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Doc's are lacking an example...How do you use bisect.insort_left)_ based on a key?

Trying to insert based on key.

bisect.insort_left(data, ('brown', 7))

puts insert at data[0].

From docs...

bisect.insort_left(a, x, lo=0, hi=len(a))

    Insert x in a in sorted order. This is equivalent to a.insert(bisect.bisect_left(a, x, lo, hi), x) assuming that a is already sorted. Keep in mind that the O(log n) search is dominated by the slow O(n) insertion step.

Sample usage:

>>> data = [('red', 5), ('blue', 1), ('yellow', 8), ('black', 0)]
>>> data.sort(key=lambda r: r[1])
>>> keys = [r[1] for r in data]         # precomputed list of keys
>>> data[bisect_left(keys, 0)]
('black', 0)
>>> data[bisect_left(keys, 1)]
('blue', 1)
>>> data[bisect_left(keys, 5)]
('red', 5)
>>> data[bisect_left(keys, 8)]
('yellow', 8)
>>>

I want to put ('brown', 7) after ('red', 5) on sorted list in data using bisect.insort_left. Right now bisect.insort_left(data, ('brown', 7)) puts ('brown', 7) at data[0]...because I am not using the keys to do insert...docs don't show to do inserts using the keys.

martineau
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Merlin
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    What is your question? – BrenBarn Dec 27 '14 at 23:44
  • Be careful with this it's an O(N) operation, check if you really need it first. Have you considered `heapq` or just calling `list.sort` before printing if list isn't sorted – jamylak Dec 27 '14 at 23:47
  • Yes, heapq does not work well for remove a node mid tree. so, this may be best for me – Merlin Dec 27 '14 at 23:50
  • @Merlin What's the general problem you are solving? – jamylak Dec 27 '14 at 23:58
  • I want to put ('brown', 7) after ('red', 5) on sorted list in data using bisect.insort_left. Right now bisect.insort_left(data, ('brown', 7)) puts ('brown', 7) at data[0]...because I am not using the keys to do insert...doc dont show to do inserts using the keys. – Merlin Dec 28 '14 at 00:05
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    The docs suggest using the [`SortedCollection` recipe](http://code.activestate.com/recipes/577197-sortedcollection/) that has support for a key-function, which the `bisect` module doesn't support. The `insert()` method in the recipe's class looks like it might do what you want (given the scaffolding the class provides). – martineau Dec 28 '14 at 00:38
  • @martineau, I am using the word 'key' loosely because that is what the docs refer to column data[1] as. – Merlin Dec 28 '14 at 00:44
  • @martineau, thanks for reference...Now really want to know how to do this.... – Merlin Dec 28 '14 at 00:48
  • You're using 'key' the way it's usually used wrt to sorting (and the way the docs and recipe are using it). For example with `list.sort()` and `sorted()` it's often given as `key=lambda x: x[1]` to allow you to specify what part of each item is the value to sort upon. Don't have the time right now, but I'll see if I can work up an example for you later... – martineau Dec 28 '14 at 00:50

6 Answers6

33

You could wrap your iterable in a class that implements __getitem__ and __len__. This allows you the opportunity to use a key with bisect_left. If you set up your class to take the iterable and a key function as arguments.

To extend this to be usable with insort_left it's required to implement the insert method. The problem here is that if you do that is that insort_left will try to insert your key argument into the list containing the objects of which the the key is a member.

An example is clearer

from bisect import bisect_left, insort_left


class KeyWrapper:
    def __init__(self, iterable, key):
        self.it = iterable
        self.key = key

    def __getitem__(self, i):
        return self.key(self.it[i])

    def __len__(self):
        return len(self.it)

    def insert(self, index, item):
        print('asked to insert %s at index%d' % (item, index))
        self.it.insert(index, {"time":item})

timetable = [{"time": "0150"}, {"time": "0250"}, {"time": "0350"}, {"time": "0450"}, {"time": "0550"}, {"time": "0650"}, {"time": "0750"}]

bslindex = bisect_left(KeyWrapper(timetable, key=lambda t: t["time"]), "0359")

islindex = insort_left(KeyWrapper(timetable, key=lambda t: t["time"]), "0359")

See how in my insert method I had to make it specific to the timetable dictionary otherwise insort_left would try insert "0359" where it should insert {"time": "0359"}?

Ways round this could be to construct a dummy object for the comparison, inherit from KeyWrapper and override insert or pass some sort of factory function to create the object. None of these ways are particularly desirable from an idiomatic python point of view.

So the easiest way is to just use the KeyWrapper with bisect_left, which returns you the insert index and then do the insert yourself. You could easily wrap this in a dedicated function.

e.g.

bslindex = bisect_left(KeyWrapper(timetable, key=lambda t: t["time"]), "0359")
timetable.insert(bslindex, {"time":"0359"})

In this case ensure you don't implement insert, so you will be immediately aware if you accidentally pass a KeyWrapper to a mutating function like insort_left which probably wouldn't do the right thing.

