在匹配元素处划分列表的 Pythonic 和有效方法是什么？

Question

This is very similar to Python: split a list based on a condition? and also https://nedbatchelder.com/blog/201306/filter_a_list_into_two_parts.html but instead of partitioning the individual elements into two lists based on a predicate, I want to divide the list into two parts at the first element that fails the predicate.

>>> divide_list(lambda x: x < 7, list(range(10)))
([0, 1, 2, 3, 4, 5, 6], [7, 8, 9])

>>> divide_list(lambda x: x < 7, [1, 3, 5, 7, 9, 5])
([1, 3, 5], [7, 9, 5])

>>> divide_list(lambda x: x < 7, [7, 9, 5])
([], [7, 9, 5])

>>> divide_list(lambda x: x < 7, [1, 3, 5])
([1, 3, 5], [])

>>> divide_list(lambda x: x['a'], [{'a': True, 'b': 1}, {'a': True}, {'a': False}])
([{'a': True, 'b': 1}, {'a': True}], [{'a': False}])

Things to note:

• the input list may not be sorted
• the input list may contain duplicate elements
• ideally we don't want to evaluate the condition multiple times (for each element, if the value is duplicated then that's ok)
• ideally it would accept an iterator as input (ie can only do a single pass over the input data)
• returning iterators is acceptable

Answer 1

I think that the naive implementation is probably best unless you actually need iterators as outputs. This could be useful if your input stream is an iterator and you don't have enough memory to materialize the whole thing at once, etc.

In that case, I think that itertools is great. My initial gut instinct was to do something like:

# broken  :-(
def divide_iter(pred, lst):
    i = iter(lst)
    yield itertools.takewhile(lst, pred)
    yield i

Unfortunately this doesn't work for a variety of reasons. Most notably, it drops an element. Even if it didn't, you could run into problems if you didn't consume the entire takewhile iterable before moving on to the next list. I think that this second problem is going to be an issue that we run into when working with iterators in general, so that's kind of a bummer, but it's the price we pay for processing things element-by-element rather than materializing entire lists at once.

Instead, let's think about grouping the items based on whether the predicate has returned true yet. Then groupby becomes a lot more appealing -- the only thing is that we need to keep track of whether the predicate has returned True. Stateful functions are not much fun so instead, we can use a class and pass a bound method as the key argument to groupby :

import itertools

class _FoundTracker(object):
    def __init__(self, predicate):
        self.predicate = predicate
        self._found = False

    def check_found(self, value):
        if self._found:
            return True
        else:
           self._found = self.predicate(value)
           return self._found

def split_iterable(iterable, predicate):
    tracker = _FoundTracker(predicate)
    for i, (k, group) in enumerate(itertools.groupby(iterable, key=tracker.check_found)):
        yield group
    if i == 0:
        yield iter(())

if __name__ == '__main__':
    for group in split_iterable(xrange(10), lambda x: x < 5):
        print(list(group))

This also has some possibly funky behavior... To demonstrate, consider:

g1, g2 = split_iterable(xrange(10), lambda x: x > 5)
print(list(g1))
print(list(g2))

You'll see that you get some really weird behavior :-). Alternatively:

g1, g2 = map(list, split_iterable(range(10), lambda x: x > 5))
print(g1)
print(g2)

should work fine.

Answer 2

A naive implementation to get things rolling:

def divide_list(pred, lst):
    before, after = [], []
    found = False
    for item in lst:
        if not found:
            if pred(item):
                before.append(item)
            else:
                found = True
        if found:
            after.append(item)
    return before, after

Answer 3

Here's my relatively efficient attempt:

from collections import Hashable

def divide_list(pred, list):
    # The predicate may be expensive, so we can
    # store elements that have already been checked
    # in a set for fast verification.
    elements_checked = set()

    # Assuming that every element of the list is of
    # the same type and the list is nonempty, we can
    # store a flag to check if an element is hashable.
    hashable = isinstance(list[0], Hashable)

    for index, element in enumerate(list):
        if hashable and element in elements_checked:
            continue

        if not pred(element):
            return list[:index], list[index:]

        if hashable:
            elements_checked.add(element)

    return list, []

If you were to benchmark this against the other answers, I reckon this will be the fastest.

I love this question by the way!

Answer 4

This is basically your naive attempt, but doesn't use a separate Boolean flag to determine when the predicate fails; it just uses a reference to first one list, then the other, to do the appending.

def divide_list(pred, lst):
     a, b = [], []
     curr = a
     for x in lst:
         if curr is a and not pred(x):
             curr = b
         curr.append(x)
     return a, b

Answer 5

Why complicated if simple possible? Already mentioned but for in my eyes not understandable reasons dropped from consideration: usage of itertools takewhile .

The code below passes all assertion tests and the function itself needs three lines of code:

from itertools import takewhile
def divide_list(pred, lstL):
    header  = list(takewhile(pred, lstL))
    trailer = lstL[len(header):]
    return header, trailer


assert divide_list(lambda x: x < 7, list(range(10))) == ([0, 1, 2, 3, 4, 5, 6], [7, 8, 9])
assert divide_list(lambda x: x < 7, [1, 3, 5, 7, 9, 5]) == ([1, 3, 5], [7, 9, 5])
assert divide_list(lambda x: x < 7, [7, 9, 5]) == ([], [7, 9, 5])
assert divide_list(lambda x: x < 7, [1, 3, 5]) == ([1, 3, 5], [])
assert divide_list(lambda x: x['a'], [{'a': True, 'b': 1}, {'a': True}, {'a': False}]) == ([{'a': True, 'b': 1}, {'a': True}], [{'a': False}])