I want to write a python program, first it asks you to enter two numbers, and then output all daffodil numbers between the two numbers, and it will continue run, until I enter a "q". I write a program, but it is wrong:
#coding=utf-8
while 1:
try:
x1=int(raw_input("please enter a number x1="))
x2=int(raw_input("please enter a number x2="))
except:
print("please enter only numbers")
continue
if x1>x2:
x1,x2=x2,x1
pass
for n in xrange(x1,x2):
i=n/100
j=n/10%10
k=n%10
if i*100+j*10+k==i+j**2+k**3:
print ("%-5d")%n
pass
Can somebody help? I think it should be simple, but I am not able to write it correctly.
I believe you've misunderstood the problem statement. Try this instead:
if i*100+j*10+k==i**3+j**3+k**3:
ref: http://en.wikipedia.org/wiki/Narcissistic_number
for n in xrange(x1,x2):
digits = map(int,str(n))
num_digits = len(digits)
if sum(map(lambda x:x**num_digits,digits)) == n:
print "%d is a magic number"%n
you will still have the issue of not being able to enter "q" since you force the input to be integers
I would like to address the quit event.
while True:
x1 = raw_input("please enter a number x1=")
x2 = raw_input("please enter a number x2=")
quit = ('q','Q')
if x1 in quit or x2 in quit:
break
else:
try:
x1, x2 = int(x1), int(x2)
except:
print("please enter only numbers")
continue
# The mathematical part... (for completeness) (not my code)
if x1>x2:
x1,x2=x2,x1
for n in xrange(x1,x2):
i=n/100
j=n/10%10
k=n%10
if i*100+j*10+k==i+j**2+k**3:
print "%-5d"%n
The pass statement is used only when you don't have anything to be executed in certain block of code. It does nothing more, so don't use it if not needed. It is there for the sake of the code looking clean & with correct indentation.
if some_thing: # don't do anything
else:
some_thing = some_thing_else
Note how the above if statement is syntactically incorrect. This is where pass comes handy. Say, you decide to write the if part later, till then you must provide pass.
if some_thing: # don't do anything
pass
else:
some_thing = some_thing_else
It's a bit tricky to know what's going on without more hints, but some issues I see right off:
You need to be consistent with indentation in Python. Your last if statement is less indented than statements above it (like the previous if and the for loop). This will cause an error. You're also using different amounts of indentation in other places, but since it's not inconsistent that's allowed (if a bad idea). Its usually best to pick one indentation standard (like four spaces) and stick with it. Often you can set your text editor to help you with this (turn on "Expand tabs to spaces" or something in the settings).
You've got two pass statements where they're unneeded or harmful. The first, after the line that has if x1>x2: x1,x2=x2,x1 is going to cause an error. You can't have an indented "suite" of code if you've put a series of simple statements on the end of your compound statement like an if. Either put the assignment on its own line, indented, or get rid of the pass. The last pass at the end of the code is not an error, just unnecessary.
You're missing a colon at the end of your try statement. Every statement in Python that introduces an indented suite ends with a colon, so it should be easy to learn where they're needed.
while True:
x1 = raw_input("please enter a number x1=")
x2 = raw_input("please enter a number x2=")
quit = ('q','Q')
if x1 in quit or x2 in quit:
break
else:
try:
x1, x2 = int(x1), int(x2)
except:
print("please enter only numbers")
continue
if x1>x2:
x1,x2=x2,x1
pass
for n in xrange(x1,x2):
i=n/100
j=n/10%10
k=n%10
if i*100+j*10+k==i+j**2+k**3:
print ("%-5d")%n
pass
i have it! thx to Ashish! it is exactly what i want! and i will quit wenn i enter q! thx a lot!
Related
I'm "newish" to python programming. I'm trying my best to make my code look nice and function well. I'm using Pycharm as my IDE. I'm doing something for myself. I play tabletop RPG's and I'm attempting to create a character creator for a game I play. I have everything working well, but Pycharm is telling me that "Expression can be simplified" and "PEP 8: E712 comparison to True should be 'if cond is not True:' or 'if not cond:'"
Here is the code in question:
fname = False
while fname != True:
new_character.firstName = input('What would you like your first name to be?\n').capitalize()
if 1 >= len(new_character.firstName) or len(new_character.firstName) > 20:
print('Name does not meet length requirements. Please try again.')
if new_character.firstName.isalpha() != True:
print('Please do not use numbers or special characters in your name. Please try again.')
if (1 < len(new_character.firstName) < 20) and (new_character.firstName.isalpha() == True):
fname = True
Pycharm is telling me that my "while fname != True:" is the part that can be simplified as well as the "if new_character.firstName.isalpha() != True:".
I've tried googling a solution for what I'm doing, but most of them are for something kinda like what I'm asking, but never with the != True portion. I've even reached out to one of my friends that's a python programmer, but I haven't heard back yet.
Again, I want to state that as it is now, the code works correctly the way it is written, I'm just wanting to understand if there is a way to make the code look cleaner/neater or do the same function and be simplified somehow.
Any pointers on how to potentially simplify those lines of code and maintain the functionality would be greatly appreciated.
Here's one way you could rewrite this code to make it easier to read, and more efficient:
# Loop until the user provides a good input
while True:
# Set a temp variable, don't constantly reassign to the new_character.firstName attribute
name = input('What would you like your first name to be?\n').capitalize()
# If the name isn't between 2 and 20 characters, start the loop over at the beginning
if not (1 < len(name) <= 20):
print('Name does not meet length requirements. Please try again.')
continue
# If the name contains anything other than letters, start the loop over at the beginning
if not name.isalpha():
print('Please do not use numbers or special characters in your name. Please try again.')
continue
# You can only reach this break if the name "passed" the two checks above
break
# Finally, assign the character name
new_character.firstName = name
One thing you could do to simplify further is to check both conditions at the same time, and print a more helpful error message that re-states the requirements explicitly:
NAME_ERROR_MESSAGE = """
Invalid name '{name}'. Your character's name
must be between 2 and 20 characters long, and
contain only letters. Please try again.
