Create function in python to find the highest of all function arguments, and return the "tag" of the value

Question

Consider the following: p1=1; p2=5; p3=7; highest=max(p1,p2,p3).

The max function would return 7. I am looking to create a similar function, which would return "p3". I have created a small function (by simple comparisons) for the above example, shown below. however I am having trouble when the number of arguments go up.

def highest(p1,p2,p3):
    if (p1>p2) and (p1>p3):
        return "p1"
    if (p2>p1) and (p2>p3):
        return "p2"
    if (p3>p1) and (p3>p1):
        return "p3"

Is there a simpler way to do this>

score 5 · Answer 1 · answered Jan 02 '11 at 06:24

5

There is no way to get the name of the variable that had the highest value in the caller (because it might be a number or a complex expression), but by using keyword arguments exclusively, you can get the name of the parameter. Something like this:

def argmax(**kwargs):
    mx = -1e+400 # overflows to -Inf
    amx = None
    for k, v in kwargs.iteritems():
        if v > mx:
            mx = v
            amx = k
    return amx

Works like this:

>>> argmax(a=1,b=2,c=3)
'c'

but the catch is, it doesn't work if any of the arguments is positional:

>>> argmax(1,2,3)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: argmax() takes exactly 0 arguments (3 given)

Depending on what you're doing, this construct might be more useful: (hat tip to http://lemire.me/blog/archives/2008/12/17/fast-argmax-in-python/ )

>>> a = [9,99,999]
>>> a.index(max(a))
2

answered Jan 02 '11 at 06:24

zwol

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The body of this function can be written more concisely: `return max(kwargs, key=kwargs.get)` – Jan 02 '11 at 13:26
`float('-inf')` gives -Inf, but you can just use `None` instead – Jochen Ritzel Jan 02 '11 at 18:03
THC4k: Good to know about `float('-inf')`. `(None < any_number) == True` seems like a language quirk that should never be used, to me. Paul, Lennart: I didn't know about the key parameter to `max` either. Agree that's a cleaner way to write it, but I'm going to leave my answer as is since it illustrates a more general technique. – zwol Jan 02 '11 at 18:54

Lennart Regebro · Accepted Answer · 2011-01-02T14:45:41.473

5

Update: Paul Hankin pointed out that max() took a key function, which I didn't know. So:

>>> def argmax(**kw):
...   return max(kw, key=kw.get)
... 
>>> argmax(foo=3, bar=5, frotz=1, kaka=-3)
'bar'

Other solutions for completeness:

In Python 2.7 and 3.x you can use dictionary comprehensions.

>>> def argmax(**kw):
...     wk = {v:k for k,v in kw.items()}
...     return wk[max(wk)]
... 
>>> argmax(foo=3, bar=5, frotz=1, kaka=-3)   
'bar'

Dictionary comprehensions are neat. :)

In earlier versions of Python you can do this:

>>> def argmax(**kw):
...     wk = dict([(v,k) for k,v in kw.items()])
...     return wk[max(wk)]
... 
>>> argmax(foo=3, bar=5, frotz=1, kaka=-3)   
'bar'

Which will work in anything after Python 2.2 or so.

edited Jan 02 '11 at 14:45

answered Jan 02 '11 at 12:53

Lennart Regebro

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How about just `return max(kw, key=kw.get)` (for Python 2.5 and later). – Jan 02 '11 at 13:24
Ah! I did not know max took a key function. Excellent! – Lennart Regebro Jan 02 '11 at 14:15
Thanks very much. THis was very helpful, and thanks @Paul Hankin for pointing out that max took a key function. – gatechgrad Jan 03 '11 at 14:13

taskinoor · Answer 3 · 2011-01-02T07:19:57.357

Obviously it dose not handle variable length arguments. If you want variable length argument then that's a different issue. If you have 10 arguments then just add them in the definition and it will return the correct argument name (not necessarily started with 'p'). But the catch is the number of arguments (3 or 5 or 10 or whatever else) is not variable. You need to know how many arguments you require.

def highest(p1,p2,p3,p4,p5):
    d = locals()
    keys = d.keys()
    max_key = keys[0]
    max_val = d[max_key]
    for i in range(1,len(keys)):
        key = keys[i]
        val = d[key]
        if val > max_val:
            max_val = val
            max_key = key

    return max_key

print highest(3,2,5,10,1)
print highest(1,5,2,2,3)
print highest(5,2,5,1,11)
print highest(3,2,1,1,2)

score 0 · Answer 4 · answered Jan 02 '11 at 07:13

The only way to keep this anything close to extensible is to take a list as argument and return the index of its highest element. You can put a p in front and begin counting at 1 if you really want to.

def highest(x):
    return 'p' + repr(x.index(max(x))+1)

Create function in python to find the highest of all function arguments, and return the "tag" of the value

4 Answers4