Passing a variable into a parameter defined by another function python - python

I am unsure of why the variable totalspeed variable is not being passed correctly to the function startgame as the startgame function is called after the gettotalspeed function.
Exerpt from call function:
gettotalspeed(party_ids)
NoOfEvents=0
startgame(party_ids,totalspeed,distance,NoOfEvents)
Functions
def gettotalspeed(party_ids):
#Get selected party members IDS
print(party_ids)
#Obtain Speeds
ids_string = ','.join(str(id) for id in party_ids)
mycursor.execute("SELECT startspeed FROM characters WHERE CharID IN ({0})".format(ids_string))
myspeeds=mycursor.fetchall()
totalspeed=0
for speedval in myspeeds:
totalspeed=totalspeed + speedval[0]
print("totalspeed is: ",totalspeed)
return totalspeed
def startgame(party_ids,totalspeed,distance,NoOfEvents):
#Check if game end
print(totalspeed)
while distance!=0:
#Travel...
distance=distance-totalspeed
NoOfEvents=NoOfEvents+1
#Generate Random Encounter
genevent(NoOfEvents)
return NoOfEvents
Error Produced:
NameError: name 'totalspeed' is not defined
Outputs (ignoring party_ids)
totalspeed is: 15

I suspect that your problem is self-evident from the main program:
gettotalspeed(party_ids)
NoOfEvents=0
startgame(party_ids,totalspeed,distance,NoOfEvents)
Of the variables you pass to your functions, only NoOfEvents is defined. party_ids, totalspeed, and distance have no definitions.
Work through a tutorial on Python scoping rules. Most of all, note that a function defines a scoping block. Variables inside the function are reclaimed when you leave the function; their names do not apply outside of that block. Your posted program has three independent totalspeed variables.

You forgot to make totalspeed a global variable like global totalspeed in your gettotalspeed() function. You might also be confused about what return does. If you wanted to do it the "proper" way, you could do totalspeed = gettotalspeed(party_ids). Hope this helps!

Related

Nested functions in python

I have a function that makes a dictionary based on file input. Then another function that makes a calculation based on the dictionary. These 2 functions are nested in a main function. I.e:
def make_dic():
data=dict()
code...
return data
def make_calculation():
for x,y in data.items()
code....
def main():
data=make_dic()
make_calculation()
My problem is that I get an error:
NameError: name 'data' is not defined
How can I get make_calculation() recognize the dictionary created in make_dic() when they are nested in main()?
Thanks.
Pass "data" as a parameter to your make_calculation function:
def make_dic():
data=dict()
code...
return data
def make_calculation(data):
for x,y in data.items()
code....
def main():
data=make_dic()
make_calculation(data)
The reason you need to do this is due to Python scoping rules (i.e. the location within a program , or "scope" of where a parameter is or is not defined).
Parameter "data" is not a global variable (using global variables is rarely recommended and should be a last resort) , and thus make_calculation(data) needs to receive the element as a parameter, which was defined elsewhere -ultimately the parameter created in make_dic()
#chepner said as much, more formally, in their comment (i.e. Python is lexically scoped). For much more on Python scoping rules see this old, but still useful ,answer : Short description of the scoping rules?
in the first function try putting global data, hope it helps.

Scopes and Customizing Open (learning Python p. 539)

I'm a bit confused about a code in the book "Learning Python", p. 539.
As far as I know assignments within a function are only in this local scope. So if I want to change a global one I first have to declare it global. But why does the following code change the builtin.open() to custom completely once called?
import builtins
def makeopen(id):
original = builtins.open
def custom(*pargs, **kargs):
print('Custom open call %r: ' % id, pargs, kargs)
return original(*pargs, **kargs)
builtins.open = custom
If I call makeopen('spam') and a F = open('text.txt') afterwards I get the custom call. So the builtin.open() has been changed in the whole script after the makeopen('spam'). Why?
And if I would make some more makeopen('xx') one builtin.open('text.txt') would print the custom call for every created makeopen. Why?
Comparing this code to
x = 99
def changing():
x = 88
changing()
print(x)
doesnt even help me. Isn't it the same but with an x instead of builtin.open()?
A variable is considered local if you assign to it anywhere in the function, unless you declare it global.
In your first piece of code, you never assign anything to builtins, so it's not considered local. You just change one of its attributes, open.
The rule is respected!
In your second piece of code, you assign something to x in x = 88, so it is considered local.
When you call makeopen, you replace the original, global open with custom. custom, when executed, prints its name and calls the original open.
If you call makeopen a second time, it will create a second, different custom function, and make the name builtins.open refer to it. When you call this function, it will print its name, then call original, which is what builtins.open referred to when it was created - and that is your first custom function, which will print its name and call the original open.
So, successive calls to makeopen create a chain of functions, and calling open will make each of them run and call its predecessor.

