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I am just testing out a few small functions for learning purposes:

def peaks(iterable):
    # returns a list of int for those values in the iterable that are bigger
    # than the value preceding and following them.
    itr = iter(iterable)
    peak =[]
    curr = next(itr)
    last = next(itr)
    try:
        while True:
            first = curr
            curr = last
            last = next(itr)
            if curr > first and curr > last:
                peak.append(curr)
    except:
        pass    
    return peak



def compress(v_iterable,b_iterable):
    #takes two iterables as parameters: it produces every value from the first iterable that
    # has its equivalent position in the second iterable representing what Python would consider
    # a True value. Terminate the iteration when either iterable terminates
    mega_it = dict(zip(v_iterable, b_iterable))
    for nxt in sorted(mega_it):
        if mega_it[nxt]:
            yield nxt


def skipper(iterable,n=0):
    #akes one iterable and one int (whose default value is 0) as parameters: it produces values from
    # the iterable, starting at the first one, skipping the number of values specified by the int
    #(skipping 0 means producing every value from the iterable)
    itr = iter(iterable)
    yield next(itr)
    while True:
        for i in range(n):
            skipped = next(itr)
        yield next(itr)

I feel like my codes are lengthy for the kind of work it does. Is there a way to make my functions cleaner or smaller?

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  • \$\begingroup\$ Could you please split this question into several questions? \$\endgroup\$ – 200_success Feb 2 '15 at 20:55
  • \$\begingroup\$ @200_success - I split the question into functions and classes. This one focuses on functions. \$\endgroup\$ – LucyBen Feb 2 '15 at 23:20
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Your comments are good, but they should all be written as docstrings instead.

When working with iterators, itertools and generator expressions are your friend. peaks() works with three iterators, so zip() is useful.

Since the function accepts an iterable, it seems fitting that it should be a generator rather than returning a list.

Python supports double-ended inequalities.

from itertools import tee

def peaks(iterable):
    """Generates values in the iterable that are greater than the value
    preceding and following them."""
    before, this, after = tee(iter(iterable), 3)
    next(this)
    next(after)
    next(after)
    for prev, curr, succ in zip(before, this, after):
        if prev < curr > succ:
            yield curr

If you wanted to keep the old interface, which returns a list, then a list comprehension would be a good way to write it.

def peaks(iterable):
    """Returns a list of values in the iterable that are greater than the
    preceding and following value."""
    xs, ys, zs = tee(iter(iterable), 3)
    next(ys)
    next(zs)
    next(zs)
    return [y for (x, y, z) in zip(xs, ys, zs) if x < y > z]

I find your compress() weird, in that it sorts the results.

You can implement the function using just a generator expression.

def compress(values, incl_excl):
    """Keeping only values whose corresponding entry in incl_excl is true,
    generates the sorted filtered list."""
    yield from sorted(val for val, include in zip(values, incl_excl) if include)

In skipper(), the two occurrences of yield next(itr) should be combined.

def skipper(iterable, n=0):
    """Generator that yields the first element, then skips n values between
    results, until the input is exhausted."""
    iterable = iter(iterable)
    while True:
        yield next(iterable)
        for _ in range(n):
            next(iterable)
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  • \$\begingroup\$ @ 200_success - sorry this might seem unrelated but when I make the changes to the function, and try to do print(peaks([0,1,-1,3,8,4,3,5,4,3,8])) it keeps giving me the memory location <generator object peaks at 0x0000000002935558> instead of [1, 8, 5]. \$\endgroup\$ – LucyBen Feb 3 '15 at 2:09
  • 1
    \$\begingroup\$ That's what I meant by "it should be a generator rather than returning a list." If you want to display a list, then write print(list(peaks([0,1,-1,…]))). \$\endgroup\$ – 200_success Feb 3 '15 at 2:13
  • \$\begingroup\$ Thanks! Is it possible to make the function cleaner without making it a generator? I ask as the program should pass a pre-designed batch test which might just have print(peaks([0,1,-1,3,8,4,3,5,4,3,8])) written. \$\endgroup\$ – LucyBen Feb 3 '15 at 2:41
  • \$\begingroup\$ Sure. I'd use a list comprehension. \$\endgroup\$ – 200_success Feb 3 '15 at 3:11
3
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skipper's functionality is covered by itertools.islice:

def skipper(iterable,n=0):
    return itertools.islice(iterable, 0, None, n+1)
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Just wanted to note that peaks could've been implemented with just two iterators. Somewhere along these lines:

def peaks(iterable):
    state = False
    for previous, following in zip(iterable, iterable[1:]):
        if state and previous > following:
            yield previous
        state = previous < following

Of course it's up to you to replace iterable[1:] with an iterator.

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0
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Code presented by 200_success avoids the problem, but I think it should be pointed out that you are ignoring all errors. A catch-all is never a good idea, it hides problems. You should use:

except StopIteration:
    pass    

Variables that are assigned to but never used are not useful. Variable _ is often considered an anonymous variable, good for replacing i. And skipped can be just skipped (pun).

for i in range(n):
    skipped = next(itr)

for _ in range(n):
    next(itr)

Alternative implementation of skipper. I would recommend asserting (checking) if arguments are sane, negative n would lead to weird unexpected results.

def skipper(iterable, n=0):
    assert n >= 0
    return (a for i,a in enumerate(iterable) if i%(n+1) == 0)
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