比提供的解决方案更快地获取排列索引和位于索引处的排列 [英] Getting permutation index and permutation at index faster than the provided solution
本文介绍了比提供的解决方案更快地获取排列索引和位于索引处的排列的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
多亏了这个answer，下面是我如何获得排列索引和一个索引的排列：
import time
def get_Cl(distinct):
Cl = []
for i in range(1, distinct + 1): # i is distincct
c = [0] * i + [1, 0]
C = [c]
for l in range(2, distinct + 1):
c = [
c[d] * d + c[d + 1] * (distinct - d)
for d in range(i + 1)
] + [0]
C.append(c)
Cl.append(C)
return Cl
def item_index(item, distinct, n_symbols, Cl):
length = len(item)
offset = 0
seen = set()
for i, di in enumerate(item):
for d in range(n_symbols):
if d == di:
break
if d in seen:
# test = Cl[distinct][length - 1 - i][len(seen)]
offset += Cl[distinct][length - 1 - i][len(seen)]
else:
offset += Cl[distinct][length - 1 - i][len(seen) + 1]
seen.add(di)
return offset
def item_at(idx, length, distinct, n_symbols, Cl):
seen = [0] * n_symbols
prefix = [0] * length
used = 0
for i in range(length):
for d in range(n_symbols):
if seen[d] != 0:
branch_count = Cl[distinct][length - 1 - i][used]
else:
branch_count = Cl[distinct][length - 1 - i][used + 1]
if branch_count <= idx:
idx -= branch_count
else:
prefix[i] = d
if seen[d] == 0:
used += 1
seen[d] = 1
break
return prefix
if __name__ == "__main__":
start_time = time.time()
Cl = get_Cl(512)
end_time = time.time()
print(f'{(end_time - start_time)} seconds for Cl')
start_time = time.time()
item = item_at(idx=432, length=512, distinct=350, n_symbols=512, Cl=Cl)
end_time = time.time()
print(f'{(end_time - start_time)} seconds for item_at')
print(item)
start_time = time.time()
print(item_index(item=item, distinct=350, n_symbols=512, Cl=Cl))
end_time = time.time()
print(f'{(end_time - start_time)} seconds for item_index')
356.3069865703583 seconds for Cl
2.5428783893585205 seconds for item_at
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 127, 128, 129, 130, 131, 132, 133, 134, 135, 136, 137, 138, 139, 140, 141, 142, 143, 144, 145, 146, 147, 148, 149, 150, 151, 152, 153, 154, 155, 156, 157, 158, 159, 160, 161, 162, 163, 164, 165, 166, 167, 168, 169, 170, 171, 172, 173, 174, 175, 176, 177, 178, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 214, 215, 216, 217, 218, 219, 220, 221, 222, 223, 224, 225, 226, 227, 228, 229, 230, 231, 232, 233, 234, 235, 236, 237, 238, 239, 240, 241, 242, 243, 244, 245, 246, 247, 248, 249, 250, 251, 252, 253, 254, 255, 256, 257, 258, 259, 260, 261, 262, 263, 264, 265, 266, 267, 268, 269, 270, 271, 272, 273, 274, 275, 276, 277, 278, 279, 280, 281, 282, 283, 284, 285, 286, 287, 288, 289, 290, 291, 292, 293, 294, 295, 296, 297, 298, 299, 300, 301, 302, 303, 304, 305, 306, 307, 308, 309, 310, 311, 312, 313, 314, 315, 316, 317, 318, 319, 320, 321, 322, 323, 324, 325, 326, 327, 328, 329, 330, 331, 332, 333, 334, 335, 336, 337, 338, 339, 340, 341, 342, 343, 344, 345, 346, 347, 348, 351, 458]
432
0.025868892669677734 seconds for item_index
它工作得很好，除非数字变得更大，然后变得非常慢。想知道是否有可能改进这个代码，比如thisAnswer，这是同一个Slow函数的改进版本，用于计算所有排列？
我在单独一行得到Cl的原因是，对于固定的distinct，item_at和item_index将有数千个调用，因此如果distinct相同，则Cl相同，因此不需要为每个item_at或item_index调用它。
