try:
input = int(input)
if input <= 0:
print "This is not a positive integer!"
else:
return input
except ValueError:
print "This is not an integer!"
input = raw_input("Enter the number again: ")
number = input_positive_integer("Enter the number to be expanded: ")
The problem stems from the fact that you're calling is_positive_integer twice. So, the first time it's called, you send it a string like 'hello', then it says it's not an integer and tells you to try again. Then you enter '20', which parses fine, and it's returned.
But then you don't save a reference to that, so it goes nowhere.
Then you call the function again, this time saving a reference to it, and it first tries the original bad string, which was still there in number. Then it complains that it's a bad input, asks you for a new one, and you provide it, terminating the program.
I spent a good hour or more looking for the answer on here. I have found a few things that help, but do not answer my question specifically. I am using Python 3.3.3. I am a novice so please be gentle.
I am trying to create a program that takes a user input, but then I need to do a check to see what datatype that input is, and then based on that datatype take a certain course of action.
Any string besides those found in this list:
valid_help_string_list = ['\'help\'', '\'HELP\'', 'help', 'HELP']
should result in the printing of:
'please enter a valid entry' or something to that effect.
Any integer (over 0 but under 500) should have float() used on it to make the rows line up.
Any float (over 0.0 but under 500.0) is valid.
For the sake of this project I am assuming nobody using this will weigh under 100 lbs or over 500.
Anything not falling within those categories should also yield the same "please enter a valid response" error message to the user.
I think it's simple enough of a project to take on for a novice. The program is meant to allow you to input your weight and then creates a pictogram based on that weight and saves it all on the next open line of the .txt file I have set up for it. Or if you want to see the legend for the pictogram, you should be able to type help in any variation found in that list.
Any help would be much appreciated.
The user input will be a string by default, so we need to check whether it could become an integer or float. As you want to turn the integers in floats anyway, there's no need to do anything complex:
def validate_input(val, min_v=100, max_v=500):
try:
val = float(val)
except ValueError:
print("Not a valid entry")
else:
if not min_v < val <= max_v:
print("Value should be between {} and {}".format(min_v, max_v))
else:
return val
return False
Now your calling loop can read:
while True:
val = input("...")
if val in valid_help_string_list:
# print help
else:
val = validate_input(val)
if val:
break
# use val
Note that this relies on the return from validate_input being either False or a number larger than 0; Python will interpret a zero return as False and not reach the break, so I recommend keeping min_v >= 0.
i have a programm that generate the list and then i ask them to tell me what they want to do from the menu and this is where my problem start i was able to get the input form the user to different function but when i try to use the if else condition it doesn't check, below are my code
def menu(x,l):
print (x)
if x == 1:
return make_table(l)
if x == 2:
y= input("enter a row (as a number) or a column (as an uppercase letter")
if y in [ "1",'2','3']:
print("Minmum is:",minimum(y,l))
if x== 3:
print ('bye')
def main():
bad_filename = True
l =[]
while bad_filename == True:
try:
filename = input("Enter the filename: ")
fp = open(filename, "r")
for f_line in fp:
f_str=f_line.strip()
f_str=f_str.split(',')
for unit_str in f_str:
unit=float(unit_str)
l.append(unit)
bad_filename = False
except IOError:
print("Error: The file was not found: ", filename)
#print(l)
condition=True
while condition==True:
print('1- open\n','2- maximum')
x=input("Enter the choice")
menu(x,l)
main()
from the bottom function i can get list and i can get the user input and i can get the data and move it in second function but it wont work after that.thank you
I think your problem is simple, and has nothing to do with how you're passing values between functions.
In main, you're reading a value from the user like this:
x=input("Enter the choice")
The input function:
… reads a line from input, converts it to a string (stripping a trailing newline), and returns that.
So, if the user types 1 at the prompt, you get back the string "1".
Now, you pass that value—perfectly correctly—to menu.
In menu, you then try to compare it to various numbers, like this:
if x == 1:
But this will never be true. A string, like "1", is never equal to a number, like 1. They're not even the same kind of value, much less the same value.
So, you need to do one of two things:
Convert the input to an number. For example, change menu(x,l) to menu(int(x), l). OR…
Write menu to expect strings. For example, change if x == 1: to if x == "1":.
You may be wondering why that print (x) didn't help you debug the problem.
print(x) prints out the end-user-friendly string representation of whatever you give it. That is, it automatically calls the str function for you. For debugging purposes, you often want to use repr instead of str, to get the programmer-friendly string representation instead of the end-user-friendly string representation.
For example, print(str("10")) will print out 10—just like print(str(10)), so you can't tell them apart. But print(repr("10")) will print out '10', unlike print(repr(10)), while prints 10, so you can tell them apart. repr can also help you spot things like strings with invisible characters in them, having special "node" objects from a parser instead of just strings, etc.
I am trying to analyse user input and only allow integers to be inputted. I have succesfully managed to only allow denominations of 100s between a certain range, but I cannot work out how to prompt the user to re-enter data if they enter a random string.
Trying to use the "try" command just results in the program getting stuck:
while True:
try:
bet=int(raw_input('Place your bets please:'))
except ValueError:
print 'l2type'
#The following receives a betting amount from the user, and then assesses whether it is a legal bet or not. If not, the user is prompted to enter a legal bet.
while True:
if bet%100==0 and 100<=bet<=20000:
print bet,"Your bet has been accepted, can you make a million?"
break
else:
print bet,"Please enter a legal bet. The table minimum is 100, with a maximum of 20000 in increments of 100."
bet = input('Place your bets please:')
You have the right approach for rejecting non-integer input, but you need to break out of the loop if your user enters valid input. Use the break statement:
while True:
try:
bet=int(raw_input('Place your bets please:'))
break # we only get here if the input was parsed successfully
except ValueError:
print 'l2type'
You will probably also want to move the range checks within the loop. If input that's out of range doesn't naturally lead to an exception, use if statements to make sure "break" is only executed if the input is completely valid.
I rather prefer my recursive version:
def ask_bet(prompt):
bet = raw_input(prompt)
# Validation. If the input is invalid, call itself recursively.
try:
bet = int(bet)
except ValueError:
return ask_bet('Huh? ')
if bet % 100 != 0:
return ask_bet('Huh? ')
if not 100 <= bet <= 20000:
return ask_bet('Huh? ')
# The input is fine so return it.
return bet
ask_bet("Place your bets please: ")
In my opinion, that is much cleaner and easier to read than while loops. You don't have to bother what are attribute values after the first iteration? Adding new validation rules is also really simple.
Generally, I try to avoid while loop in favour of recursive version. Of course, not all the time, since it's slower and the stack is not infinite.