Heap#

La structure heap ou tas en français est utilisée pour trier. Elle peut également servir à obtenir les k premiers éléments d’une liste.

Un tas est peut être considéré comme un tableau $T$ qui vérifie une condition assez simple, pour tout indice $i$ , alors $T[i] \geqslant \max(T[2i+1], T[2i+2])$ . On en déduit nécessairement que le premier élément du tableau est le plus grand. Maintenant comment transformer un tableau en un autre qui respecte cette contrainte ?

[2]:

%matplotlib inline

Transformer en tas#

[3]:

def swap(tab, i, j):
    "Echange deux éléments."
    tab[i], tab[j] = tab[j], tab[i]


def entas(heap):
    "Organise un ensemble selon un tas."
    modif = 1
    while modif > 0:
        modif = 0
        i = len(heap) - 1
        while i > 0:
            root = (i - 1) // 2
            if heap[root] < heap[i]:
                swap(heap, root, i)
                modif += 1
            i -= 1
    return heap


ens = [1, 2, 3, 4, 7, 10, 5, 6, 11, 12, 3]
entas(ens)

[3]:

[12, 11, 5, 10, 7, 3, 1, 6, 4, 3, 2]

Comme ce n’est pas facile de vérifer que c’est un tas, on le dessine.

Dessiner un tas#

[4]:

from teachpyx.tools import draw_diagram


def dessine_tas(heap):
    rows = ["blockdiag {"]
    for i, v in enumerate(heap):
        if i * 2 + 1 < len(heap):
            rows.append(
                '"[{}]={}" -> "[{}]={}";'.format(i, heap[i], i * 2 + 1, heap[i * 2 + 1])
            )
            if i * 2 + 2 < len(heap):
                rows.append(
                    '"[{}]={}" -> "[{}]={}";'.format(
                        i, heap[i], i * 2 + 2, heap[i * 2 + 2]
                    )
                )
    rows.append("}")
    return draw_diagram("\n".join(rows))


ens = [1, 2, 3, 4, 7, 10, 5, 6, 11, 12, 3]
dessine_tas(entas(ens))

[4]:

../../_images/practice_py-base_nbheap_7_0.png

Le nombre entre crochets est la position, l’autre nombre est la valeur à cette position. Cette représentation fait apparaître une structure d’arbre binaire.

Première version#

[5]:

def swap(tab, i, j):
    "Echange deux éléments."
    tab[i], tab[j] = tab[j], tab[i]


def _heapify_max_bottom(heap):
    "Organise un ensemble selon un tas."
    modif = 1
    while modif > 0:
        modif = 0
        i = len(heap) - 1
        while i > 0:
            root = (i - 1) // 2
            if heap[root] < heap[i]:
                swap(heap, root, i)
                modif += 1
            i -= 1


def _heapify_max_up(heap):
    "Organise un ensemble selon un tas."
    i = 0
    while True:
        left = 2 * i + 1
        right = left + 1
        if right < len(heap):
            if heap[left] > heap[i] >= heap[right]:
                swap(heap, i, left)
                i = left
            elif heap[right] > heap[i]:
                swap(heap, i, right)
                i = right
            else:
                break
        elif left < len(heap) and heap[left] > heap[i]:
            swap(heap, i, left)
            i = left
        else:
            break


def topk_min(ens, k):
    "Retourne les k plus petits éléments d'un ensemble."

    heap = ens[:k]
    _heapify_max_bottom(heap)

    for el in ens[k:]:
        if el < heap[0]:
            heap[0] = el
            _heapify_max_up(heap)
    return heap


ens = [1, 2, 3, 4, 7, 10, 5, 6, 11, 12, 3]
for k in range(1, len(ens) - 1):
    print(k, topk_min(ens, k))

1 [1]
2 [2, 1]
3 [3, 2, 1]
4 [3, 3, 1, 2]
5 [4, 3, 1, 3, 2]
6 [5, 4, 3, 3, 2, 1]
7 [5, 6, 3, 4, 2, 3, 1]
8 [5, 7, 3, 6, 2, 3, 1, 4]
9 [5, 10, 3, 7, 2, 3, 1, 6, 4]

