# Function application optimization.

Discussion in 'Python' started by Jacek Generowicz, Dec 12, 2003.

1. ### Jacek GenerowiczGuest

Given

fncs = [func1, func2, ..., funcN]
args = [arg1, arg2, ..., argN]

How should one spell

results = map(lambda f,a: f(a), fncs, args)

in order to get the result most quickly ?

Unfortunately "apply" takes a tuple of arguments, and there is no
"funcall"[*] in Python.

[*] def funcall(fn, *args):
return fn(*args)

Jacek Generowicz, Dec 12, 2003

2. ### Bengt RichterGuest

On 12 Dec 2003 10:47:50 +0100, Jacek Generowicz <> wrote:

>Given
>
> fncs = [func1, func2, ..., funcN]
> args = [arg1, arg2, ..., argN]
>
>How should one spell
>
> results = map(lambda f,a: f(a), fncs, args)
>
>in order to get the result most quickly ?
>
>
>
>Unfortunately "apply" takes a tuple of arguments, and there is no
>"funcall"[*] in Python.
>
>
>[*] def funcall(fn, *args):
> return fn(*args)
>

>>> fncs = [lambda x,f='func_%s(%%s)'%i:f%x for i in xrange(4)]
>>> args = 'zero one two three'.split()
>>> map(lambda f,a: f(a), fncs, args)

['func_0(zero)', 'func_1(one)', 'func_2(two)', 'func_3(three)']

I'd probably try a list comprehension

>>> [f(a) for f,a in zip(fncs,args)]

['func_0(zero)', 'func_1(one)', 'func_2(two)', 'func_3(three)']

Regards,
Bengt Richter

Bengt Richter, Dec 12, 2003

3. ### Duncan BoothGuest

Jacek Generowicz <> wrote in
news::

> Given
>
> fncs = [func1, func2, ..., funcN]
> args = [arg1, arg2, ..., argN]
>
> How should one spell
>
> results = map(lambda f,a: f(a), fncs, args)
>
> in order to get the result most quickly ?
>

Well, the way you wrote it isn't actually bad. The obvious alternative
(using a list comprehension) is slightly slower, but there isn't an awful
lot to choos between any of them, I suspect most of the time goes in the
actual f(a) function call. Any of these methods runs at about 300,000
function calls/second on my slow laptop.

>>> setup = """import itertools

def fn1(a): pass
fns = [fn1] * 100
args = [0] * 100
"""
>>> stmt1 = "result = [ f(a) for (f,a) in itertools.izip(fns, args) ]"
>>> stmt2 = "result = [ f(a) for (f,a) in zip(fns, args) ]"
>>> t = timeit.Timer(stmt1, setup)
>>> t.repeat(3,10000)

[3.5571303056673855, 3.5537404893639177, 3.5594278043718077]
>>> t = timeit.Timer(stmt2, setup)
>>> t.repeat(3,10000)

[3.893281967400867, 3.87834794645687, 3.8829105375124868]
>>> setup = """import itertools

def fn1(a): pass
fns = [fn1] * 1000
args = [0] * 1000
"""
>>> t = timeit.Timer(stmt1, setup)
>>> t.repeat(3,1000)

[3.3503928571927304, 3.3343195853104248, 3.3495254285111287]
>>> t = timeit.Timer(stmt2, setup)
>>> t.repeat(3,1000)

[3.8062683944467608, 3.7946001516952492, 3.7881063096007779]
>>> stmt3 = "results = map(lambda f,a: f(a), fns, args)"
>>> t = timeit.Timer(stmt3, setup)
>>> t.repeat(3,1000)

[3.3275902384241363, 3.3010907810909202, 3.3174872784110789]

The f(a) call is taking about half the time in any of these methods, so you
aren't going to get very much improvement whatever you do to the loop.

--
Duncan Booth
int month(char *p){return(124864/((p[0]+p[1]-p[2]&0x1f)+1)%12)["\5\x8\3"
"\6\7\xb\1\x9\xa\2\0\4"];} // Who said my code was obscure?

Duncan Booth, Dec 12, 2003
4. ### Bruno DesthuilliersGuest

Jacek Generowicz wrote:
> Given
>
> fncs = [func1, func2, ..., funcN]
> args = [arg1, arg2, ..., argN]
>
> How should one spell
>
> results = map(lambda f,a: f(a), fncs, args)
>
> in order to get the result most quickly ?
>

I dont know if it will be faster (timeit might tell you this), but what
about something as simple stupid as:

results = []
nbfuns = len(fncs)
for i in range(len):
results.append(fncs(args))

Note that you must have at least len(fncs) args.