"""
while True:
name = input('What would you like your first name to be?\n').capitalize()
if (1 < len(name) <= 20) and name.isalpha():
new_character.firstName = name
break
print(NAME_ERROR_MESSAGE.format(name=name)
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 looked on stackoverflow to solve my problem but from all explanations about loops and checks I don't understand why my code isn't working.
So I want to build a dictionary (totally new to Python btw) and I read that I can also check if the input is in the dicitonary module but that is actually not what I want to do here. I just want to see if the raw_input contains at least one number in the string (not if the string only contains numbers) and if the length of the input string is at least 2.
If the input passes those checks it should move on (the rest of this dictionary will come later. For now I only want to understand what I did wrong with my check)
Here's my code, help would be very much appreciated!
def check():
if any(char.isdigit() for char in original):
print ("Please avoid entering numbers. Try a word!")
enter_word()
elif len(original)<1:
print ("Oops, you didn't enter anything. Try again!")
enter_word()
else:
print ("Alright, trying to translate:")
print ("%s") %(original)
def enter_word():
original = raw_input("Enter a word:").lower()
check()
enter_word()
Edit: Works now perfectly with the following code:
def check(original):
if any(char.isdigit() for char in original):
print "Please avoid entering numbers. Try a word!"
enter_word()
elif len(original) < 1:
print "Oops, you didn't enter anything. Try again!"
enter_word()
else:
print "Alright, trying to translate:"
print "{}".format(original)
def enter_word():
original = raw_input("Enter a word:").lower()
check(original)
enter_word()
You need to pass the input original to your check() function:
def check(original):
if any(char.isdigit() for char in original):
print("Please avoid entering numbers. Try a word!")
enter_word()
elif len(original) < 1:
print("Oops, you didn't enter anything. Try again!")
enter_word()
else:
print("Alright, trying to translate:")
print("{}".format(original))
def enter_word():
original = input("Enter a word:").lower()
check(original)
enter_word()
In addition to this, you had some syntax errors in your code. Since you used print() instead of print I assume you are using Python3. However, to read user input you used raw_input() which was the way to do in Python2 and became input() in Python3. I fixed this. Another thing I fixed was the string formatting of the print() statement in the else branch. You might take a look at the string format mini-language.
Question: What would be the issue if python continues to ask for the same input over and over again, and won't advance to the end of the program?
Where do you want to go? X
And how many days will you be staying in X? 1
And how many days will you be staying in X? 2
And how many days will you be staying in X? 164
And how many days will you be staying in X? 59
...
Here's the relevant part of the code:
# Import modules
import destinations
import currency
save_itinerary = True
main_function = True
while (main_function):
# Determine length of stay
while True:
try:
length_of_stay = int(input("And how many days will you be staying in " + destinations.destination[0] + "? "))
# Check for non-positive input
if (length_of_stay <= 0):
print("Please enter a positive number of days.")
continue
except ValueError:
print("The value you entered is invalid. Only numerical values are valid.")
break
else:
break
The reason your code is looping forever is that you have two nested while loops, and you never break out of the outer one. You do use break statements to exit the inner loop, but the condition for the outer loop is never changed, and you never execute a break at the right level to exit it.
Here's what I think a better version of your code would be:
# get rid of the outer while loop, which was never ending
while True:
try:
length_of_stay = int(input("And how many days will you be staying in " + destinations.destination[0] + "? "))
if (length_of_stay <= 0):
print("Please enter a positive number of days.")
continue
except ValueError:
print("The value you entered is invalid. Only numerical values are valid.")
# don't break here, you want to stay in the loop!
else:
break
I've used comments to indicate my changes.
You could also move the else: break block up and indent it so that it is attached to the if statement, rather than the try/except statements (and then get rid of the unnecessary continue statement). That's makes the flow a bit more obvious, though there's not really anything wrong with how it is now.
I start my python script asking the user what they want to do?
def askUser():
choice = input("Do you want to: \n(1) Go to stack overflow \n(2) Import from phone \n(3) Import from camcorder \n(4) Import from camcorder?");
print ("You entered: %s " % choice);
I would then like to:
Confirm the user has entered something valid - single digit from 1 - 4.
Jump to corresponding function based on import. Something like a switch case statement.
Any tips on how to do this in a pythonic way?
Firstly, semi-colons are not needed in python :) (yay).
Use a dictionary. Also, to get an input that will almost certainly be between 1-4, use a while loop to keep on asking for input until 1-4 is given:
def askUser():
while True:
try:
choice = int(input("Do you want to: \n(1) Go to stack overflow \n(2) Import from phone \n(3) Import from camcorder \n(4) Import from camcorder?"))
except ValueError:
print("Please input a number")
continue
if 0 < choice < 5:
break
else:
print("That is not between 1 and 4! Try again:")
print ("You entered: {} ".format(choice)) # Good to use format instead of string formatting with %
mydict = {1:go_to_stackoverflow, 2:import_from_phone, 3:import_from_camcorder, 4:import_from_camcorder}
mydict[choice]()
We use the try/except statements here to show if the input was not a number. If it wasn't, we use continue to start the while-loop from the beginning.
.get() gets the value from mydict with the input you give. As it returns a function, we put () afterwards to call the function.