Compressing "n"-time object member call

Question

Is there any non-explicit for way to call a member n times upon an object?

I was thinking about some map/reduce/lambda approach, but I couldn't figure out a way to do this -- if it's possible.

Just to add context, I'm using BeautifulSoup, and I'm extracting some elements from an html table; I extract some elements, and then, the last one.

Since I have:

# First value
print value.text

# Second value
value = value.nextSibling
print value.text

# Ninth value
for i in xrange(1, 7):
    value = value.nextSibling
print value.text

I was wondering if there's any lambda approach -- or something else -- that would allow me to do this:

# Ninth value
((value = value.nextSibling) for i in xrange(1, 7))
print value.text

P.S.: No, there's no problem whatsoever with the for approach, except I really enjoy one-liner solutions, and this would fit really nice in my code.

What does `nextSibling` return in the case there is no next sibling? — Silas Ray, Jan 31 '13 at 22:03
@sr2222 you can consider `nextSibling` is always to return a valid value; the issue I'm trying to point rather about the code writing -- working is a mere consequence (: — Rubens, Jan 31 '13 at 22:06
So this isn't a "handle n-length list" problem, it's a "one-liner for specific slice" question. — Silas Ray, Jan 31 '13 at 22:07
@sr2222 I hope the title edit suits better the problem itself. Thanks for pointing it out. — Rubens, Jan 31 '13 at 22:10

score 7 · Accepted Answer · answered Jan 31 '13 at 22:02

7

I have a strong preference for the loop, but you could use reduce:

>>> class Foo(object):
...     def __init__(self):
...        self.count = 0
...     def callme(self):
...        self.count += 1
...        return self
... 
>>> a = Foo()
>>> reduce(lambda x,y:x.callme(),range(7),a)
<__main__.Foo object at 0xec390>
>>> a.count
7

answered Jan 31 '13 at 22:02

mgilson

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whoa, looks very neat; let me try it! – Rubens Jan 31 '13 at 22:04
1

@StoryTeller -- I'm happy to see alternatives to satisfy OP's criteria. As I state up front, I wouldn't use this. I'd use a loop. – mgilson Jan 31 '13 at 22:05
1

@StoryTeller: The OP explicitly said he wants a one-liner, even though he knows it isn't idiomatic. So it's a perfectly reasonable answer. – abarnert Jan 31 '13 at 22:05
1

Wouldn't using map be easier? As in `map(lambda _: a.callme(), range(7))` – l4mpi Jan 31 '13 at 22:05
@l4mpi: No, that will return a list of 7 copies of `a.callme()`, not `a.callme().callme()…callme()`. – abarnert Jan 31 '13 at 22:06
@l4mpi -- Nope. That calls the `callme` method on the same object 7 times -- It was just easist to have my method return `self` so it could chain an infinite number of `.callme().callme().callme()...` – mgilson Jan 31 '13 at 22:06
1

Ah, of course... I overlooked the fact that the returned object was used for successive calls as it doesn't make a difference in this example. – l4mpi Jan 31 '13 at 22:08
Yes well, my point is that it isn't really a *one-liner*. You have to write quite a few lines of preperation. If it's for a single use, I stand by what I said. – StoryTeller - Unslander Monica Jan 31 '13 at 22:10
2

@StoryTeller: What lines of preparation? He's just created a simple example to test his one-liner with. – abarnert Jan 31 '13 at 22:11
1

@StoryTeller the actual solution is just the `reduce` line, everything else is just an example. – l4mpi Jan 31 '13 at 22:12
Never read code examples before youv'e had coffee. My apologies @mgilson – StoryTeller - Unslander Monica Jan 31 '13 at 22:13
Well, among all the very amazing answers, I'm accepting yours, as I'm some what lambdaddicted (: Thank you all for the answers! – Rubens Feb 01 '13 at 00:34

score 4 · Answer 2 · edited May 23 '17 at 12:18

4

You want a one-liner equivalent of this:

for i in xrange(1, 7):
    value = value.nextSibling

This is one line:

for i in xrange(1, 7): value = value.nextSibling

If you're looking for something more functional, what you really want is a compose function, so you can compose callme() (or attrgetter('my_prop'), or whatever) 7 times.