HTH,
Bruno

Bruno Desthuilliers, Dec 12, 2003
5. ### Jacek GenerowiczGuest

(Bengt Richter) writes:

> I'd probably try a list comprehension
>
> >>> [f(a) for f,a in zip(fncs,args)]

> ['func_0(zero)', 'func_1(one)', 'func_2(two)', 'func_3(three)']

Yup, been there, done that, it's slower. (I guess I should have
mentioned that in the original post.)

Jacek Generowicz, Dec 12, 2003
6. ### Peter OttenGuest

Jacek Generowicz wrote:

> Given
>
> fncs = [func1, func2, ..., funcN]
> args = [arg1, arg2, ..., argN]
>
> How should one spell
>
> results = map(lambda f,a: f(a), fncs, args)
>
> in order to get the result most quickly ?
>

If you can afford to destroy the args list, a tiny speed gain might be in
for you (not verified):

for index, func in enumerate(fncs):
args[index] = func[index]

Peter

Peter Otten, Dec 12, 2003
7. ### John RothGuest

"Jacek Generowicz" <> wrote in message
news:...
> Given
>
> fncs = [func1, func2, ..., funcN]
> args = [arg1, arg2, ..., argN]
>
> How should one spell
>
> results = map(lambda f,a: f(a), fncs, args)
>
> in order to get the result most quickly ?
>
>
>
> Unfortunately "apply" takes a tuple of arguments, and there is no
> "funcall"[*] in Python.
>
>
> [*] def funcall(fn, *args):
> return fn(*args)

Building on a couple of other responses:

Untested code:

fncs = [func1, func2, ..., funcN]
args = [arg1, arg2, ..., argN]
results = []
for function, arguement in zip(fncs, args):
results.append(function(arguement))

Notice the use of zip() to put the two lists together.
I haven't timed it, but removing an extra layer of
function call has got to be faster. Likewise, zip
and tuple unpacking is most likely going to be
faster than indexing every time through the loop.
The append, on the other hand, might slow things
down a bit.

John Roth

>

John Roth, Dec 12, 2003
8. ### Duncan BoothGuest

"John Roth" <> wrote in
news::

> Building on a couple of other responses:
>
> Untested code:
>
> fncs = [func1, func2, ..., funcN]
> args = [arg1, arg2, ..., argN]
> results = []
> for function, arguement in zip(fncs, args):
> results.append(function(arguement))
>
> Notice the use of zip() to put the two lists together.
> I haven't timed it, but removing an extra layer of
> function call has got to be faster. Likewise, zip
> and tuple unpacking is most likely going to be
> faster than indexing every time through the loop.
> The append, on the other hand, might slow things
> down a bit.

Yes, getting rid of the append does indeed speed things up. On the same
system as I posted timings for the list comprehension, the fastest way I've
found so far is to get rid of the appends by preallocating the list, and to
go for the plain old simple technique of writing a for loop out explicitly.
Using 'enumerate' to avoid the lookup on one of the input lists is nearly
as fast, but not quite.

>>> setup = """import itertools

def fn1(a): pass
fns = [fn1] * 1000
args = [0] * 1000
"""

>>> stmt1 = """result = args[:]

for i in range(len(fns)):
result = fns(args)
"""
>>> min(timeit.Timer(stmt1, setup).repeat(3,1000))

2.9747384094916924
>>> stmt3 = "results = map(lambda f,a: f(a), fns, args)"
>>> min(timeit.Timer(stmt3, setup).repeat(3,1000))

3.3257092731055309
>>>

--
Duncan Booth
int month(char *p){return(124864/((p[0]+p[1]-p[2]&0x1f)+1)%12)["\5\x8\3"
"\6\7\xb\1\x9\xa\2\0\4"];} // Who said my code was obscure?

Duncan Booth, Dec 12, 2003
9. ### Peter OttenGuest

Peter Otten wrote:

> for index, func in enumerate(fncs):
> args[index] = func[index]

Oops,
args[index] = func(args[index])

And it's much slower than result.append(func(args[index]), too :-(

Peter

Peter Otten, Dec 12, 2003