I'm trying to take a list of tuples that contain the name of individual functions and its arguments and instantiate them..this will be used for a multi-threading application.
import pandas as pd
df = pd.DataFrame()
def func1(df,arg2,arg3):
print 'func1'
print type(df)
print arg2,arg4
def func2(arg1,df,arg3,arg4):
print 'func2'
print arg1
print type(df)
print arg3,arg4
list_of_func_tups = [(func1,df,'two','three'),(func2,'one',df,'three','four')]
for tup in list_of_func_tups:
f = tup[0]
args = tup[1:]
results = f(args)
But the problem is that args is still seen as a tuple, hence, the error stating that only one argument was given.
TypeError: func1() takes exactly 3 arguments (1 given)
Is there a way that I can somehow get args seen as its respective individual arguments? Maybe a type of for loop or something? Anythin?
Let's define a function:
def f(arg1, arg2, arg3):
return arg1 + arg2 + arg3
As Willem mentions, you can flatten a tuple or list with an asterix.
f(*(1, 2, 3))
You can use the double asterix for dictionaries
f(**{'arg1': 1, 'arg2': 2, 'arg3': 3})
What you are looking for is a way to feed the tuple as positional arguments. You can do this by using a leading asterisk (*):
result = f(*args)
This is documented in the documentation of function calls:
If the syntax *expression appears in the function call,
expression must evaluate to an iterable. Elements from this iterable are treated as if they were additional positional
arguments; if there are positional arguments x1,..., xN, and
expression evaluates to a sequence y1, ..., yM, this is equivalent to
a call with M+N positional arguments x1, ..., xN, y1, ..., yM.
Related
Can anyone explain the difference when unpacking the dictionary using single or double asterisk? You can mention their difference when used in function parameters, only if it is relevant here, which I don't think so.
However, there may be some relevance, because they share the same asterisk syntax.
def foo(a,b)
return a+b
tmp = {1:2,3:4}
foo(*tmp) #you get 4
foo(**tmp) #typeError: keyword should be string. Why it bothers to check the type of keyword?
Besides, why the key of dictionary is not allowed to be non-string when passed as function arguments in THIS situation? Are there any exceptions? Why they design Python in this way, is it because the compiler can't deduce the types in here or something?
When dictionaries are iterated as lists the iteration takes the keys of it, for example
for key in tmp:
print(key)
is the same as
for key in tmp.keys():
print(key)
in this case, unpacking as *tmp is equivalent to *tmp.keys(), ignoring the values. If you want to use the values you can use *tmp.values().
Double asterisk is used for when you define a function with keyword parameters such as
def foo(a, b):
or
def foo(**kwargs):
here you can store the parameters in a dictionary and pass it as **tmp. In the first case keys must be strings with the names of the parameter defined in the function firm. And in the second case you can work with kwargs as a dictionary inside the function.
def foo(a,b)
return a+b
tmp = {1:2,3:4}
foo(*tmp) #you get 4
foo(**tmp)
In this case:
foo(*tmp) mean foo(1, 3)
foo(**tmp) mean foo(1=2, 3=4), which will raise an error since 1 can't be an argument. Arg must be strings and (thanks # Alexander Reynolds for pointing this out) must start with underscore or alphabetical character. An argument must be a valid Python identifier. This mean you can't even do something like this:
def foo(1=2, 3=4):
<your code>
or
def foo('1'=2, '3'=4):
<your code>
See python_basic_syntax for more details.
It is a Extended Iterable Unpacking.
>>> def add(a=0, b=0):
... return a + b
...
>>> d = {'a': 2, 'b': 3}
>>> add(**d)#corresponding to add(a=2,b=3)
5
For single *,
def add(a=0, b=0):
... return a + b
...
>>> d = {'a': 2, 'b': 3}
>>> add(*d)#corresponding to add(a='a',b='b')
ab
Learn more here.
