i tried calling .copy on it and then passing it in the function. that didn't work.
when i tried coppying in the function itself it still changed the original list.
the function is in another file
main.py
win_condition.check_r_win(board)
win_condition.py
def check_r_win(board):
board = _board.copy()
for col in board:
while(len(col) <= ROWS):
col.append("-")
I don't really get what you are trying here. Python's list is a mutable object type. If you pass a object reference of a list to a function and change the list within this function, it also gets changed outside of the function scope.
Related
Imagine I have a dict.
d = ['a': 1 , 'b':3]
I'm having a hard time to understand the difference between d.get and d.get().
I know that d.get() get the value from the key, like this:
print(d.get('a') )
output: 1
But when I write d.get, it shows this:
print(d.get)
output: <built-in method get of dict object at .........>
What is 'd.get' doing in my code?
I'm using python 3X
A method is literally just an attribute of an object that happens to be of type <class function>. The output you see is essentially what happens when you try to call print() on any function object, and is essentially a concise string representation that python creates for the function.
Actually calling a function is done with parentheses: d.get('a'), which means to execute the behavior the function refers to. It doesn't especially matter where the function is, though: I could do the following, and it would still work:
d = {'a': 1 , 'b':3}
freefunc = d.get
freefunc('a')
This is what the term "first class functions" refers to, when people compare python to something like Java. An entire function can be encapsulated in a variable and treated no differently than any other variable or attribute.
The short answer? There is no difference between the two methods. They are the same exact method.
The difference in your code is at when you write .get() you call the method, but when you write .get you just get a pointer (or location in the memory, to be exact) for that method, to call it later on if needed.
In the first scenario, you are calling print on the result of executing get('a'), which in this case is 1.
In your second scenario, you are calling print on the get function itself, instead of on an execution of it, which evaluates to its documentation, i.e. <built-in method get of dict object at... etc.
If I make a code like:
lists = ["a='1'", "b='2'", "c=a+b"]
returned_list = []
for x in lists:
exec(x)
print(c)
It works, and It print "12". but, If I use exec() in function:
lists = ["a='1'", "b='2'", "c=a+b"]
def test(lst):
for x in lists:
exec(x)
print(c)
test(lists)
It returns NameError: name 'c' is not defined. How could I use exec() in function?
When you assign a new variable in a function, you are actually assigning a variable in a scope which will be closed after the function is closed.
Imagine it as a bubble with an item inside, which after the bubble blows, the item blows and disappears as well. It means, using exec() in a function would create a temporary local variable. But since functions have a predefined code, adding new variables to them without changing the code directly, would not be possible. in that case we need to use global keyword for each new variable in exec to make the variable save in the main and not in function. Therefor, your list would like this:
lists = ["global a\na='1'"]
also I'm not quite sure if you like the output of a+b be 12, if not, you can just remove the single quotes around each number such as "a=1" to make them integers
for further information check this and this
I have a dataset and I want to make a function that does the .get_dummies() so I can use it in a pipeline for specific columns.
When I run dataset = pd.get_dummies(dataset, columns=['Embarked','Sex'], drop_first=True)
alone it works, as in, when I run df.head() I can still see the dummified columns but when I have a function like this,
def dummies(df):
df = pd.get_dummies(df, columns=['Embarked','Sex'], drop_first=True)
return df
Once I run dummies(dataset) it shows me the dummified columsn in that same cell but when I try to dataset.head() it isn't dummified anymore.
What am I doing wrong?
thanks.
You should assign the result of the function to df, call the function like:
dataset=dummies(dataset)
function inside them have their own independent namespace for variable defined there either in the signature or inside
for example
a = 0
def fun(a):
a=23
return a
fun(a)
print("a is",a) #a is 0
here you might think that a will have the value 23 at the end, but that is not the case because the a inside of fun is not the same a outside, when you call fun(a) what happens is that you pass into the function a reference to the real object that is somewhere in memory so the a inside will have the same reference and thus the same value.
With a=23 you're changing what this a points to, which in this example is 23.
And with fun(a) the function itself return a value, but without this being saved somewhere that result get lost.
