我有一个集合列表,我希望对 n 个不同的样本进行采样,每个样本包含每个集合中的一个项目。 我不想要的是按顺序排列,因此,例如,我将从第一组中获得必须具有相同项目的所有样本。我也不想创建所有笛卡尔积,因为这在效率方面可能是不可能的...... 知道如何做吗?或者甚至是近似这种行为的东西?
不起作用的示例:
(prod for i, prod in zip(range(n), itertools.product(*list_of_sets)))
上述所有解决方案在迭代结束时都浪费了大量资源来过滤重复结果。这就是为什么我想到了一种从开始到结束具有(几乎)线性速度的方法。
这个想法是:(仅在你的头脑中)给标准阶笛卡尔积的每个结果一个索引。例如,对于
A
xB
xC
和 2000
x1
x2
= 4000
元素:
0: (A[0], B[0], C[0])
1: (A[1], B[0], C[0])
...
1999: (A[1999], B[0], C[0])
2000: (A[0], B[0], C[1])
...
3999: (A[1999], B[0], C[1])
done.
所以还有一些问题需要解决:
2000*1*2=4000
,下面的每个数字都是有效的索引。n
,只需使用random.sample(xrange(numer_of_indices), n)
。但是,如果您还不知道样本大小(更一般的情况),则必须动态生成索引以免浪费内存。在这种情况下,您只需使用 index = random.randint(0, k - 1)
生成 k = numer_of_indices
即可获取第一个索引,使用 k = number_of_indices - n
获取第 n
结果。只需检查我下面的代码(请注意,我在那里使用单边链表来存储已完成的索引。它使插入操作为 O(1) 操作,我们在这里需要大量插入)。i
。那么 i % 2000
将是结果的 A
的索引。现在 i // 2000
可以递归地处理为剩余因子的笛卡尔积的索引。这就是我想出的代码:
def random_order_cartesian_product(*factors):
amount = functools.reduce(lambda prod, factor: prod * len(factor), factors, 1)
index_linked_list = [None, None]
for max_index in reversed(range(amount)):
index = random.randint(0, max_index)
index_link = index_linked_list
while index_link[1] is not None and index_link[1][0] <= index:
index += 1
index_link = index_link[1]
index_link[1] = [index, index_link[1]]
items = []
for factor in factors:
items.append(factor[index % len(factor)])
index //= len(factor)
yield items
以下生成器函数生成非重复样本。仅当生成的样本数量远小于可能样本的数量时,它才能高效工作。它还要求集合的元素是可散列的:
def samples(list_of_sets):
list_of_lists = list(map(list, list_of_sets)) # choice only works on sequences
seen = set() # keep track of seen samples
while True:
x = tuple(map(random.choice, list_of_lists)) # tuple is hashable
if x not in seen:
seen.add(x)
yield x
>>> lst = [{'b', 'a'}, {'c', 'd'}, {'f', 'e'}, {'g', 'h'}]
>>> gen = samples(lst)
>>> next(gen)
('b', 'c', 'f', 'g')
>>> next(gen)
('a', 'c', 'e', 'g')
>>> next(gen)
('b', 'd', 'f', 'h')
>>> next(gen)
('a', 'c', 'f', 'g')
您可以使用
sample
库中的 random
:
import random
[[random.sample(x,1)[0] for x in list_of_sets] for _ in range(n)]
例如:
list_of_sets = [{1,2,3}, {4,5,6}, {1,4,7}]
n = 3
可能的输出是:
[[2, 4, 7], [1, 4, 7], [1, 6, 1]]
编辑:
如果我们想避免重复,我们可以使用
while
循环并将结果收集到 set
。此外,您可以检查 n
是否有效,并返回无效 n
值的笛卡尔积:
chosen = set()
if 0 < n < reduce(lambda a,b: a*b,[len(x) for x in list_of_sets]):
while len(chosen) < n:
chosen.add(tuple([random.sample(x,1)[0] for x in list_of_sets]))
else:
chosen = itertools.product(*list_of_sets)
这是一个完整的版本,带有示例和一些修改,以便于理解和使用:
import functools
import random
def random_order_cartesian_product(factors):
amount = functools.reduce(lambda prod, factor: prod * len(factor), factors, 1)
print(amount)
print(len(factors[0]))
index_linked_list = [None, None]
for max_index in reversed(range(amount)):
index = random.randint(0, max_index)
index_link = index_linked_list
while index_link[1] is not None and index_link[1][0] <= index:
index += 1
index_link = index_link[1]
index_link[1] = [index, index_link[1]]
items = []
for factor in factors:
items.append(factor[index % len(factor)])
index //= len(factor)
yield items
factors=[
[1,2,3],
[4,5,6],
[7,8,9]
]
n = 5
all = random_order_cartesian_product(factors)
count = 0
for comb in all:
print(comb)
count += 1
if count == n:
break
我们使用素数及其一个模 n 的原根创建一个序列,该序列在一个区间内访问每个数字一次。 更具体地说,我们正在寻找 整数乘法群模 n 的生成器。
我们必须选择比乘积稍大的素数prod([len(i) for i in iterables)]
,因此我们必须考虑到索引错误的情况。
import random
from math import gcd
import math
from math import prod
from typing import Iterable
def next_prime(number):
if number < 0:
raise ValueError('Negative numbers can not be primes')
if number <= 1:
return 2
if number % 2 == 0:
number -= 1
while True:
number += 2
max_check = int(math.sqrt(number)) + 2
for divider in range(3, max_check, 2):
if number % divider == 0:
break
else:
return number
def is_primitive_root(a, n):
phi = n - 1
factors = set()
for i in range(2, int(phi ** 0.5) + 1):
if phi % i == 0:
factors.add(i)
factors.add(phi // i)
for factor in factors:
if pow(a, factor, n) == 1:
return False
return True
def find_random_primitive_root(n):
while True:
a = random.randint(2, n - 1)
if gcd(a, n) == 1 and is_primitive_root(a, n):
return a
class CoordinateMapper:
"""
A class that maps linear indices to multi-dimensional coordinates within a specified shape.