To use your example data

from bisect import bisect_left


class KeyWrapper:
    def __init__(self, iterable, key):
        self.it = iterable
        self.key = key

    def __getitem__(self, i):
        return self.key(self.it[i])

    def __len__(self):
        return len(self.it)

data = [('red', 5), ('blue', 1), ('yellow', 8), ('black', 0)]
data.sort(key=lambda c: c[1])

newcol = ('brown', 7)

bslindex = bisect_left(KeyWrapper(data, key=lambda c: c[1]), newcol[1])
data.insert(bslindex, newcol)

print(data)

Here is the class with proper typing:

from typing import TypeVar, Generic, Sequence, Callable


T = TypeVar('T')
V = TypeVar('V')


class KeyWrapper(Generic[T, V]):
    def __init__(self, iterable: Sequence[T], key: Callable[[T], V]):
        self.it = iterable
        self.key = key

    def __getitem__(self, i: int) -> V:
        return self.key(self.it[i])

    def __len__(self) -> int:
        return len(self.it)

Chris
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Paul Rooney
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    This is excellent and didn't get the love it deserves. It's concise and much more efficient than any other alternative I've seen. If you know that `data` is already sorted in the correct order, there's no need to calculate the key for each element. The whole point of a binary search is to get `O(log n)` instead of `O(n)`. What's the point if you have to calculate the key for each element first? – Eric Duminil Oct 16 '17 at 12:05
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    Second the comment about how excellent this answer is. My use case is a big query result from django with everything sorted by certain fields. I want to break the result into chunks based on when the most significant of the fields change, and all I really need is the indices of when this happens. bisect + the KeyWrapper approach lets me do exactly this very efficiently. – aggieNick02 Aug 15 '19 at 13:54
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    This was fixed in Python 3.10. Added key param to bisect_left, etc. – dilit Nov 20 '21 at 04:33
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    Interesting solution. You basically create a Sequence, may I suggest the class is a sublass of Sequence? – J Agustin Barrachina May 05 '22 at 13:04
  • Thanks I'll take a look at adding it. – Paul Rooney May 05 '22 at 22:48
20

This does essentially the same thing the SortedCollection recipe does that the bisect documentation mentions in its See also: section at the end, but unlike the insert() method in the recipe, the function shown supports a key-function.

What's being done is a separate sorted keys list is maintained in parallel with the sorted data list to improve performance (it's faster than creating the keys list before each insertion, but keeping it around and updating it isn't strictly required). The ActiveState recipe encapsulated this for you within a class, but in the code below they're just two separate independent lists being passed around (so it'd be easier for them to get out of sync than it would be if they were both held in an instance of the recipe's class).

from bisect import bisect_left

def insert(seq, keys, item, keyfunc=lambda v: v):
    """Insert an item into a sorted list using a separate corresponding
       sorted keys list and a keyfunc() to extract the key from each item.

    Based on insert() method in SortedCollection recipe:
    http://code.activestate.com/recipes/577197-sortedcollection/
    """
    k = keyfunc(item)  # Get key.
    i = bisect_left(keys, k)  # Determine where to insert item.
    keys.insert(i, k)  # Insert key of item to keys list.
    seq.insert(i, item)  # Insert the item itself in the corresponding place.

# Initialize the sorted data and keys lists.
data = [('red', 5), ('blue', 1), ('yellow', 8), ('black', 0)]
data.sort(key=lambda r: r[1]) # Sort data by key value
keys = [r[1] for r in data]   # Initialize keys list
print(data)  # -> [('black', 0), ('blue', 1), ('red', 5), ('yellow', 8)]

insert(data, keys, ('brown', 7), keyfunc=lambda x: x[1])
print(data)  # -> [('black', 0), ('blue', 1), ('red', 5), ('brown', 7), ('yellow', 8)]

Follow-on question:
    Can bisect.insort_left be used?

No, you can't simply use the bisect.insort_left() function to do this because it wasn't written in a way that supports a key-function—instead it just compares the whole item passed to it to insert, x, with one of the whole items in the array in its if a[mid] < x: statement. You can see what I mean by looking at the source for the bisect module in Lib/bisect.py.

Here's the relevant excerpt:

def insort_left(a, x, lo=0, hi=None):
    """Insert item x in list a, and keep it sorted assuming a is sorted.

    If x is already in a, insert it to the left of the leftmost x.

    Optional args lo (default 0) and hi (default len(a)) bound the
    slice of a to be searched.
    """

    if lo < 0:
        raise ValueError('lo must be non-negative')
    if hi is None:
        hi = len(a)
    while lo < hi:
        mid = (lo+hi)//2
        if a[mid] < x: lo = mid+1
        else: hi = mid
    a.insert(lo, x)

You could modify the above to accept an optional key-function argument and use it:

def my_insort_left(a, x, lo=0, hi=None, keyfunc=lambda v: v):
    x_key = keyfunc(x)  # Get comparison value.
    . . .
        if keyfunc(a[mid]) < x_key: # Compare key values.
            lo = mid+1
    . . .

