For example, I'd like to do something like: greet(,'hola'), where greet is:
def greet(person='stranger', greeting='hello')
This would help greatly for testing while writing code
Upon calling a function you can use the variable names to make it even more clear what variable will assume which value. At the same time, if defaults are provided in the function definition, skipping variables when calling the function does not raise any errors. So, in short you can just do this:
def greet(person='stranger', greeting='hello')
print('{} {}'.format(greeting, person))
return
greet(greeting='hola') # same as greet(person='stranger', greeting='hola')
# returns 'hola stranger'
Note that, as I said above this would not work if for example your function definition was like this:
def greet(person, greeting)
print('{} {}'.format(greeting, person))
return
Since in this case, Python would complain saying that it does not know what to do with person; no default is supplied..
And by the way, the problem you are describing is most likely the very reason defaults are used in the first place
Without knowing the other parameters, and only knowing that the parameter you want to change is in second position you could use the inspect module to get function signature & associated default values.
Then make a copy of the default values list and change the one at the index you want:
import inspect
def greet(person='stranger', greeting='hello'):
print(person,greeting)
argspec = inspect.getargspec(greet)
defaults = list(argspec.defaults)
defaults[1] = "hola" # change second default parameter
greet(**dict(zip(argspec.args,defaults)))
Assuming that all parameters have default values (else it shifts the lists an that fails) that prints:
stranger hola
Related
I've been looking up the information about why people should not use empty list as a function argument and pass none type as an argument and came up with a few questions.
Here is my code:
def add_employee(emp, emp_list=None):
if emp_list is None:
emp_list = []
emp_list.append(emp)
print(emp_list)
And here is code without second argument's type specified:
def add_employee(emp, emp_list):
emp_list.append(emp)
return emp_list
When I do not define emp_list as an empty list or none I can not utilize function's deafualt argument behavior: I can't call it like add_employee('Mark'), I had to add second variable to pass. Why is it good to have that backup default behaviour? Why couldn't I just leave it as emp_list.
Here is a great explanation of why you should avoid using mutable arguments in your defaults. Or at least use them sparingly: link
The general gist of it that the list will be created for the first time (in a location in memory) when you define the function. So as python reads the code and interprets it, you will end up creating that list once on the first 'read' of the function.
What this means is that you are not creating a fresh list each time you call that function, only when you define it.
To illustrate I will use the example from the link I shared above:
def some_func(default_arg=[]):
default_arg.append("some_string")
return default_arg
>>> some_func()
['some_string']
>>> some_func()
['some_string', 'some_string']
>>> some_func([])
['some_string']
>>> some_func()
['some_string', 'some_string', 'some_string']
If I understood your question correctly, you are asking why you're better off defining the emp_list explicitly rather than having it outside the function. In my mind it boils down to encapsulation. You're essentially trying to make sure that your function doesn't change the behavior of anything outside of its scope so you're forced to pass it things directly and be explicit. In practice if you have a variable outside of the scope named emp_list, it is absolutely fine to just append to it as long as you understand what the expected behavior is.
If you pass a list in the first bit of code as the emp_list, then the variable emp_list will contain your list a. The if statement will check if list a is None and since that check fails, it will skip the next line of assigning it a fresh empty list.
Thanks for reading my question. As I'm still new to Python, I would like to ask about the () in Python.
def addOne(myFunc):
def addOneInside():
return myFunc() + 1
return addOneInside # <-----here is the question
#addOne
def oldFunc():
return 3
print oldFunc()
Please note that on line four, although the programme returns a function, it does not need parentheses(). Why does it NOT turn out with an error for syntax error? Thank you very much for your answers in advance!
The parentheses are used to run a function, but without them the name still refers to the function just like a variable.
return myFunc() + 1
This will evaluate the myFunc function, add 1 to its value and then return that value. The brackets are needed in order to get the function to run and return a numeric value.
return addOneInside
This is not actually running addOneInside, it is merely returning the function as a variable. You could assign this to another name and store it for later use. You could theoretically do this:
plusOne = addOneInside
plusOne()
And it will actually call the addOneInside function.
The particular instance in your initial question is known as a Decorator, and it's a way for you to perform code on the parameters being passed to your function. Your example is not very practical, but I can modify it to show a simple use case.
Let's say that you want to only have positive numbers passed to your function. If myFunc is passed a negative number, you want it to be changed to 0. You can manage this with a decorator like this.
def addOne(myFunc):
def addOneInside(num):
if num < 0:
num = 0
return myFunc(num)
return addOneInside # <-----here is the question
#addOne
def oldFunc(number):
return number
To explain, the #addOne is the decorator syntax, and it's attaching the addOneInside function to be called on the argument/s of oldFunc whenever you call it. So now here's some sample output:
oldFunc(-12)
>>> 0
oldFunc(12)
>>> 12
So now you could add logic to oldFunc that operates independently of the parameter parsing logic. You could also relatively easily change what parameters are permitted. Maybe there's also a maximum cap to hit, or you want it to log or note that the value shouldn't be negative. You can also apply this decorator to multiple functions and it will perform the same on all of them.
This blogpost explained a lot for me, so if this information is too brief to be clear, try reading the long detailed explanation there.
Your indentation in function addOne() was incorrect (I have fixed it), but I don't think that this was your problem.
