I wasn't entirely sure how to word this question, so I'll stick with explaining my specific goal.
I'm trying to implement an 'eat' option to a user_input function for a text adventure. I want it to check if the verb is 'eat', check if the item is in the player's inventory, check if the item is consumable or not, and then - my goal - show that the player eats it (and only show it once, see below). The 'eat' option has no functionality as of yet, I'm only testing print statements.
I'm totally aware that the code below is not the most efficient way to handle user input (I'm still tweaking ways to handle input that is unexpected), so I'm 100% open to criticism. This was just my initial way of handling it and I'd like to try to make it work. I have a few other verbs ('go', 'look', 'take', etc.) that work, I'm just having a lot of trouble with 'eat'.
def user_input(self):
player = Player()
while True:
user_input = input("> ").lower()
words = user_input.split(" ")
wordlist = list(words)
verb = wordlist[0]
if len(wordlist) == 2:
noun = wordlist[1]
if noun not in items:
print("Somehow '{}' isn't the right word.".format(noun))
continue
else:
pass
# The above works fine because I don't have
# an issue with any of the other verbs regarding
# that bit of code.
# There's more code between these blocks to:
# handle if the user enters a noun that is > 1 word,
# define a few more verbs,
# then,
if verb == "eat":
for item in player.inventory:
if item.name == noun:
if isinstance(item, Consumable):
print("You eat the {}.".format(noun))
break
else:
print("You can't eat that!")
break
else:
print("You don't have '{}'.".format(noun))
I had to use the for loop (at least, I think I had to) because I'm iterating over a list that has objects in them, not strings, so I couldn't just use
if noun in player.inventory:
(I still tried it a million times though, it took forever to come up with a solution for that problem). Here's my specific example for the code above:
class Fists(Weapons):
def __init__(self):
self.name = "fists"
# more instance variables for Fists()
class LeatherWallet(Classy):
def __init__(self):
self.name = "leather wallet"
# more ...
class Pizza(Consumable):
def __init__(self):
self.name = "pizza"
# more ...
class Player():
def __init__(self):
self.inventory = [Fists(), LeatherWallet(), Pizza()]
# other stuff
*eat fists
You can't eat that!
*eat leather wallet
You don't have 'leather wallet'.
You can't eat that!
*eat pizza
You don't have 'pizza'.
You don't have 'pizza'.
You eat the pizza.
From the looks of it, it has got to be a simple fix, because it's clear what's happening as it iterates over the list. I just don't know how to (or if you can) wrangle a for loop to make it check conditions first then print later. At the risk of sounding dumb, I turn to you all for help!
Thank you, and please let me know if I can make this question/my goal any clearer.
Edit:
Tried to make the goal a little clearer in opening paragraphs.
You're not saying what you want the output to be, but my best guess is you don't want to say "you don't have X" more than once.
One way to do this is to use a flag, initialized to false, set to true when you find the item. When you get out of the loop the flag will tell you if you found it.
Something like this:
if verb == "eat":
found = false
for item in player.inventory:
if item.name == noun:
found = true
if isinstance(item, Consumable):
print("You eat the {}.".format(noun))
break
else:
print("You can't eat that!")
break
if !found:
print("You don't have '{}'.".format(noun))
for loops have else clauses that are run if the loop doesn't exit with a break. The idea is that the loop is looking for an item, will break when it finds it, and the else clause handles the default case where you don't find what you're looking for. The variables used inside the loop are still valid after the loop ends so the found (or defaulted) variable can be used in the next bit of code.