...and call it like this:

my_insort_left(data, ('brown', 7), keyfunc=lambda v: v[1])

Actually, if you're going to write a custom function, for the sake of more efficiency at the expense of unneeded generality, you could dispense with the adding of a generic key function argument and just hardcode everything to operate the way needed with the data format you have. This will avoid the overhead of repeated calls to a key-function while doing the insertions.

def my_insort_left(a, x, lo=0, hi=None):
    x_key = x[1]   # Key on second element of each item in sequence.
    . . .
        if a[mid][1] < x_key: lo = mid+1  # Compare second element to key.
    . . .

...called this way without passing keyfunc:

my_insort_left(data, ('brown', 7))
martineau
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  • THANKS, can bisect.insort_left be used? – Merlin Dec 28 '14 at 02:15
  • You could probably use it to insert the key of the new item into the `keys` list, but not the item itself into the `data` list (because it doesn't support a key-function and would use the whole item as the key, and since the item is a tuple it would sort by the string value in it first). – martineau Dec 28 '14 at 02:27
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    Since `data` is a list, and lists are mutable sequences, you can insert items into them at arbitrary indexes using [`data.insert(i, x)`](https://docs.python.org/2/library/stdtypes.html?highlight=sequence#mutable-sequence-types). The code in the `insert()` function in my answer does this twice, once to insert the key value into the `keys` list, and again to also insert the whole item at the same relative position of the `data` list. Why are you so adamant about using `bisect.insort_left()` -- is using it a homework assignment or something? – martineau Dec 28 '14 at 05:16
  • its not homework!!! I just want to avoid sorting long lists containing tuples repeatedly, Hoping this method would work... – Merlin Dec 29 '14 at 01:46
  • In that case you can use my original answer or make a trivial change to the `bisect.insort_left()` source shown in the follow-up I added -- it'd take literally the changing of a couple of lines plus the adding of an argument to the function `def` to make it do what you want. Note that while using either avoids repeatedly sorting the list, it's still expensive, [O(n)](http://en.wikipedia.org/wiki/Search_data_structure#Asymptotic_amortized_worst-case_analysis), to insert elements into sorted arrays. – martineau Dec 29 '14 at 08:56
  • It's perfectly possible to use `bisect.insort_left`. See the other [answer](https://stackoverflow.com/a/39501468/6419007). – Eric Duminil Oct 16 '17 at 12:00
10

Add comparison methods to your class

Sometimes this is the least painful way, especially if you already have a class and just want to sort by a key from it:

#!/usr/bin/env python3

import bisect
import functools

@functools.total_ordering
class MyData:
    def __init__(self, color, number):
        self.color = color
        self.number = number
    def __lt__(self, other):
        return self.number < other.number
    def __str__(self):
        return '{} {}'.format(self.color, self.number)

mydatas = [
    MyData('red', 5),
    MyData('blue', 1),
    MyData('yellow', 8),
    MyData('black', 0),
]
mydatas_sorted = []
for mydata in mydatas:
    bisect.insort(mydatas_sorted, mydata)
for mydata in mydatas_sorted:
    print(mydata)

Output:

black 0
blue 1
red 5
yellow 8

See also: "Enabling" comparison for classes

Tested in Python 3.5.2.

Upstream requests/patches

I get the feeling this is going to happen sooner or later ;-)

Ciro Santilli OurBigBook.com
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  • This makes a lot of sense and involves a lot less messing around. It seems like the way to go unless you can't or wont change the class. – Paul Rooney Jun 17 '19 at 02:49
9

As of Python 3.10, all the binary search helpers in the bisect module now accept a key argument:

key specifies a key function of one argument that is used to extract a comparison key from each input element. The default value is None (compare the elements directly).

Therefore, you can pass the same function you used to sort the data:

>>> import bisect
>>> data = [('red', 5), ('blue', 1), ('yellow', 8), ('black', 0)]
>>> data.sort(key=lambda r: r[1])
>>> data
[('black', 0), ('blue', 1), ('red', 5), ('yellow', 8)]
>>> bisect.insort_left(data, ('brown', 7), key=lambda r: r[1])
>>> data
[('black', 0), ('blue', 1), ('red', 5), ('brown', 7), ('yellow', 8)]
Eugene Yarmash
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7

If your goal is to mantain a list sorted by key, performing usual operations like bisect insert, delete and update, I think sortedcontainers should suit your needs as well, and you'll avoid O(n) inserts.

mmj
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    Specific to this question: sortedcontainers.SortedList includes [bisect_key* methods](http://www.grantjenks.com/docs/sortedcontainers/sortedlistwithkey.html#sortedcontainers.SortedListWithKey.L.bisect_key) – GrantJ Jun 30 '16 at 03:09
  • @GrantJ link from comment is dead – Xeverous Nov 01 '22 at 15:49
  • Updated link, sortedcontainers.SortedKeyList includes bisect_key_left and right: https://grantjenks.com/docs/sortedcontainers/sortedlist.html#sortedcontainers.SortedKeyList.bisect_key_left – GrantJ Nov 03 '22 at 01:09
3

From python version 3.10, the key argument has been added.

It will be something like:

import bisect
bisect.bisect_left(('brown', 7), data, key=lambda r: r[1])

Sources:

J Agustin Barrachina
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