Python のリストの内部実装は配列であり、これは線形リストです。リスト内の要素を検索するには、時間計算量が O(n) の list.index() メソッドを使用できます。大量のデータの場合、二分探索を最適化に使用できます。二分探索ではオブジェクトを順序付けする必要があり、その基本原則は次のとおりです:
1. 配列の中央の要素が検索対象の要素である場合、検索プロセスは終了します。
2. 特定の要素が中央の要素より大きいか小さい場合、中央の要素より大きいか小さい配列の半分を検索し、以前と同様に中央の要素から比較を開始します。 3. あるステップで配列が空の場合、それは見つからないことを意味します。 二分探索もアルゴリズムの比較ごとに探索範囲が半分に減り、時間計算量はO(logn)になります。 再帰とループをそれぞれ使用して二分探索を実装します:def binary_search_recursion(lst, value, low, high): if high < low: return None mid = (low + high) / 2 if lst[mid] > value: return binary_search_recursion(lst, value, low, mid-1) elif lst[mid] < value: return binary_search_recursion(lst, value, mid+1, high) else: return mid def binary_search_loop(lst,value): low, high = 0, len(lst)-1 while low <= high: mid = (low + high) / 2 if lst[mid] < value: low = mid + 1 elif lst[mid] > value: high = mid - 1 else: return mid return None
if __name__ == "__main__": import random lst = [random.randint(0, 10000) for _ in xrange(100000)] lst.sort() def test_recursion(): binary_search_recursion(lst, 999, 0, len(lst)-1) def test_loop(): binary_search_loop(lst, 999) import timeit t1 = timeit.Timer("test_recursion()", setup="from __main__ import test_recursion") t2 = timeit.Timer("test_loop()", setup="from __main__ import test_loop") print "Recursion:", t1.timeit() print "Loop:", t2.timeit()
Recursion: 3.12596702576 Loop: 2.08254289627
import bisect import random random.seed(1) print'New Pos Contents' print'--- --- --------' l = [] for i in range(1, 15): r = random.randint(1, 100) position = bisect.bisect(l, r) bisect.insort(l, r) print'%3d %3d' % (r, position), l
New Pos Contents --- --- -------- 14 0 [14] 85 1 [14, 85] 77 1 [14, 77, 85] 26 1 [14, 26, 77, 85] 50 2 [14, 26, 50, 77, 85] 45 2 [14, 26, 45, 50, 77, 85] 66 4 [14, 26, 45, 50, 66, 77, 85] 79 6 [14, 26, 45, 50, 66, 77, 79, 85] 10 0 [10, 14, 26, 45, 50, 66, 77, 79, 85] 3 0 [3, 10, 14, 26, 45, 50, 66, 77, 79, 85] 84 9 [3, 10, 14, 26, 45, 50, 66, 77, 79, 84, 85] 44 4 [3, 10, 14, 26, 44, 45, 50, 66, 77, 79, 84, 85] 77 9 [3, 10, 14, 26, 44, 45, 50, 66, 77, 77, 79, 84, 85] 1 0 [1, 3, 10, 14, 26, 44, 45, 50, 66, 77, 77, 79, 84, 85]
def grade(score,breakpoints=[60, 70, 80, 90], grades='FDCBA'): i = bisect.bisect(breakpoints, score) return grades[i] print [grade(score) for score in [33, 99, 77, 70, 89, 90, 100]]
['F', 'A', 'C', 'C', 'B', 'A', 'A']
def binary_search_bisect(lst, x): from bisect import bisect_left i = bisect_left(lst, x) if i != len(lst) and lst[i] == x: return i return None
Recursion: 4.00940990448 Loop: 2.6583480835 Bisect: 1.74922895432
>>> import numpy as np >>> from bisect import bisect_left, bisect_right >>> data = [2, 4, 7, 9] >>> bisect_left(data, 4) 1 >>> np.searchsorted(data, 4) 1 >>> bisect_right(data, 4) 2 >>> np.searchsorted(data, 4, side='right') 2
In [20]: %timeit -n 100 bisect_left(data, 99999) 100 loops, best of 3: 670 ns per loop In [21]: %timeit -n 100 np.searchsorted(data, 99999) 100 loops, best of 3: 56.9 ms per loop In [22]: %timeit -n 100 bisect_left(data, 8888) 100 loops, best of 3: 961 ns per loop In [23]: %timeit -n 100 np.searchsorted(data, 8888) 100 loops, best of 3: 57.6 ms per loop In [24]: %timeit -n 100 bisect_left(data, 777777) 100 loops, best of 3: 670 ns per loop In [25]: %timeit -n 100 np.searchsorted(data, 777777) 100 loops, best of 3: 58.4 ms per loop
In [30]: data_ndarray = np.arange(0, 1000000) In [31]: %timeit np.searchsorted(data_ndarray, 99999) The slowest run took 16.04 times longer than the fastest. This could mean that an intermediate result is being cached. 1000000 loops, best of 3: 996 ns per loop In [32]: %timeit np.searchsorted(data_ndarray, 8888) The slowest run took 18.22 times longer than the fastest. This could mean that an intermediate result is being cached. 1000000 loops, best of 3: 994 ns per loop In [33]: %timeit np.searchsorted(data_ndarray, 777777) The slowest run took 31.32 times longer than the fastest. This could mean that an intermediate result is being cached. 1000000 loops, best of 3: 990 ns per loop
>>> np.searchsorted([1,2,3,4,5], 3) 2 >>> np.searchsorted([1,2,3,4,5], 3, side='right') 3 >>> np.searchsorted([1,2,3,4,5], [-10, 10, 2, 3]) array([0, 5, 1, 2])