#🔒 help with code
topaz elk
topaz elk
so in the line 55applyFunctionToFolders(top_level_folders, removeParentFolderPrefix)
i want to pass function parameters for the removeParentFolderPrefix()
because as you u can see in line 44 , it takes a prefix parameter
i dont know how exactly to proceed
or what my thought process should be in problems like this
unborn timber
you want partial application
topaz elk
whats that?
unborn timber
basically, creating a new function that would accept less arguments, forward them to call the original function and send some additional parameters
e.g.
f(x, y) = z
g(x) = f(x, 0)
that might be too abstract of an example but i hope its understandable
topaz elk
i do not understand how u can put ( , ) on the left hand side
unborn timber
it is pseudocode, more like math
!e
def f(x, y):
return x + y
print((lambda x: f(x, 10))(2))
scarlet fossilBOT
!e
```py
def f(x, y):
return x + y
print((lambda x: f(x, 10))(2))
```
:white_check_mark: Your 3.12 eval job has completed with return code 0.
12
will need to learn how lambda works
unborn timber
it is not required, you can use def just fine here too
topaz elk
lambda is like a function blueprint?
unborn timber
no, it is like def but: anonymous, is an expression, and limited to evaluating a single expression (so no multiple statements)
unborn timber
basically, you'd make a function thats like removeParentFolderPrefix, but only takes the path, and the prefix is already known and it would just call the original removeParentFolderPrefix with that path and prefix
though in applyFunctionToFolders you dont actually pass the folder path to the callable? and you use generic types without parametrizing them, e.g. just list. list of what? you can specify e.g. list[Path], same with just Callable: callable with what parameters and what does it return? Callable[[int, str], float] would be something like def f(x: int, y: str) -> float:
topaz elk
good idea! i will make it list[Path] since thats more suitable
so the reason i didnt make Callable more specific is , if i want to use a different function?
its a good idea right?
unborn timber
but the signature stays the same
topaz elk
does it?
unborn timber
yes, you will always be passing a folder path to the callable in the applyFunctionToFolders
unborn timber
lambda arguments: expression is the same as f in
def f(arguments):
return expression
it is just a way to define a function in an expression, it is never required and can always be replaced with a def
topaz elk
so
def add(first , second):
return first+second
lambda first, second : first+second
like that?
unborn timber
yes, those functions are equivalent, though the lambda version doesnt bind it to any name
unborn timber
to not have multiple statements, it is basically a non-reason, just something you can do. it is mostly used like in your case where you want to pass a function to another function, and you can express its body with just one expression so you can use it instead of def'ing a function, having to think of a name, ...
topaz elk
i see!
so
def mainFunction(x,y)
return x+y
def additionalFunc(x):
mainFunction(x,10)
#and i use it like this?
ParentFunction(additionalFunc(100))
wait what
unborn timber
you didnt make it return, and in this case it wasnt particularly useful - the reason for the partial application was that you wanted to pass a function but with the prefix argument set, not apply it fully
unborn timber
ParentFunction(additionalFunc)
unborn timber
but then how would i pass arguments?
the thing we went over was doing just that, but maybe you need an example
topaz elk
def greet(name: str, greeting: str) -> str:
return f"{greeting}, {name}!"
def process_greeting(func, *args):
# Call the passed function with the provided arguments
return func(*args)
result = process_greeting(greet, "Alice", "Hello")
print(result) # Output: "Hello, Alice!"
remember your original problem
you have apply_to_folders(top_level_folders, function_that_takes_a_folder_path)
you want to provide removeParentFolderPrefix as the function_that_takes_a_folder_path, but it takes 2 arguments: the path and a prefix, so you create another function ,that takes just the folder path, and forwards it to original function with some already known prefix
topaz elk
right and i want to somehow pass argument to the function_that_takes_a_folder_path
i dont understand the last part
unborn timber
!e
def f(xs, g):
return [g(x) for x in xs]
def g(x, y):
return x + y
print(f([1, 2, 3], lambda x: g(x, 5)))
:white_check_mark: Your 3.12 eval job has completed with return code 0.
[6, 7, 8]
f(xs, g) expects g to have 1 argument, but in our case it had 2, and we wanted to call it with y = 5 each time, so we made a function that takes just 1 argument, and calls g with that argument and a constant 5 as the y argument
topaz elk
thats three functions?
unborn timber
yes
topaz elk
i also have 3 functions, applyfunction(), removePrefix(), and one that i need to make
unborn timber
yes, you'd make a function that takes just the folder path, and calls removeprefix with that folder path and some predefined prefix
(and make applyfunction actually call it with the folder path, currently you're not providing arguments to it, just doing folderOperation())
topaz elk
thank you for explaining this to me! i will try to make it now
wait
applyFunctionToFolders(top_level_folders, lambda x : removeFolderPrefix(x,"my Prefix"))
like this?
unborn timber
yep
topaz elk
thank you!
scarlet fossilBOT
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