Même chose avec les indices au lieu des valeurs#

[6]:

def _heapify_max_bottom_position(ens, pos):
    "Organise un ensemble selon un tas."
    modif = 1
    while modif > 0:
        modif = 0
        i = len(pos) - 1
        while i > 0:
            root = (i - 1) // 2
            if ens[pos[root]] < ens[pos[i]]:
                swap(pos, root, i)
                modif += 1
            i -= 1


def _heapify_max_up_position(ens, pos):
    "Organise un ensemble selon un tas."
    i = 0
    while True:
        left = 2 * i + 1
        right = left + 1
        if right < len(pos):
            if ens[pos[left]] > ens[pos[i]] >= ens[pos[right]]:
                swap(pos, i, left)
                i = left
            elif ens[pos[right]] > ens[pos[i]]:
                swap(pos, i, right)
                i = right
            else:
                break
        elif left < len(pos) and ens[pos[left]] > ens[pos[i]]:
            swap(pos, i, left)
            i = left
        else:
            break


def topk_min_position(ens, k):
    "Retourne les positions des k plus petits éléments d'un ensemble."

    pos = list(range(k))
    _heapify_max_bottom_position(ens, pos)

    for i, el in enumerate(ens[k:]):
        if el < ens[pos[0]]:
            pos[0] = k + i
            _heapify_max_up_position(ens, pos)
    return pos


ens = [1, 2, 3, 7, 10, 4, 5, 6, 11, 12, 3]
for k in range(1, len(ens) - 1):
    pos = topk_min_position(ens, k)
    print(k, pos, [ens[i] for i in pos])

1 [0] [1]
2 [1, 0] [2, 1]
3 [2, 1, 0] [3, 2, 1]
4 [10, 2, 0, 1] [3, 3, 1, 2]
5 [5, 10, 0, 2, 1] [4, 3, 1, 3, 2]
6 [6, 5, 2, 10, 1, 0] [5, 4, 3, 3, 2, 1]
7 [5, 7, 10, 6, 1, 2, 0] [4, 6, 3, 5, 2, 3, 1]
8 [5, 3, 10, 7, 1, 2, 0, 6] [4, 7, 3, 6, 2, 3, 1, 5]
9 [5, 4, 10, 3, 1, 2, 0, 7, 6] [4, 10, 3, 7, 2, 3, 1, 6, 5]

[7]:

import numpy.random as rnd

X = rnd.randn(10000)

%timeit topk_min(X, 20)

3.87 ms ± 553 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

[8]:

%timeit topk_min_position(X, 20)

3.4 ms ± 372 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

Coût de l’algorithme#

[9]:

from tqdm import tqdm
from pandas import DataFrame
from teachpyx.ext_test_case import measure_time

rows = []
for n in tqdm(list(range(1000, 20001, 1000))):
    X = rnd.randn(n)
    res = measure_time(
        "topk_min_position(X, 100)",
        {"X": X, "topk_min_position": topk_min_position},
        div_by_number=True,
        number=10,
    )
    res["size"] = n
    rows.append(res)

df = DataFrame(rows)
df.head()

100%|██████████| 20/20 [00:09<00:00,  2.01it/s]

[9]:

	average	deviation	min_exec	max_exec	repeat	number	ttime	context_size	warmup_time	size
0	0.001400	0.000271	0.001090	0.001892	10	10	0.014000	232	0.001138	1000
1	0.002110	0.000517	0.001622	0.003214	10	10	0.021104	232	0.001633	2000
2	0.001967	0.000299	0.001747	0.002545	10	10	0.019669	232	0.002969	3000
3	0.002692	0.000611	0.002081	0.003779	10	10	0.026925	232	0.003983	4000
4	0.003096	0.000276	0.002660	0.003578	10	10	0.030960	232	0.003305	5000

[10]:

import matplotlib.pyplot as plt

df[["size", "average"]].set_index("size").plot()
plt.title("Coût topk en fonction de la taille du tableau");

../../_images/practice_py-base_nbheap_17_0.png

A peu près linéaire comme attendu.