edited May 23 '17 at 12:18

Community

1
1

answered Jan 31 '13 at 22:07

abarnert

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geez, man! didn't know I could do that (: let me try this one too! – Rubens Jan 31 '13 at 22:08
3

As a side note -- You cheated ;-). – mgilson Jan 31 '13 at 22:08
@mgilson: There is no cheating in code golf. (And I upvoted your comment so you'll get a higher score, which counts as losing in golf, right?) – abarnert Jan 31 '13 at 22:10

score 2 · Answer 3 · answered Jan 31 '13 at 22:18

In case of BS you can use nextSiblingGenerator() with itertools.islice to get the nth sibling. It would also handle situations where there is no nth element.

from itertools import islice

nth = 7
next(islice(elem.nextSiblingGenerator(), nth, None), None)

score 2 · Answer 4 · answered Jan 31 '13 at 22:49

2

Disclaimer: eval is evil.

value = eval('value' + ('.nextSibling' * 7))

answered Jan 31 '13 at 22:49

rodrigo

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yep, that's amazingly evil (: – Rubens Jan 31 '13 at 23:05

rodrigo · Answer 5 · 2013-01-31T22:45:31.963

Ah! But reduce is not available in Python3, at least not as a built in.

So here is my try, portable to Python2/3 and based on the OP failed attempt:

[globals().update(value=value.nextSibling) for i in range(7)]

That assumes that value is a global variable. If value happens to be a member variable, then write instead:

[self.__dict__.update(value=value.nextSibling) for i in range(7)]

You cannot use locals() because the list comprehension creates a nested local scope, so the real locals() is not directly available. However, you can capture it with a bit of work:

(lambda loc : [loc.update(x=x.nextSibling) for i in range(7)])(locals())

Or easier if you don't mind duplicating the number of lines:

loc = locals()
[loc.update(value=value.nextSibling) for i in range(7)]

Or if you really fancy one-liners:

loc = locals() ; [loc.update(value=value.nextSibling) for i in range(7)]

Yes, Python can use ; too 8-)

UPDATE:

Another fancy variation, now with map instead of the list comprehension:

list(map(lambda d : d.update(value=value.nextSibling), 7 * [locals()]))

Note the clever use of list multiplication to capture the current locals() and create the initial iterable at the same time.

whoa, man! I'll try this one too. And now I'm quite afraid I don't know which answer to choose ^^ — Rubens, Jan 31 '13 at 22:43
well you can just put `try: from functools import reduce; except: pass` at the top, and then the code is portable with Python 2 and 3 — newacct, Feb 01 '13 at 09:20
I'm pretty sure that this is `locals` abuse -- especially since it's [**not guaranteed**](http://docs.python.org/2/library/functions.html#locals) that modifying locals will modify the current namespace (in which case you'd just end up with a list of `None`, and no mutated `value`) — mgilson, Feb 01 '13 at 14:55

score 1 · Answer 6 · answered Feb 01 '13 at 09:27

1

The most direct way to write it would be:

value = reduce(lambda x, _: x.nextSibling, xrange(1,7), value)

answered Feb 01 '13 at 09:27

newacct

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very nice one, too; I guess this is somehow a shortening on the vars from the accepted answer. Just a question: wouldn't `reduce(lambda x, _: x.nextSibling, xrange(1,7), value)` suffice, as the assignment is already performed in the reduce process? – Rubens Feb 01 '13 at 09:33
@Rubens: what do you mean assignment? How would that be possible? Python is pass by value. A function cannot assign to the caller's variables. – newacct Feb 01 '13 at 09:34

Compressing "n"-time object member call

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