Args:
dims (Iterable[int]): An iterable representing the dimensions of the desired shape.
Example Usage:
shape = (2, 3, 5, 4)
mapper = CoordinateMapper(shape)
coordinates = mapper.map(10)
print(coordinates) # Output: [0, 1, 2, 2]
"""
def __init__(self, dims: Iterable[int]):
self.moduli = [prod(dims[i:]) for i in range(len(dims))]
self.divisors = [prod(dims[i + 1:]) for i in range(len(dims))]
def map(self, n: int):
return [(n % self.moduli[i]) // self.divisors[i] for i in range(len(self.moduli))]
def sampler(l):
close_prime = next_prime(l)
state = root = find_random_primitive_root(close_prime)
while state > l:
state = (state * root) % close_prime # Inlining the computation leads to a 20% speed up
yield state - 1
for i in range(l - 1):
state = (state * root) % close_prime
while state > l:
state = (state * root) % close_prime
yield state - 1
def _unique_combinations(*iterables):
cartesian_product_cardinality = prod([len(i) for i in iterables])
coordinate_mapper = CoordinateMapper([len(i) for i in iterables])
sequence = sampler(cartesian_product_cardinality)
for state in sequence:
yield tuple(iterable[coord] for iterable, coord in zip(iterables, coordinate_mapper.map(state)))
from itertools import product
a = [1, 2, 3, 5]
b = ["a", "b", "c", "d"]
u = _unique_combinations(a, b)
assert sorted(u) == sorted(product(a, b))
我开始对各种方法进行基准测试。除了 @matmarbon 之外,我无法获得任何解决方案来运行而不会出现断言错误:
from itertools import product
import time
approaches= {
'prime_roots':_unique_combinations,
'matmarbon':random_order_cartesian_product,
'itertools.product':itertools.product,
}
a = list(range(10))
b = list(range(10))
for name, approach in approaches.items():
assert sorted(u)==sorted(product(a,b))
对于我对以下 2 种算法进行了基准测试,以 itertools 作为基线
import pandas as pd
import timeit
import matplotlib.pyplot as plt
def benchmark(approaches, list_lengths, num_repetitions):
results = []
for approach, function in approaches.items():
for length in list_lengths:
a = list(range(length))
b = list(range(length))
def f():
for item in function(a,b):
pass
execution_time = timeit.timeit(f, number=num_repetitions)
entry = {
'Approach': approach,
'List Length': length,
'Execution Time': execution_time
}
print(entry)
results.append(entry)
results_df = pd.DataFrame(results)
# Plot the benchmark results
plt.figure(figsize=(10, 6))
for approach in approaches.keys():
approach_results = results_df[results_df['Approach'] == approach]
plt.plot(approach_results['List Length'], approach_results['Execution Time'], marker='o', label=approach)
plt.xlabel('List Length')
plt.ylabel('Execution Time (seconds)')
plt.title('Benchmark Results')
plt.grid(True)
plt.legend()
plt.show()
list_lengths = [10,20,30,40,50,60,70,80,90,100]
num_repetitions = 3
benchmark(approaches, list_lengths, num_repetitions)
看来 @matmarbon 的算法正确的是 O(k^n)
。 质根在
O(n^k)
中执行,对于
k~len(iterables)
(假设迭代大小稍微均匀)在内存方面,素根方法获胜只是因为只需要
O(1)
内存并且不存储任何内容。从分布角度来看,素根方法实际上并不是随机的,而是一个难以预测的确定性序列。在实践中,序列应该足够“随机”。
归功于这个
堆栈溢出答案,它启发了解决方案。
random.sample
可以完成工作。也许还有一种更好的方法可以避免重复重采样,直到存在足够多的非重复采样,这是迄今为止我拥有的最好的方法。
import operator
import random
def get_cart_product(list_of_sets, n=None):
max_products_num = reduce(operator.mul, [len(cluster) for cluster in list_of_sets], 1)
if n is not None and n < max_products_num:
refs = set()
while len(refs) < n:
refs.add(tuple(random.sample(cluster, 1)[0] for cluster in list_of_sets))
return refs
return (prod for i, prod in zip(range(n), itertools.product(*list_of_sets)))
return itertools.product(*list_of_sets)
请注意,代码假设有一个冻结集列表,否则应进行
random.sample(cluster, 1)[0]
的转换。
def random_cartesian_product(*lists, seed: Optional[int] = None, n: int):
rnd = random.Random(seed)
cartesian_idxs: Set[Tuple[int, ...]] = set()
list_lens: List[int] = [len(l) for l in lists]
max_count: int = 1
for l_len in list_lens:
max_count *= l_len
if max_count < n:
raise ValueError(f'At most {max_count} cartesian product elements can be created.')
while len(cartesian_idxs) < n:
rnd_idx: Tuple[int, ...] = tuple(
rnd.randint(0, l_len - 1)
for l_len in list_lens
)
if rnd_idx not in cartesian_idxs:
cartesian_idxs.add(rnd_idx)
elem = []
for l_idx, l in zip(rnd_idx, lists):
elem.append(l[l_idx])
yield elem
这可以迭代或一次性使用:
list(random_sample_cartesian_product(['a', 'b', 'c'], [1,2], ['x'], n=6, seed=None))
[['b', 1, 'x'],
['a', 1, 'x'],
['a', 2, 'x'],
['c', 1, 'x'],
['c', 2, 'x'],
['b', 2, 'x']]