I'm making a basic text-based hangman game, and I'm stumped as to why inpLetter is returning None from letterVerify()? It's expected to return a string response so I'm unsure why it isn't.
def letterVerify(prompt):
try:
inp = input(prompt)
verify = str.count(inp)
if (verify > 1) or (verify < 1):
print("Please enter one character.")
if inp not in alphabet or alphabetCaps:
print("Please enter a letter from the English alphabet.")
else:
return inp
except:
print("Please enter a letter from the English alphabet.")
inpLetter = letterVerify("Enter a letter you think is in this word. ")
The problem is in line 4, verify = str.count(inp). count() function returns the number of times a particular character has appeared in a string. It is used as
string = 'abcabcabcabcabc'
string.count('a')
This returns 5. Your code could not find this error because of the try-except block. Whenever this error was raised, it was caught by the exception block, and hence made it difficult to debug. You can rather try using the len() function. This will do the same work you want. It will check if the length of input is 1 or not. Modifications to your code are as
def letterVerify(prompt):
inp = input(prompt)
verify = len(inp)
if verify > 1 or verify < 1:
print('Please enter 1 character')
elif not inp.isalpha():
print('Please enter a letter from English alphabet')
else:
return inp
inpLetter = letterVerify('Enter a letter you think is in this word')
Also, you did not specify what alphabet and alphabetCaps are. I have used isalpha() method, that checks only for alphabetical characters (what you needed). Hope this resolves your issue.
I made your function a little terse for better readability.
First, you should avoid naming your function in camelCase, this violates PEP8. Rather you should use snake_case while naming functions and variables.
Also, simple length checking will do the verify trick. str.count returns the number of times a letter appears in a string, you don't want that here.
You don't need the exception handler here. Python will internally make your inputs str type.
You can lower your string and make comparison with the built in string.ascii_lowercase property. This will eliminate redundant control flow logic.
Here is the function, refactored:
def letter_verify(prompt):
inp = input(prompt)
# remove extra whitespace
inp = inp.strip()
# simple length checking with len function will do the trick
if len(inp) != 1:
print("Please enter one character.")
# just make your input to lowercase and then compare
# this will help you avoid comparison with two cases
elif inp.lower() not in string.ascii_lowercase:
print("Please enter a letter from the English alphabet.")
else:
return inp
Related
I'm very new to programming, and more new to Python, and I'm having a hard time figuring this out. I'm trying to make a simple text hangman game, and I'm trying to validate player1's input. Here's what I have so far.
a = 0
int = ["1","2","3","4","5","6","7","8","9","0"]
while a == 0:
print("Please choose any word by typing it here:")
d = input()
d = list(d)
print(d)
if any(int) == any(d):
print("Invalid input. Please choose a non-number with no spaces or special characters.")
I can't figure why, whether I include a number in my response or not, it always meets my any(int) == any(d) condition. I know I must just be misunderstanding how any() works.
You are right: the issue is with how you use the any() function. any() returns either True or False depending on whether or not any of the elements in its iterable argument are True, in other words if they exist and are not False or an empty list or empty string or 0. This will always evaluate as True for int because int does contain an element that is True and the same will happen for d unless you do not give any input when the input() function prompts you (you hit return without typing anything). What your conditional is basically asking is the following:
if True==True
To fix this, just change your code to the following:
a = 0
int = ["1","2","3","4","5","6","7","8","9","0"]
while a == 0:
print("Please choose any word by typing it here:")
d = input()
print(d)
for letter in d:
if letter in int:
print("Invalid input. Please choose a non-number with no spaces or special characters.")
break
The easiest solution, however, does not involve the int list at all. There is actually a special method in python that can gauge whether or not a string has a number (or special character) in it, and this method, .isalpha() can be used to streamline this process even further. Here's how to implement this solution:
while True:
d = input("Please choose any word by typing it here: \n")
print(d)
if not d.isalpha():
print("Invalid input. Please choose a non-number with no spaces or special characters.")
Try this:
a = 0
ints = ["1","2","3","4","5","6","7","8","9","0"]
while a == 0:
print("Please choose any word by typing it here:")
word = input()
word_list = list(word)
if any(letter in ints for letter in word_list):
print("Invalid input. Please choose a non-number with no spaces or special characters.")
I'm using python3 on mac.