[11]:

from tqdm import tqdm
from teachpyx.ext_test_case import measure_time

rows = []
X = rnd.randn(10000)
for k in tqdm(list(range(500, 2001, 150))):
    res = measure_time(
        "topk_min_position(X, k)",
        {"X": X, "topk_min_position": topk_min_position, "k": k},
        div_by_number=True,
        number=5,
    )
    res["k"] = k
    rows.append(res)

df = DataFrame(rows)
df.head()

100%|██████████| 11/11 [00:08<00:00,  1.33it/s]

[11]:

	average	deviation	min_exec	max_exec	repeat	number	ttime	context_size	warmup_time	k
0	0.008493	0.001842	0.006837	0.013038	10	5	0.084926	232	0.014053	500
1	0.009114	0.002111	0.007600	0.013480	10	5	0.091137	232	0.022466	650
2	0.011651	0.003198	0.008757	0.017215	10	5	0.116511	232	0.010779	800
3	0.011374	0.001973	0.009725	0.016044	10	5	0.113739	232	0.012545	950
4	0.016104	0.003647	0.011865	0.022728	10	5	0.161042	232	0.018794	1100

[12]:

df[["k", "average"]].set_index("k").plot()
plt.title("Coût topk en fonction de k");

../../_images/practice_py-base_nbheap_20_0.png

Pas évident, au pire en $O(n\ln n)$ , au mieux en $O(n)$ .

Version simplifiée#

A-t-on vraiment besoin de _heapify_max_bottom_position ?

[13]:

def _heapify_max_up_position_simple(ens, pos, first):
    "Organise un ensemble selon un tas."
    i = first
    while True:
        left = 2 * i + 1
        right = left + 1
        if right < len(pos):
            if ens[pos[left]] > ens[pos[i]] >= ens[pos[right]]:
                swap(pos, i, left)
                i = left
            elif ens[pos[right]] > ens[pos[i]]:
                swap(pos, i, right)
                i = right
            else:
                break
        elif left < len(pos) and ens[pos[left]] > ens[pos[i]]:
            swap(pos, i, left)
            i = left
        else:
            break


def topk_min_position_simple(ens, k):
    "Retourne les positions des k plus petits éléments d'un ensemble."

    pos = list(range(k))
    pos[k - 1] = 0

    for i in range(1, k):
        pos[k - i - 1] = i
        _heapify_max_up_position_simple(ens, pos, k - i - 1)

    for i, el in enumerate(ens[k:]):
        if el < ens[pos[0]]:
            pos[0] = k + i
            _heapify_max_up_position_simple(ens, pos, 0)
    return pos


ens = [1, 2, 3, 7, 10, 4, 5, 6, 11, 12, 3]
for k in range(1, len(ens) - 1):
    pos = topk_min_position_simple(ens, k)
    print(k, pos, [ens[i] for i in pos])

1 [0] [1]
2 [1, 0] [2, 1]
3 [2, 1, 0] [3, 2, 1]
4 [10, 2, 1, 0] [3, 3, 2, 1]
5 [5, 10, 2, 1, 0] [4, 3, 3, 2, 1]
6 [5, 6, 10, 2, 1, 0] [4, 5, 3, 3, 2, 1]
7 [6, 7, 10, 5, 2, 1, 0] [5, 6, 3, 4, 3, 2, 1]
8 [5, 4, 10, 7, 6, 2, 1, 0] [4, 10, 3, 6, 5, 3, 2, 1]
9 [3, 4, 6, 5, 7, 10, 2, 1, 0] [7, 10, 5, 4, 6, 3, 3, 2, 1]

[14]:

X = rnd.randn(10000)

%timeit topk_min_position_simple(X, 20)

3.79 ms ± 1.13 ms per loop (mean ± std. dev. of 7 runs, 100 loops each)

[ ]:

[ ]:

Notebook on github