更新：来自答案的测试结果
0.008994340896606445 seconds for item_at
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 127, 128, 129, 130, 131, 132, 133, 134, 135, 136, 137, 138, 139, 140, 141, 142, 143, 144, 145, 146, 147, 148, 149, 150, 151, 152, 153, 154, 155, 156, 157, 158, 159, 160, 161, 162, 163, 164, 165, 166, 167, 168, 169, 170, 171, 172, 173, 174, 175, 176, 177, 178, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 214, 215, 216, 217, 218, 219, 220, 221, 222, 223, 224, 225, 226, 227, 228, 229, 230, 231, 232, 233, 234, 235, 236, 237, 238, 239, 240, 241, 242, 243, 244, 245, 246, 247, 248, 249, 250, 251, 252, 253, 254, 255, 256, 257, 258, 259, 260, 261, 262, 263, 264, 265, 266, 267, 268, 269, 270, 271, 272, 273, 274, 275, 276, 277, 278, 279, 280, 281, 282, 283, 284, 285, 286, 287, 288, 289, 290, 291, 292, 293, 294, 295, 296, 297, 298, 299, 300, 301, 302, 303, 304, 305, 306, 307, 308, 309, 310, 311, 312, 313, 314, 315, 316, 317, 318, 319, 320, 321, 322, 323, 324, 325, 326, 327, 328, 329, 330, 331, 332, 333, 334, 335, 336, 337, 338, 339, 340, 341, 342, 343, 347, 348, 344, 345, 346, 349]
432
0.006995677947998047 seconds for item_index
推荐答案
在此回答中，我将演示两个可以提高item_at和item_index速度的修改。
在开始之前，我们先初始化CL表，以便使用distinct=200
def get_Cl(length, distinct):
i = distinct
c = [0] * i + [1, 0]
C = [c]
for l in range(2, length+1):
c = [
c[d] * d + c[d + 1] * (i - d)
for d in range(i + 1)
] + [0]
C.append(c)
return C;
Cl = {200:get_Cl(300, 200)}
修改为item_index
注意，item_index的内部循环只是通过不依赖于d in seen但不在d自身中的值来递增offset。如果我们事先知道d in seen将是True的次数。因此，让我们以一种跟踪数组d之前看到的值的数量seen_before[d]的方式重写代码。
import numpy as np
def item_index_bs(item, distinct, n_symbols, Cl):
length = len(item)
offset = 0
seen = set()
seen_before = np.zeros(n_symbols, dtype=np.uint64)
for i, di in enumerate(item):
offset += Cl[distinct][length - 1 - i][len(seen)] * int(seen_before[di])
+ Cl[distinct][length - 1 - i][len(seen) + 1] * int(di - seen_before[di]);
if di not in seen:
seen.add(di)
seen_before[di+1:] += 1;
return offset
这可以用
测试pp = item_at(256, 300, 200, 300, Cl)
item_index_factored(pp, 200, 300, Cl) # 1.8ms
item_index(pp, 200, 300, Cl) # 5.39ms
修改为item_at
对于item_at，我们不能像item_index那样简单地对术语进行分组，但我们可能会跳过一些迭代，假设idx减少了a，否则减少了b，所以它最多减少max(a,b)，至少需要idx//max(a,b)才能找到要使用的数字。然后我们通过将a和b乘以它们各自的系数来进行更新。
def item_at_skip(idx, length, distinct, n_symbols, Cl):
seen = [0] * n_symbols;
prefix = [0] * length
used = 0
for i in range(length):
a = Cl[distinct][length - 1 - i][used];
b = Cl[distinct][length - 1 - i][used + 1]
c = idx // max(a,b) # d will be at least c
ac = sum(seen[:c]) # the number of time a is subtracted
idx -= a * ac + b * (c - ac);
for d in range(c, n_symbols):
if seen[d] != 0:
branch_count = a
else:
branch_count = b
if branch_count <= idx:
idx -= branch_count
else:
prefix[i] = d
if seen[d] == 0:
used += 1
seen[d] = 1
break
return prefix
assert item_at_skip(10**200, 300, 200, 300, Cl) == item_at(10**200, 300, 200, 300, Cl)
item_at_skip(10**200, 300, 200, 300, Cl) # 3.16ms
item_at(10**200, 300, 200, 300, Cl) # 6.25ms
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