In your case, if you move that final else clause out one level, it will only run if the item is not found. In your example output, the two "you don't have 'pizza1" lines will no longer be printed.
for item in player.inventory:
if item.name == noun:
if isinstance(item, Consumable):
print("You eat the {}.".format(noun))
break
else:
print("You can't eat that!")
break
else:
print("You don't have '{}'.".format(noun))
item = None # <-- you can give a default value if you want to
# use the variable later
Looping through the list is okay until the list gets large. Then you are better off indexing the list with a dict for faster lookup.
self.inventory = dict((item.name, item)
for item in (Fists(), LeatherWallet(), Pizza()))
Then the for loop is replaced with
try:
item = player.inventory[noun]
if isinstance(item, Consumable):
print("You eat the {}.".format(noun))
else:
print("You can't eat that!")
except KeyError:
print("You don't have '{}'.".format(noun))
So, another approach is to just make this work, right?
if noun in player.inventory:
...
You have (at least) two ways to do it.
List comprehension
The first and easy one is using a list comprehension:
inventory_items = [item.name for name in player.inventory]
if noun in inventory_items:
...
That's it.
Custom class
The second one is creating an Inventory class inheriting from list and overriding the __contains__ method so you can compare them however you want.
Something like:
class Inventory(list):
def __contains__(self, item):
"""
Override `list.__contains__`.
"""
# First, check whether `item` is in the inventory.
#
# This is what a normal `list` would do.
if list.__contains__(self, item):
return True
# If we couldn't find the item in the list searching
# the "normal" way, then try comparing comparing it
# against the `name` attribute of the items in our
# inventory.
for inventory_item in self:
# Ensure we can do `inventory_item.name` so we don't get
# an AttributeError.
if 'name' not in dir(inventory_item):
continue
if inventory_item.name == item:
return True
return False
Then you instantiate your inventory like:
self.inventory = Inventory([Fists(), LeatherWallet(), Pizza()])
Remember that Inventory inherits from list, so if you can do list(x) then you should also be able to do Inventory(x) and get a very similar result (depending on how much you override in your class).
You could also get creative and make all items inherit from an Item class where you define the __eq__ method and make Pizza() == 'pizza' to simplify the Inventory comparisons.
Related
I'm writing a program where you can play blackjack with a computer. I've implemented a class, a "main" and a bunch of others functions. The problem is that almost all of my them are build to get the proper input from the user. For instance:
def get_the_answer():
while True:
answer = input("Write 'reveal' to reveal cards, 'add' to add one more: ")
if answer in ['reveal', 'add']:
break
else:
continue
return answer
Or this one, which also has some prints in it:
def start_and_greet():
print('', colored('Hi! Shall we play Blackjack?', 'yellow'), '', sep='\n')
print(colored(blackjack()))
print()
while True:
answer = input("Write 'start' / 'exit': ")
if answer in ['start', 'exit']:
break
else:
continue
return answer
If I put all my inputs and prints in "main" function, there will be almost no additional functions besides "main", so this won't be acceptable by the task.
The main reason why I'm worrying about it is that I don't understand how one should test this...
Thank you in advance for your help!
I know that it is not recommended to use a function with print / input, although I saw how David Malan (CS50P course) uses the similar one in lectures:
def get_number():
while True:
n = int(input("What's n? "))
if n > 0:
break
return n
I have a question on how to check a user's input and make sure they are returning a specific string. Currently, the function when called will ask the user for their input. However, if they choose a string that is not part of the function, the else statement will execute and continue the code. I am trying to figure out how to loop this function, until a user inputs one of the strings that the function is looking for. Can anyone help me with this? I am new to python and would appreciate any help.
def dwarf_class_definer(dwarf_class):
if dwarf_class == "Thane":
print("'Ahhh Nobility'")
elif dwarf_class == "Mekanik":
print("'Interesting a Mechanic'")
elif dwarf_class == "Ancestrite":
print("'A spiritualist. I see...'")
elif dwarf_class == "Prisoner":
print("'Never met a gen-u-ine 'Last chancer.'")
elif dwarf_class == "Civilian":
print("'ehhh a civilian? Wut you doing here?'")
else:
print("You aren't choosing a valid class.")
dwarf_class = input("Which Class will you choose?: ")
dwarf_class_definer(dwarf_class)
A while loop will keep going until you tell it not to anymore. You can see when an expected value is supplied, the break command will terminate the while loop. A dictionary can also make your code a lot cleaner and easier to maintain compared to a bunch of if statements.
