从函数返回的解压缩可变长度列表

Question

Ok so I'm defining a function that takes a variable number of inputs and clamps each of them

def clamp(*args):
    return [ max(min(arg, 0.8), 0.2) for arg in args]

I like the readability of this function:

a = 0.12
b = 0.45
c = 0.992

A,B,C = clamp(a,b,c)

print A,B,C

>> 0.2, 0.45, 0.8

This takes advantage of python's automatic unpacking of lists to tuples. The trouble I am having is that if I only give one argument, python doesn't unpack it from the list and I get a list instead of a float, which is annoying.

print clamp(a)

>> [0.2]

My solution so far is to check the length of the list and index it if there is only one element:

def clamp(*args):
    result = [ max(0.2, min(0.8,arg)) for arg in args]
    return result if len(result) > 1 else result[0]

a = 0.12
print clamp(a)

>> [0.2]

My question is, is there a more idiomatic way to do this?

Answer 1

I'm not very conviced but here is an alternative solution

>>> clam = lambda a: max(min(a, 0.8), 0.2)

>>> def clamp(a, *args):
...     if args:
...        return [ clam(arg) for arg in (a,)+args]
...     else:
...        return clam(a)
... 
>>> clamp(123, 123)
[0.8, 0.8]
>>> clamp(123)
0.8

Answer 2

You can force it to unpack a single element by adding a comma after the name. Not ideal but here you go:

A, = clamp(a)

Answer 3

def clamp(*args):
    rt = [max(min(arg, 0.8), 0.2) for arg in args]
    return rt if len(rt) > 1 else rt[0]

or else, remember how you make a tuple of one element. Put a comma after A to force unpack for a single variable.

A, = clamp(0.12) # with your old function which always returns list

Answer 4

def clamp(*args):
    lst = [max(min(arg, 0.8), 0.2) for arg in args]
    if len(lst)==1:
        return lst[0]
    return lst

Answer 5

I think when you call it with one element, you can add a , after the name. a, = clamp(1) . In this case, you don't have to modify your function and the automatic unpacking still happens.

>>> a, b, c = [1,2,3]
>>> a
1
>>> b
2
>>> c
3
>>> a, = [1]
>>> a
1
>>>