案例 k = 2：np.triu_indices
我已经k = 2使用上述功能的许多变体测试了案例perfplot。毫无疑问，赢家是，np.triu_indices我现在看到，np.dtype([('', np.intp)] * 2)即使对于像igraph.EdgeList.
from itertools import combinations, chain
from scipy.special import comb
import igraph as ig #graph library build on C
import networkx as nx #graph library, pure Python
def _combs(n):
return np.array(list(combinations(range(n),2)))
def _combs_fromiter(n): #@Jaime
indices = np.arange(n)
dt = np.dtype([('', np.intp)]*2)
indices = np.fromiter(combinations(indices, 2), dt)
indices = indices.view(np.intp).reshape(-1, 2)
return indices
def _combs_fromiterplus(n):
dt = np.dtype([('', np.intp)]*2)
indices = np.fromiter(combinations(range(n), 2), dt)
indices = indices.view(np.intp).reshape(-1, 2)
return indices
def _numpy(n): #@endolith
return np.transpose(np.triu_indices(n,1))
def _igraph(n):
return np.array(ig.Graph(n).complementer(False).get_edgelist())
def _igraph_fromiter(n):
dt = np.dtype([('', np.intp)]*2)
indices = np.fromiter(ig.Graph(n).complementer(False).get_edgelist(), dt)
indices = indices.view(np.intp).reshape(-1, 2)
return indices
def _nx(n):
G = nx.Graph()
G.add_nodes_from(range(n))
return np.array(list(nx.complement(G).edges))
def _nx_fromiter(n):
G = nx.Graph()
G.add_nodes_from(range(n))
dt = np.dtype([('', np.intp)]*2)
indices = np.fromiter(nx.complement(G).edges, dt)
indices = indices.view(np.intp).reshape(-1, 2)
return indices
def _comb_index(n): #@HYRY
count = comb(n, 2, exact=True)
index = np.fromiter(chain.from_iterable(combinations(range(n), 2)),
int, count=count*2)
return index.reshape(-1, 2)
fig = plt.figure(figsize=(15, 10))
plt.grid(True, which="both")
out = perfplot.bench(
setup = lambda x: x,
kernels = [_numpy, _combs, _combs_fromiter, _combs_fromiterplus,
_comb_index, _igraph, _igraph_fromiter, _nx, _nx_fromiter],
n_range = [2 ** k for k in range(12)],
xlabel = 'combinations(n, 2)',
title = 'testing combinations',
show_progress = False,
equality_check = False)
out.show()
想知道为什么np.triu_indices不能扩展到更多维度？
情况 2 ≤ k ≤ 4：（triu_indices在此处实现）= 最高 2 倍加速
np.triu_indices 实际上可能是案例的赢家k = 3，即使k = 4我们实现了一个通用的方法。此方法的当前版本等效于：
def triu_indices(n, k):
x = np.less.outer(np.arange(n), np.arange(-k+1, n-k+1))
return np.nonzero(x)
它为两个序列 0,1,...,n-1 构建关系x < y的矩阵表示，并找到它们不为零的单元格的位置。对于 3D 情况，我们需要添加额外的维度和相交关系x < y和y < z。对于下一维过程是相同的，但这会导致巨大的内存过载，因为需要 n^k 个二进制单元，并且其中只有 C(n, k) 达到 True 值。内存使用量和性能增长了 O(n!)，因此该算法itertools.combinations仅在 k 值较小时表现出色。这最好实际用于案例k=2和k=3
def C(n, k): #huge memory overload...
if k==0:
return np.array([])
if k==1:
return np.arange(1,n+1)
elif k==2:
return np.less.outer(np.arange(n), np.arange(n))
else:
x = C(n, k-1)
X = np.repeat(x[None, :, :], len(x), axis=0)
Y = np.repeat(x[:, :, None], len(x), axis=2)
return X&Y
def C_indices(n, k):
return np.transpose(np.nonzero(C(n,k)))
让我们用perfplot结帐：
import matplotlib.pyplot as plt
import numpy as np
import perfplot
from itertools import chain, combinations
from scipy.special import comb
def C(n, k): # huge memory overload...
if k == 0:
return np.array([])
if k == 1:
return np.arange(1, n + 1)
elif k == 2:
return np.less.outer(np.arange(n), np.arange(n))
else:
x = C(n, k - 1)
X = np.repeat(x[None, :, :], len(x), axis=0)
Y = np.repeat(x[:, :, None], len(x), axis=2)
return X & Y
def C_indices(data):
n, k = data
return np.transpose(np.nonzero(C(n, k)))
def comb_index(data):
n, k = data
count = comb(n, k, exact=True)
index = np.fromiter(chain.from_iterable(combinations(range(n), k)),
int, count=count * k)
return index.reshape(-1, k)
def build_args(k):
return {'setup': lambda x: (x, k),
'kernels': [comb_index, C_indices],
'n_range': [2 ** x for x in range(2, {2: 10, 3:10, 4:7, 5:6}[k])],
'xlabel': f'N',
'title': f'test of case C(N,{k})',
'show_progress': True,
'equality_check': lambda x, y: np.array_equal(x, y)}
outs = [perfplot.bench(**build_args(n)) for n in (2, 3, 4, 5)]
fig = plt.figure(figsize=(20, 20))
for i in range(len(outs)):
ax = fig.add_subplot(2, 2, i + 1)
ax.grid(True, which="both")
outs[i].plot()
plt.show()
k=2因此，对于(相当于k=3` ) 实现了最佳性能提升，np.triu_indices) and for 它几乎快了两倍。
案例 k > 3：（numpy_combinations在此处实现）= 高达 2.5 倍的加速
在这个问题之后（感谢@Divakar），我设法找到了一种基于前一列和帕斯卡三角形计算特定列值的方法。它还没有尽可能地优化，但结果确实很有希望。开始了：
from scipy.linalg import pascal
def stretch(a, k):
l = a.sum()+len(a)*(-k)
out = np.full(l, -1, dtype=int)
out[0] = a[0]-1
idx = (a-k).cumsum()[:-1]
out[idx] = a[1:]-1-k
return out.cumsum()
def numpy_combinations(n, k):
#n, k = data #benchmark version
n, k = data
x = np.array([n])
P = pascal(n).astype(int)
C = []
for b in range(k-1,-1,-1):
x = stretch(x, b)
r = P[b][x - b]
C.append(np.repeat(x, r))
return n - 1 - np.array(C).T
基准测试结果是：
# script is the same as in previous example except this part
def build_args(k):
return {'setup': lambda x: (k, x),
'kernels': [comb_index, numpy_combinations],
'n_range': [x for x in range(1, k)],
'xlabel': f'N',
'title': f'test of case C({k}, k)',
'show_progress': True,
'equality_check': False}
outs = [perfplot.bench(**build_args(n)) for n in (12, 15, 17, 23, 25, 28)]
fig = plt.figure(figsize=(20, 20))
for i in range(len(outs)):
ax = fig.add_subplot(2, 3, i + 1)
ax.grid(True, which="both")
outs[i].plot()
plt.show()
尽管它仍然无法与之抗争，itertools.combinations但n < 15在其他情况下它是一个新的赢家。最后但并非最不重要的一点是，numpy当组合数量变得非常大时，展示它的力量。它能够在处理大约 40'000'000 个大小为 14 的项目的 C(28, 14) 组合时存活下来