以简单的方式将多个参数从数组传递到函数

Question

Well I really don't know how to describe this in words but I've an easy example:

def foo(x,y,z):
     r=x+....

a = [1,2,3]
foo(a[0], a[1], a[2])

Can I do the same without calling a[n] several times? but keeping "x,y,z" definition?

Answer 1

Use the unpacking operator !

>>> def foo(x,y,z):
...     return x+y+z
... 
>>> a = [1,2,3]
>>> foo(*a)
6

From the documentation

If they are not available separately, write the function call with the *-operator to unpack the arguments out of a list or tuple:

 >>> args = [3, 6] >>> range(*args) # call with arguments unpacked from a list [3, 4, 5] 

Answer 2

Yeah use unpacking operator to pass any arbitrary arguments to your function :

def foo(*args):
     for i in args : # you can iterate over args or get its elements by indexing
         #do stuff

a = [1,2,3]
foo(*a)

And if you just pass 3 argument you can use unpacking in call time.

def foo(x,y,z):
     r=x+....

a = [1,2,3]
foo(*a)

Answer 3

Try this

def foo(x,y,z):
     r=x+....

a = [1,2,3]
foo(*a)