I'm currently doing a project. However, I was trying to use "while = True" to continuously use the program until a condition is met. Please, tell me what am I missing in my code. Thanks!
import json
import difflib
from difflib import get_close_matches
data = json.load(open("project1/data.json"))
word = input("Enter a word or enter 'END' to quit: ")
def keyword(word):
word = word.lower()
while type(word) == str:
if word in data:
return data[word]
elif word == 'END'.lower():
break
elif len(get_close_matches(word, data.keys())) > 0:
correction = input("Did you mean %s insted? Enter Yes of No: " % get_close_matches(word, data.keys())[0])
if correction == "Yes".lower():
return data[get_close_matches(word, data.keys())[0]]
elif correction == "No".lower():
return "This word doesn't exist. Plese enter again. "
else:
return "Please enter 'Yes' or 'No: "
else:
return "This word doesn't exist. Please enter again."
print("Thanks!")
output = (keyword(word))
if type(output) == list:
for item in output:
print(item)
else:
print(output)
I think this might be the setup you are looking for.
def keyword(word):
if word in data:
return data[word]
elif len(get_close_matches(word, data.keys())):
correction = input(f"Did you mean {get_close_matches(word, data.keys())[0]} instead? y/n: ")
if correction == 'y':
return data[get_close_matches(word, data.keys())[0]]
elif correction == 'n':
return "This word doesn't exist. Please enter again."
else:
return "Please type 'y' or 'n': "
else:
return "This word doesn't exist. Please enter again."
while True:
word = input("Enter a word: ").lower()
if word == 'end':
print("Thanks!")
break
else:
print(keyword(word))
Looking at the source code and your question, it seems like what you want to achieve is basically to continuously accept input from the user until the user enters something like 'end'. One way to go about this is to separate out the while-loop logic from the function. The overarching while-loop logic is at the bottom half of the code, where we continuously accept input from the user until the user inputs some lower or upper case variant of 'end'. If this condition is not met, we proceed to printing out the result of the function call keyword(word).
Minimal modifications were made to the original keyword() function, but here are a few changes worthy of note:
The while type(word) == str is unnecessary, since the result stored from the input() function will always be a string. In other words, the condition will always return True.
Having return statements within a while loop defeats the purpose of a loop, since the loop will only be executed once. After returning the specified value, the function will exit out of the loop. This is why we need to separate out the loop logic from the function.
Although %s works, it's a relic of C. This might be a matter of personal choice, but I find f-strings to be much more pythonic.
You are using the worng condition.
type((3,4))== list
is False. You must use
type((3,4)) == tuple
I want to take user input and compare it to a list of key words through a function, if any of the words input by the user match a key word, the condition is met and breaks the loop. If none of the words match a key word, then the console asks for input again. I have been manipulating this loop and have either gotten it to continuously ask for input no matter if a key word is met or validate every word input. Any advice on how to correct it would be great appreciated.
def validated_response(user_complaint):
valid_list = user_complaint.split()
while True:
if user_complaint == "stop":
break
for valid in valid_list:
if valid.lower() not in user_complaint.lower():
print("Response not recognized, please try again")
input("Enter response: ")
continue
else:
print("response validated: ")
break
return True
This function will continue getting user input until the input matches "kwrd1", "kwrd2", or "kwrd3":
def get_input():
keywords = ['kwrd1', 'kwrd2', 'kwrd3']
user_input = None
while True:
user_input = input()
if user_input in keywords:
break
return user_input
If you are matching it against a python keyword, there is a builtin keyword module:
import keyword
def get_input():
user_input = None
while True:
user_input = input()
if keyword.iskeyword(user_input):
break
return user_input
You are always reaching the else statement if the first element in valid_list is not a substring of the user_complaint string. That means you're always breaking out of the for loop and reentering the infinite while loop. Try this instead:
def validated_response(user_complaint):
valid_list = user_complaint.split()
if user_complaint == "stop":
return
inp = input("Enter response: ")
while inp.lower() not in valid_list:
inp = input("Enter response: ")
The provided code has a number of problems. The example does also not show how the function would be called, but I assumed you'd want to call it with a text that has all the keywords you're looking for.
The first problem is you are calling input, but not storing its return value, so you're not actually collecting the user input.
Secondly, you're comparing the various parts of valid_list to the contents of user_complaint.lower(), but that means you're compare a string to the characters in another string - not what you want.
Thirdly, you're asking for new input in a single clause of a condition, inside your loop, so this will lead to messages printing repeatedly and the user having to enter new text before all the comparisons are done.
Finally, you're mixing continue, break and return in a way that doesn't work. continue tells Python to move on to the next cycle of a loop, skipping any remaining code in the current cycle. break tells Python to get out of the current block (in this case the loop). return tells Python to exit the function altogether and return a provided value (or None if none is provided).