To update the variable outside you need to reassigned to the result of the function
a = 0
def fun(a):
a=23
return a
a = fun(a)
print("a is",a) #a is 23
which in your case it would be dataset=dummies(dataset)
If you want that your function make changes in-place to the object it receive, you can't use =, you need to use something that the object itself provide to allow modifications in place, for example
this would not work
a = []
def fun2(a):
a=[23]
return a
fun2(a)
print("a is",a) #a is []
but this would
a = []
def fun2(a):
a.append(23)
return a
fun2(a)
print("a is",a) #a is [23]
because we are using a in-place modification method that the object provided, in this example that would be the append method form list
But such modification in place can result in unforeseen result, specially if the object being modify is shared between processes, so I rather recomend the previous approach
I am new to Python and am at a lost as to what I'm doing wrong. I am trying to use the fqdn variable that is being returned to the caller which is main() but I'm getting NameError: name 'fqdn' is not defined
I'm betting this is some type of global variable statement issue or something like that, but I've been researching this and can't figure it out.
If a function from a module returns a value, and the caller is main(), shouldn't main() be able to use that returned value???
Here's the layout:
asset.py
def import_asset_list():
# Open the file that contains FQDNs
openfile = open(r"FQDN-test.txt")
if openfile.mode == 'r':
# Remove CR from end of each item
fqdn = openfile.read().splitlines()
# Add https to the beginning of every item in list
fqdn = ["https://" + item for item in fqdn]
openfile.close()
return fqdn
tscan.py
def main():
import asset
asset.import_asset_list()
# Iterate through list
for i in fqdn:
if SCHEDULED_SCAN == 1:
create_scheduled_scan(fqdn)
launch_scan(sid)
check_status_scan(uuid)
else:
create_scan(fqdn)
launch_scan(sid)
check_status_scan(uuid)
Short Explanation
Yes, main() should be able to use the returned value, but it's only the value that is returned, not the variable name. You have to define a variable of your own name, to receive the value, and use that instead.
Long Explanation
The name of a variable inside any function is simply a "label" valid only within the scope of this function. A function is an abstraction which means "Give me some input(s), and I will give you some output(s)". Within the function, you need to reference the inputs somehow and, potentially, assign some additional variables to perform whatever it is you would like to. These variable names have no meaning whatsoever outside the function, other than to, at most, convey some information as to the intended use of the function.
When a function returns a value, it does not return the "name" of the variable. Only the value (or the reference in memory) of the variable. You can define your own variable at the point where you call the function, give it your own name and assign to it the returned result of the function, so you simply have to write:
def main():
import asset
my_asset_list = asset.import_asset_list()
# Iterate through list
for i in my_asset_list:
if SCHEDULED_SCAN == 1:
create_scheduled_scan(my_asset_list)
launch_scan(sid)
check_status_scan(uuid)
else:
create_scan(my_asset_list)
launch_scan(sid)
check_status_scan(uuid)
I don't know where the uuid and the sid variables are defined.
To make sure you have understood this properly, remember:
You can have multiple functions in the same file, and use identically-named variables within all those functions, this will be no problem because a variable (with its name) only exists within each specific function scope.
Variable names do not "cross" the boundaries of the scope, only variable values/references and to do this, a special construct is used, i.e. the return [something] statement.
I have created three functions. The first function is used in the other two functions but I am passing it a hardcoded filepath. I want to be able to pass this in as a parameter, but I seem to be getting an issue.
Essentially, given a file_path, my function will get the first item in the list and then the second item.
So far my code is as follows :
def sort_files(file_path):
"""Sort files in ascending order"""
files = os.listdir(file_path)
return sorted(files, reverse=True)
def current_day():
"""Get the current day file"""
return sort_files(file_path)[0]
def previous_day():
"""Get the previous day file"""
return sort_files(file_path)[1]
If you want a function to accept an argument, you need to define it as doing so by specifying the parameter name it will be known as in the function (as you did with sort_files).
How are you executing the call to the current_day and previous_day. You should make them as function that can take a parameter.
Also please post the code that you are using to execute the whole setup.