I think the ** double asterisk in function parameter and unpacking dictionary means intuitively in this way:
#suppose you have this function
def foo(a,**b):
print(a)
for x in b:
print(x,"...",b[x])
#suppose you call this function in the following form
foo(whatever,m=1,n=2)
#the m=1 syntax actually means assign parameter by name, like foo(a = whatever, m = 1, n = 2)
#so you can also do foo(whatever,**{"m":1,"n":2})
#the reason for this syntax is you actually do
**b is m=1,n=2 #something like pattern matching mechanism
so b is {"m":1,"n":2}, note "m" and "n" are now in string form
#the function is actually this:
def foo(a,**b): # b = {"m":1,"n":2}
print(a)
for x in b: #for x in b.keys(), thanks to #vlizana answer
print(x,"...",b[x])
All the syntax make sense now. And it is the same for single asterisk. It is only worth noting that if you use single asterisk to unpack dictionary, you are actually trying to unpack it in a list way, and only key of dictionary are unpacked.
[https://docs.python.org/3/reference/expressions.html#calls]
A consequence of this is that although the *expression syntax may appear after explicit keyword arguments, it is processed before the keyword arguments (and any **expression arguments – see below). So:
def f(a, b):
print(a, b)
f(b=1, *(2,))
f(a=1, *(2,))
#Traceback (most recent call last):
# File "<stdin>", line 1, in <module>
#TypeError: f() got multiple values for keyword argument 'a'
f(1, *(2,))
Suppose I have a function like:
def myfun(a, b, c):
return (a * 2, b + c, c + b)
Given a tuple some_tuple = (1, "foo", "bar"), how can I use some_tuple to call myfun, to get the result (2, "foobar", "barfoo")
I know could define myfun so that it accepts the tuple directly, but I want to call the existing myfun.
See also: What do the * (star) and ** (double star) operators mean in a function call?.
myfun(*some_tuple) does exactly what you request. The * operator simply unpacks the tuple (or any iterable) and passes them as the positional arguments to the function. Read more about unpacking arguments.
Note that you can also expand part of argument list:
myfun(1, *("foo", "bar"))
Take a look at the Python tutorial section 4.7.3 and 4.7.4.
It talks about passing tuples as arguments.
I would also consider using named parameters (and passing a dictionary) instead of using a tuple and passing a sequence. I find the use of positional arguments to be a bad practice when the positions are not intuitive or there are multiple parameters.
This is the functional programming method. It lifts the tuple expansion feature out of syntax sugar:
apply_tuple = lambda f, t: f(*t)
Redefine apply_tuple via curry to save a lot of partial calls in the long run:
from toolz import curry
apply_tuple = curry(apply_tuple)
Example usage:
from operator import add, eq
from toolz import thread_last
thread_last(
[(1,2), (3,4)],
(map, apply_tuple(add)),
list,
(eq, [3, 7])
)
# Prints 'True'
Similar to #Dominykas's answer, this is a decorator that converts multiargument-accepting functions into tuple-accepting functions:
apply_tuple = lambda f: lambda args: f(*args)
Example 1:
def add(a, b):
return a + b
three = apply_tuple(add)((1, 2))
Example 2:
#apply_tuple
def add(a, b):
return a + b
three = add((1, 2))
>>> class Test(object):
>>> def test(self,*arg):
>>> print(arg[0],arg[1])
>>> p = Test()
>>> t = 2,3
>>> p.test(t)
gives me IndexError: tuple index out of range
why is that? and how do i get the value for that tuple?
You passed in just one argument (the whole tuple (2, 3)), so only arg[0] exists; if you meant the tuple values to be separate arguments, apply them with the *args call syntax:
p.test(*t)
The alternative is to not use the *arg catchall argument in your function definition:
def test(self, arg):
Now your function has two normal positional arguments, self and arg. You can only pass in one argument, and if that is your tuple, arg[0] and arg[1] will work as expected.
