计算范德蒙德矩阵的有效方法 [英] Efficient way to compute the Vandermonde matrix
问题描述
我正在为相当大的一维数组计算 Vandermonde matrix .一种自然而干净的方法是使用 np.vander() .但是，我发现这是大约.比基于列表理解的方法慢 2.5倍.
I'm calculating Vandermonde matrix for a fairly large 1D array. The natural and clean way to do this is using np.vander(). However, I found that this is approx. 2.5x slower than a list comprehension based approach.
In [43]: x = np.arange(5000)
In [44]: N = 4
In [45]: %timeit np.vander(x, N, increasing=True)
155 µs ± 205 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
# one of the listed approaches from the documentation
In [46]: %timeit np.flip(np.column_stack([x**(N-1-i) for i in range(N)]), axis=1)
65.3 µs ± 235 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
In [47]: np.all(np.vander(x, N, increasing=True) == np.flip(np.column_stack([x**(N-1-i) for i in range(N)]), axis=1))
Out[47]: True
I'm trying to understand where the bottleneck is and the reason why does the implementation of native np.vander() is ~ 2.5x slower.
效率对我的实施至关重要.因此，也欢迎更快的选择！
Efficiency matters for my implementation. So, even faster alternatives are also welcome!
推荐答案
这里有更多方法，其中一些方法(在我的计算机上)比迄今为止发布的方法要快得多.
Here are some more methods some of which are quite a bit faster (on my computer) than what has been posted so far.
我认为最重要的发现是，它实际上在很大程度上取决于您想要多少个学位.幂运算(我相信对小整数指数特别适用)仅对较小的指数范围有意义.指数越多，基于乘法的方法票价就越好.
The most important observation I think is that it really depends a lot on how many degrees you want. Exponentiation (which I believe is special cased for small integer exponents) only makes sense for small exponent ranges. The more exponents the better multiplication based approaches fare.
我想强调一个基于multiply.accumulate的方法(ma)，它类似于numpy的内置方法，但是速度更快(不是因为我跳过了支票-nnc，numpy-no-checks证明了这一点) .除了最小的指数范围外，对我来说实际上是最快的.
I'd like to highlight a multiply.accumulate based method (ma) which is similar to numpy's builtin approach but faster (and not because I skimped on checks - nnc, numpy-no-checks demonstrates this). For all but the smallest exponent ranges it is actually the fastest for me.
出于我不了解的原因，据我所知，numpy实现会执行三项操作，这些操作据我所知都是缓慢而不必要的:(1)它使基本向量有很多副本. (2)使它们不连续. (3)它完成了原位累积，我相信会进行缓冲.
For reasons I do not understand, the numpy implementation does three things that are to the best of my knowledge slow and unnecessary: (1) It makes quite a few copies of the base vector. (2) It makes them non-contiguous. (3) It does the accumulation in-place which I believe forces buffering.
我想提到的另一件事是，对于小范围的整数(基本上是ma的手动版本)，最快的速度因简单的预防性提升而降低了两倍以上到更大的dtype(safe_e_1可能有点用词不当).
Another thing I'd like to mention is that the fastest for small ranges of ints (out_e_1 essentially a manual version of ma), is slowed down by a factor of more than two by the simple precaution of promoting to a larger dtype (safe_e_1 arguably a bit of a misnomer).
基于广播的方法称为bc_*，其中*表示广播轴(b表示基，e表示exp)'cheat'表示结果不连续.
The broadcasting based methods are called bc_* where * indicates the broadcast axis (b for base, e for exp) 'cheat' means the result is noncontiguous.