dwarf_classes = {
"Thane": "'Ahhh Nobility'",
"Mekanik": "'Interesting a Mechanic'",
"Ancestrite": "'A spiritualist. I see...'",
"Prisoner": "'Never met a gen-u-ine 'Last chancer.'",
"Civilian": "'ehhh a civilian? Wut you doing here?'",
}
while True:
dwarf_class = input("Which Class will you choose?: ")
if dwarf_class in dwarf_classes.keys():
print(dwarf_classes[dwarf_class])
break
print("You aren't choosing a valid class.")
example:
$ python3 so.py
Which Class will you choose?: Python!
You aren't choosing a valid class.
Which Class will you choose?: Prisoner
'Never met a gen-u-ine 'Last chancer.'
What I want is this - I have a list of names created from user input. now i have to come up with a way for the user to edit a name by entering the name that they what to edit and then obviously edit it into what they want and store it in the list.
if it helps heres everything I have so far. def edit() is where im struggling.
def mainMenu():
print("\nMAIN MENU")
print("1. Display Members:")
print("2. Add A Member(s):")
print("3. Remove A Member:")
print("4. Edit Member:")
print("5. Exit:")
selection = int(input("\nEnter Choice: "))
if selection == 1:
display()
elif selection == 2:
add()
elif selection == 3:
remove()
elif selection == 4:
edit()
elif selection == 5:
exit()
else:
print("Invalid choice, enter 1-5.")
mainMenu()
def display():
#displaying roster...
print(roster)
mainMenu()
def add():
#adding team members...
size = int(input("How many players are you adding?"))
global roster
roster = [0] * size
for i in range(size):
roster[i] = input("Enter members name: ")
roster.append(roster)
mainMenu()
def remove():
#removing a team member...
roster.remove(input("Enter member to be removed: "))
mainMenu()
def edit():
#edit a team member...
roster.insert(input("Enter Name to be edited: "))
mainMenu()
mainMenu()
Removing and adding elements are pretty easy in python because they are directly supported by the language. Each of them can be translated into only one instruction.
When something doesn't seem very obvious, such as the editing functionality you are trying to implement, try breaking it down to things that can be expressed as a simple operation that holds one one line (even if not in order).
To find the answer I thought this: somewhere in my code, I want to type roster[ind_name_to_edit] = new_name.
I knew then that before typing this, I would want to find the value of ind_name_to_edit. This can be done by roster.index(name_to_edit). And you already know how to get the name to be edited and the name to edit ;)
If you're still unsure how to do what you want to do, re-read this answer and see the documentation of the index method of list in python3 and maybe some examples here.
N.B: If your list is supposed to be sorted in some way, you should implement your own search algorithm instead of using index, and you should consider re-sorting the list after the edit. I know it's a long shot but just in case.
I am currently taking some computer science courses in school and have come to a dead end and need a little help. Like the title says, I need of create a list of Class objects from a file with objects that have a duplicate not added to the list, I was able to successfully do this with a python set() but apparently that isn't allowed for this particular assignment, I have tried various other ways but can't seem to get it working without using a set. I believe the point of this assignment is comparing data structures in python and using the slowest method possible as it also has to be timed. my code using the set() will be provided.
import time
class Students:
def __init__(self, LName, FName, ssn, email, age):
self.LName = LName
self.FName = FName
self.ssn = ssn
self.email = email
self.age = age
def getssn(self):
return self.ssn
def main():
t1 = time.time()
f = open('InsertNames.txt', 'r')
studentlist = []
seen = set()
for line in f:
parsed = line.split(' ')
parsed = [i.strip() for i in parsed]
if parsed[2] not in seen:
studentlist.append(Students(parsed[0], parsed[1], parsed[2], parsed[3], parsed[4]))
seen.add(parsed[2])
else:
print(parsed[2], 'already in list, not added')
f.close()
print('final list length: ', len(studentlist))
t2 = time.time()
print('time = ', t2-t1)
main()
A note, that the only duplicates to be checked for are those of the .ssn attribute and the duplicate should not be added to the list. Is there a way to check what is already in the list by that specific attribute before adding it?
edit: Forgot to mention only 1 list allowed in memory.