Using your demo class:
>>> class Test(object):
>>> def test(self,*arg):
>>> print(arg[0],arg[1])
When doing this:
>>> p = Test()
>>> t = 2,3
>>> p.test(t)
arg will have a value of [(1,2),]
When doing this:
>>> p = Test()
>>> t = 2,3
>>> p.test(*t)
arg will have a value of [1,2]
The * in the function means that all remaining arguments (non-keyword) are put into a list for you.
In the first case you send (1,2) has a single argument. In the second case the tuple is made into individual arguments using the * thus you send in 1 and 2.
For complete documentation on this refer to this Python article:
http://docs.python.org/2/reference/expressions.html#calls
Let's say I have a method definition like this:
def myMethod(a, b, c, d, e)
Then, I have a variable and a tuple like this:
myVariable = 1
myTuple = (2, 3, 4, 5)
Is there a way I can pass explode the tuple so that I can pass its members as parameters? Something like this (although I know this won't work as the entire tuple is considered the second parameter):
myMethod(myVariable, myTuple)
I'd like to avoid referencing each tuple member individually if possible...
You are looking for the argument unpacking operator *:
myMethod(myVariable, *myTuple)
From the Python documentation:
The reverse situation occurs when the
arguments are already in a list or
tuple but need to be unpacked for a
function call requiring separate
positional arguments. For instance,
the built-in range() function expects
separate start and stop arguments. If
they are not available separately,
write the function call with the
*-operator to unpack the arguments out of a list or tuple:
>>> range(3, 6) # normal call with separate arguments
[3, 4, 5]
>>> args = [3, 6]
>>> range(*args) # call with arguments unpacked from a list
[3, 4, 5]
In the same fashion, dictionaries can
deliver keyword arguments with the
**-operator:
>>> def parrot(voltage, state='a stiff', action='voom'):
... print "-- This parrot wouldn't", action,
... print "if you put", voltage, "volts through it.",
... print "E's", state, "!"
...
>>> d = {"voltage": "four million", "state": "bleedin' demised", "action": "VOOM"}
>>> parrot(**d)
-- This parrot wouldn't VOOM if you put four million volts through it. E's bleedin' demised !
So, Python functions can return multiple values. It struck me that it would be convenient (though a bit less readable) if the following were possible.
a = [[1,2],[3,4]]
def cord():
return 1, 1
def printa(y,x):
print a[y][x]
printa(cord())
...but it's not. I'm aware that you can do the same thing by dumping both return values into temporary variables, but it doesn't seem as elegant. I could also rewrite the last line as "printa(cord()[0], cord()[1])", but that would execute cord() twice.
Is there an elegant, efficient way to do this? Or should I just see that quote about premature optimization and forget about this?
printa(*cord())
The * here is an argument expansion operator... well I forget what it's technically called, but in this context it takes a list or tuple and expands it out so the function sees each list/tuple element as a separate argument.
It's basically the reverse of the * you might use to capture all non-keyword arguments in a function definition:
def fn(*args):
# args is now a tuple of the non-keyworded arguments
print args
fn(1, 2, 3, 4, 5)
prints (1, 2, 3, 4, 5)
fn(*[1, 2, 3, 4, 5])
does the same.
Try this:
>>> def cord():
... return (1, 1)
...
>>> def printa(y, x):
... print a[y][x]
...
>>> a=[[1,2],[3,4]]
>>> printa(*cord())
4
The star basically says "use the elements of this collection as positional arguments." You can do the same with a dict for keyword arguments using two stars:
>>> a = {'a' : 2, 'b' : 3}
>>> def foo(a, b):
... print a, b
...
>>> foo(**a)
2 3
Actually, Python doesn't really return multiple values, it returns one value which can be multiple values packed into a tuple. Which means that you need to "unpack" the returned value in order to have multiples.
A statement like
x,y = cord()
does that, but directly using the return value as you did in
printa(cord())
doesn't, that's why you need to use the asterisk. Perhaps a nice term for it might be "implicit tuple unpacking" or "tuple unpacking without assignment".