If you are using Python3, then print is a function and must be called like this:
print(oldFunc())
I have written one method in python as follows :
def report(self,year,month):
print year,month
emptylist=[]
parameter month is optional. Sometimes it's NOT passed from the caller function but sometimes it is passed.How is this different from the following code ?
def func(self,*args,**kwargs):print args,kwargs I know this has simple answer but I'm new to python and programming stuff. Please help me out.
If month has a default value, you can try
def report(self, year, month='default_value'):
print year, month
emptylist = []
The value of month gets overwritten if passed
In addition to Ashoka's answer, it's normally a good idea to use None as the default value in the method's signature and assign the true default value with in function body. This makes it easier to wrap the function, pass arguments, and it also avoids problems with mutable default values.
def func(kwarg=None):
if kwarg is None:
kwarg = 'default'
print(kwarg)
func()
func(kwarg='Hello')
I am running into a problem writing recursive member functions in Python. I can't initialize the default value of a function parameter to be the same value as a member variable. I am guessing that Python doesn't support that capability as it says self isn't defined at the time I'm trying to assign the parameter. While I can code around it, the lack of function overloading in Python knocks out one obvious solution I would try.
For example, trying to recursively print a linked list I get the following code for my display function;
def display(self,head = -1):
if head == -1:
head = self.head
if not head:
return
print head,
self.display(head.link)
While this code works, it is ugly.
The main function looks like this:
def main():
l = List();
l.insert(3);
l.insert(40);
l.insert(43);
l.insert(45);
l.insert(65);
l.insert(76);
l.display()
if __name__ == "__main__":
main()
If I could set the display function parameter to default to self.head if it is called without parameters then it would look much nicer. I initially tried to create two versions of the function, one that takes two parameters and one that takes one but as I said, Python doesn't support overloading. I could pass in an argument list and check for the number of arguments but that would be pretty ugly as well (it would make it look like Perl!). The trouble is, if I put the line
head = self.head
inside the function body, it will be called during every recursive call, that's definitely not the behavior I need. None is also a valid value for the head variable so I can't pass that in as a default value. I am using -1 to basically know that I'm in the initial function call and not a recursive call. I realize I could write two functions, one driving the other but I'd rather have it all self contained in one recursive function. I'm pretty sure I'm missing some basic pythonic principle here, could someone help me out with the pythonic approach to the problem?
Thanks!
I don't really see what's wrong with your code. If you chose a falsy default value for head, you could do: head = head or self.head which is more concise.
Otherwise, this is pretty much what you have to do to handle default arguments. Alternatively, use kwargs:
def display(self,**kwargs):
head = kwargs.get("head", self.head)
if not head:
return
print head,
self.display(head=head.link) # you should always name an optional argument,
# and you must name it if **kwargs is used.
I need help with parameteres. Do both of these function definitions do the exact same thing for print_twice?
def print_twice(lol):
print lol
print lol
def print_twice(michael):
print michael
print michael
If yes, then I'm guessing the word used for the parameter doesn't matter, correct?
The word we use for the parameter does matter. It is important that the word you use:
is meaningful and clearly explains what the argument is for,
does not override some variable name from the external scope.
Importance of meaningful arguments' names
The name you use for argument is important, because the names of the arguments, their default values and the function name are the things developers using your function first see, even without the need to look into function documentation (eg. by using help(your_function)). Just use IDLE to define your function and then try to use it - when writing it, the IDLE will show you possible arguments.
So please, give them meaningful names that will make using your function easier and will not require looking into the documentation.
Overriding variables from outer scopes
When it comes to the second point, just look at this example:
def show_elements(elements):
"""Shows elements of the passed argument
"""
for element in elements:
print element
which works ok, but if you replace elements with eg. list, you will override list within this specific scope:
def show_elements(list):
"""Shows elements of the passed argument
"""
for element in list:
print element
and then if you would like to use list eg. for building a list, or converting from other type into list, then you will have problems. list is a builtin and you should not override it. Similar is true also about the other variables from the scopes surrounding the function.
Historically, when Python was resolving variable names by first looking into local scope, then global and builtin scopes, skipping all nonlocal ones (eg. scope from the function in which our function was defined), enclosing scope's variables were passed that way:
def funca():
local_val1 = 'some value1'
local_val2 = 'some value2'
def funcb(local_val1=local_val1):
# local_val1 is accessible here, even though local_val2 is not
...
...
But since the above is no longer true, you will need to take surrounding scopes into account, thus using non-conflicting name is important.
Yes they do. The name of a parameter is irrelevant, although good programming practices mandate that the name should be clear and self-explanatory
That's correct, the name of the parameter doesn't matter.
yes that is correct its just a variable name ...
That is correct. The word used for the parameter in the function definition is only a name, and does not refer to anything external.
Programming is supposed to be logical. So, let's think logically. If you write "a" on a paper twice, you get "a" two times. Same with "b." So you're doing the same function with letters. But what if you reassigned a value to a, each time a went through the function. I mean, what if a was a number, then IMO the closest you could get is something like this:
def func1(a, b):
a = input("Enter a number: ")
b = input("Enter another number: ")
b *= b
a *= a
print func1(a)
print func1(a)
print func1(b)
print func1(b)
Now, when I try to compile this specific code online, I get an error but I think something like this will work for the sake of trying to do what you're doing if done correctly? It's a good experiment, and I would imagine some usage in it.