Python: cannot return a value assigned inside of a function outside of that function to later use

def program(n):
name = input("What is your name? >")
return name
print(name)
I have a code that i am trying to execute very similar to this. When executing it, it will not return the variable 'name' i used in the function in order to use that variable outside the function. Why is this?
I am super new to coding by the way so please excuse me if i made a stupid mistake.
When you run your program, you need to assign the result (i.e. whatever is returned in your program, to another variable). Example:
def get_name():
name = input('Name please! ')
return name
name = get_name()
print('Hello ' + name)
Pssst.. I took your function parameter n away since it was not being used for anything. If you're using it inside your actual program, you should keep it :)
For a bit of a more in-depth explanation...
Variables that are declared inside your neat little function over there can't be seen once you come out of it (though there are some exceptions that we don't need to get into right now). If you're interested in how this works, it's known as "variable scope."
To execute the content of a function you need to make a call to the function and assign the return value to some variable. To fix your example, you would do:
def get_name():
name = input("What is your name? >")
return name
name = get_name()
print(name)
I have changed the function name from program() to get_name() seeing as program() is a ambiguous name for a function.
This snippet will make a call to the get_name() function and assign the return value to the variable name. It is important to note, that the name variable inside the function is actually a different variable to the one that we are assigning to outside the function. Note: I have removed the argument n from get_name() since it was not being used.

Name Error when passing a parameter?

I am writing a program and keep on getting trouble with the error:
File "C:/Users//Documents///", line 47, in <module>
print(operations(sorted_list_desc))
NameError: name 'sorted_list_desc' is not defined
Even though you can not see the file I am working with, I have provided my code. I am supposed to loop the years and randomly choose an operator for each of the years. However, I keep on getting the error above. Any ideas why I keep on getting it?
import csv
#List to store different operators
list_operators = ["+","-","*","/"]
#Function to read file of list
def read_file():
#List to store data
sorted_list_desc = []
year = csv_file.readline()
population = csv_file.readline()
return sorted_list_desc
print(read_file())
def operations(sorted_list_desc):
for i in range(len(sorted_list_desc)):
operator = random.choice(list_operator)
return operator
print(operations(sorted_list_desc))
##
sorted_list_desc is generated by read_file(). Change your last line to:
print(operations(read_file()))
The line:
print(read_file())
does not magically created a global object withe name sorted_list_desc.
You need to hand it over explicitly to operations().
Alternatively, you could write:
sorted_list_desc = read_file()
print(operations(sorted_list_desc))
You have only assigned sorted_list_desc inside your function read_file(). Thus, your error is telling you that sorted_list_desc is not assigned to anything outside your function's scope. See Python scoping rules, and notice that you don't even need to pass sorted_list_desc as a parameter because it is assigned inside your function anyways.
Looks like sorted_list_desc is defined within the scope of your read_file() function and thus you can't reference it outside of that function. You can call print(operations(sorted_list_desc)) within 'read_file()', or you can define sorted_list_desc = [] at a global scope.

Python Parameter Confusion

I'm a beginner, writing a Python Blackjack script, and got confused about whether or not a function (dealPlayer) needs a parameter. It works either way, with a parameter or without. I'm not sure if I've had a brain fart, or I've not learned something along the way. Here's the code:
import random
dealer = []
player = []
c = ""
deck = [2,2,2,2,3,3,3,3,4,4,4,4,5,5,5,5,6,6,6,6,7,7,7,7,8,8,8,8,
9,9,9,9,10,10,10,10,10,10,10,10,10,10,10,10,11]
def dealPlayer(deck):
cardOne = random.choice(deck)
cardTwo = random.choice(deck)
player.append(cardOne)
player.append(cardTwo)
deck.remove(cardOne)
deck.remove(cardTwo)
The question is, do I need (deck) as a parameter in the function? It works with or without (deck) as a parameter. I've gone back over different tutorials, and other's code, but I'm still confused. Thanks for any help.
The reason your code works with or without deck as a parameter is because there is a global variable named deck, so when you reference deck inside your function, the function will first look for the local variable (the parameter) and then if it doesn't find it, it will look for the global variable.
It's best to refactor your code to not use global variables at all -- define deck initially inside a function and then pass that as a result or argument to other functions as needed. If you don't want to do that, then at least make sure your argument does not shadow (have the same name as) the global variable, to avoid confusion further on. Or remove the argument entirely and use the global variable only, if that's appropriate for your program.
did i get you right that if your function is:
def dealPlayer():
the code still works? this should raise a undefined deck error. EDIT: this was wrong of course its global. And just works without it. but thats a bad practice.
def dealPlayer():
deck = []
this should raise a Index Error.
cardOne = random.choice()
This raises a TypeError.

Categories

Resources