Here is an example that more or less follows the structure you set up, but with all of the problems remedied:
def validated_response(keywords):
valid_list = keywords.split()
while True:
user_input = input('Enter response: ').lower().split()
if user_input == ['stop']:
return False
for valid in valid_list:
if valid.lower() in user_input:
print('response validated: ')
return True
print('Response not recognized, please try again')
print(validated_response('trigger test'))
How would I go about telling the user when they've got the correct letter in a list? Only way I know of is to insert the index, but that doesn't feel very flexible, especially when the words vary in length.
import random
possibleWords = [["apple"], ["grapefruit"], ["pear"]]
randomWord = random.choice(possibleWords)
anotherWord = ''.join(randomWord)
finalWord = list(anotherWord)
maxTries = list(range(0, 11))
attemptsMade = 0
triesLeft = 10
print("Hangman!")
print("\nYou got {} tries before he dies!".format(maxTries[10]))
print("There's {} possible letters.".format(len(finalWord)))
for tries in maxTries:
userChoice = input("> ")
if userChoice == finalWord[0]:
print("You got the first letter correct! It is {}.".format(finalWord[0]))
else:
print("Ouch! Wrong letter! {} tries remaining.".format(triesLeft))
attemptsMade += 1
triesLeft -= 1
Talking about characters in lists, or - what i think is more likely in your case - chars in words you can just check for
if userChoice in finalWord:
# [...] do stuff here
and further on use the index function to determine the position (or positions if multiple occurence).
finalWord.index(userChoice)
You could sure also go the way using index() function directly and work your way using the return values.
Use Python's "in" keyword to check if something is within a list/iterable.
if userChoice in finalWord:
Though for this, I'd just use regex or list comprehension to get the indexes while you are at it, since you might want them for the game.
char_indexes = [i for (i, l) in enumerate(finalWord) if l == userChoice]
if len(char_indexes):
Use a set for the letters that are in the word, and whenever the player guesses a letter, check if the letter is still in the set. If it’s not, it was a wrong letter; if it is, then remove that letter and just continue. If the set is empty at some point, then the player guessed all letters of the word.
Something to get you started:
def hangman (word):
letters = set(word.lower())
attempts = 5
while attempts > 0:
guess = input('Guess a character ')
if guess[0].lower() in letters:
print('That was correct!')
letters.remove(guess[0])
else:
print('That was not correct!')
attempts -= 1
if not letters:
print('You solved the word:', word)
return
hangman('grapefruit')
import re
def check_input():
while True:
try:
sequence = raw_input("Please input:")
if sequence = [a,t,c,g]: # checking for valid input
continue
else:
print("invalid input, sequence coule only contain the "
"following letters (a,t,c,g)"):
check_input()
I basically want the script to check the user's input whether it contains only these four letters (a,t,c,g). If their input contains anything other than that four letters, it could print that second statement and prompt the user to input again. I saw there are similar questions and I already tried to change my script according to those posts but it still gives me the invalid syntax error at the if < sequence position. Anyone knows what's wrong here?
You need to iterate over every letter in the input and check if it is in the set of allowed letters, like this:
sequence = raw_input("Please input:")
for letter in sequence:
if letter not in "atcg":
print("invalid input, sequence coule only contain the following letters (a,t,c,g)")
When you discover that the sequence is invalid, you could choose to end the check by using a break statement after the print, or you could count how many invalid letters are allowed.
Your function must check and give user whether True or False:
def check_input(word):
result = True
for letter in sequene:
if letter in 'atcg':
continue
else:
result = False
break
return result
check_input('atgc')
Error Message:
if check_input('agct'):
continue
else:
print "error message"
You could also use the filter command:
def checkInp():
seq = raw_input("Please input sequence:")
if not filter(lambda m: m not in 'ATGC', seq) == '':
print 'Invalid input characters in sequence: ' + filter(lambda m: m not in 'ATGC', seq)
print 'Pleas'
check_input()
else: return seq
sequence, once input by the user will be a string, so you would need to iterate over each character in the sequence and use in to verify the existence of the character in the accepted characters string. String comparisons in Python are also case sensitive, so you need to match the case of the input to your expected string. I've used uppercase based on your sample input.
def check_input():
sequence = input("Please input:")
sequence.upper()
for letter in sequence:
if letter in 'ATCG':
continue
else:
print("invalid input, sequence could only contain the
following letters: a, t, c or g.")