You can write
if not any(s.ssn==parsed[2] for s in studentlist):
without committing to this comparison as the meaning of ==. At this level of work, you probably are expected to write out the loop and set a flag yourself rather than use a generator expression.
Since you already took the time to write a class representing a student and since ssn is a unique identifier for the instances, consider writing an __eq__ method for that class.
def __eq__(self, other):
return self.ssn == other.ssn
This will make your life easier when you want to compare two students, and in your case make a list (specifically not a set) of students.
Then your code would look something like:
with open('InsertNames.txt') as f:
for line in f:
student = Student(*line.strip().split())
if student not in student_list:
student_list.append(student)
Explanation
Opening a file with with statement makes your code more clean and
gives it the ability to handle errors and do cleanups correctly. And
since 'r' is a default for open it doesn't need to be there.
You should strip the line before splitting it just to handle some
edge cases but this is not obligatory.
split's default argument is ' ' so again it isn't necessary.
Just to clarify the meaning of this item is that the absence of a parameter make the split use whitespaces. It does not mean that a single space character is the default.
Creating the student before adding it to the list sounds like too
much overhead for this simple use but since there is only one
__init__ method called it is not that bad. The plus side of this
is that it makes the code more readable with the not in statement.
The in statement (and also not in of course) checks if the
object is in that list with the __eq__ method of that object.
Since you implemented that method it can check the in statement
for your Student class instances.
Only if the student doesn't exist in the list, it will be added.
One final thing, there is no creation of a list here other than the return value of split and the student_list you created.
This question references info from my previous question:
Text Game - If statement based of input text - Python
So, now I have this:
#Choice Number1
def introchoice():
print()
print("Do you 'Hesitate? or do you 'Walk forward")
print()
def Hesitate():
print()
print("You hesistate, startled by the sudden illumination of the room. Focusing on the old man who has his back turned to you. He gestures for you to come closer. \n ''Come in, Come in, don't be frightened. I'm but a frail old man'' he says.")
print()
#
def Walk():
print()
print("DEFAULT")
print()
#
def pick():
while True:
Input = input("")
if Input == "Hesitate":
Hesitate()
break
if Input == "Walk":
Walk()
break
#
#
pick()
#-#-#-#-#-#-#-#-#-#-#-#-#
#Clean-up
#-#-#-#-#-#-#-#-#-#-#-#-#
Now what I want to do is this;
def pick():
while True:
Input = input("")
if Input == "Hesitate":
Hesitate()
break
if Input == "Walk":
Walk()
break
if Input is not "Walk" or "Hesitate":
print("INVALID")
break
#
#
pick()
#-#-#-#-#-#-#-#-#-#-#-#-#
#Clean-up
#-#-#-#-#-#-#-#-#-#-#-#-#
Now that I have the game determine specific text inputs, I want it to be able to detect if the input was not one of the choices. That way, as typed in the above code, If the input text is not either "Walk" or "hesitate", print Text "INVALID"
How would I do this exactly?
I guess you want to still receiving input if it is "invalid", so the break statements have to be inside the ifs. Otherwise, the loop will only iterate one time.
Also, I would recommend you to use a if-elif-else structure so your code looks more organized.
You can't use is or is not in this case, because these operators are used to check if the objects are the same (same reference). Use the operators == and != to check for equality.
while True:
my_input = input("> ")
if my_input == "Hesitate":
hesitate()
break
elif my_input == "Walk":
walk()
break
else:
print("INVALID")
Notes:
I would recommend you to follow Python naming conventions. Names of variables and methods should start in lowercase.