计时(最好3次):
rep=100 n_b=5000 n_e=4 b_tp=<class 'numpy.int32'> e_tp=<class 'numpy.int32'>
vander 0.16699657 ms
bc_b 0.09595204 ms
bc_e 0.07959786 ms
ma 0.10755240 ms
nnc 0.16459018 ms
out_e_1 0.02037535 ms
out_e_2 0.02656622 ms
safe_e_1 0.04652272 ms
safe_e_2 0.04081079 ms
cheat bc_e_cheat 0.04668466 ms
rep=100 n_b=5000 n_e=8 b_tp=<class 'numpy.int32'> e_tp=<class 'numpy.int32'>
vander 0.25086462 ms
bc_b apparently failed
bc_e apparently failed
ma 0.15843041 ms
nnc 0.24713077 ms
out_e_1 apparently failed
out_e_2 apparently failed
safe_e_1 0.15970622 ms
safe_e_2 0.19672418 ms
bc_e_cheat apparently failed
rep=100 n_b=5000 n_e=4 b_tp=<class 'float'> e_tp=<class 'numpy.int32'>
vander 0.16225773 ms
bc_b 0.53315020 ms
bc_e 0.56200830 ms
ma 0.07626799 ms
nnc 0.16059748 ms
out_e_1 0.03653416 ms
out_e_2 0.04043702 ms
safe_e_1 0.04060494 ms
safe_e_2 0.04104209 ms
cheat bc_e_cheat 0.52966076 ms
rep=100 n_b=5000 n_e=8 b_tp=<class 'float'> e_tp=<class 'numpy.int32'>
vander 0.24542852 ms
bc_b 2.03353578 ms
bc_e 2.04281270 ms
ma 0.11075758 ms
nnc 0.24212880 ms
out_e_1 0.14809043 ms
out_e_2 0.19261359 ms
safe_e_1 0.15206112 ms
safe_e_2 0.19308420 ms
cheat bc_e_cheat 1.99176601 ms
代码:
import numpy as np
import types
from timeit import repeat
prom={np.dtype(np.int32): np.dtype(np.int64), np.dtype(float): np.dtype(float)}
def RI(k, N, dt, top=100):
return np.random.randint(0, top if top else N, (k, N)).astype(dt)
def RA(k, N, dt, top=None):
return np.add.outer(np.zeros((k,), int), np.arange(N)%(top if top else N)).astype(dt)
def RU(k, N, dt, top=100):
return (np.random.random((k, N))*(top if top else N)).astype(dt)
def data(k, N_b, N_e, dt_b, dt_e, b_fun=RI, e_fun=RA):
b = list(b_fun(k, N_b, dt_b))
e = list(e_fun(k, N_e, dt_e))
return b, e
def f_vander(b, e):
return np.vander(b, len(e), increasing=True)
def f_bc_b(b, e):
return b[:, None]**e
def f_bc_e(b, e):
return np.ascontiguousarray((b**e[:, None]).T)
def f_ma(b, e):
out = np.empty((len(b), len(e)), prom[b.dtype])
out[:, 0] = 1
np.multiply.accumulate(np.broadcast_to(b, (len(e)-1, len(b))), axis=0, out=out[:, 1:].T)
return out
def f_nnc(b, e):
out = np.empty((len(b), len(e)), prom[b.dtype])
out[:, 0] = 1
out[:, 1:] = b[:, None]
np.multiply.accumulate(out[:, 1:], out=out[:, 1:], axis=1)
return out
def f_out_e_1(b, e):
out = np.empty((len(b), len(e)), b.dtype)
out[:, 0] = 1
out[:, 1] = b
out[:, 2] = c = b*b
for i in range(3, len(e)):
c*=b
out[:, i] = c
return out
def f_out_e_2(b, e):
out = np.empty((len(b), len(e)), b.dtype)
out[:, 0] = 1
out[:, 1] = b
out[:, 2] = b*b
for i in range(3, len(e)):
out[:, i] = out[:, i-1] * b
return out
def f_safe_e_1(b, e):
out = np.empty((len(b), len(e)), prom[b.dtype])
out[:, 0] = 1
out[:, 1] = b
out[:, 2] = c = (b*b).astype(prom[b.dtype])
for i in range(3, len(e)):
c*=b
out[:, i] = c
return out
def f_safe_e_2(b, e):
out = np.empty((len(b), len(e)), prom[b.dtype])
out[:, 0] = 1
out[:, 1] = b
out[:, 2] = b*b
for i in range(3, len(e)):
out[:, i] = out[:, i-1] * b
return out
def f_bc_e_cheat(b, e):
return (b**e[:, None]).T
for params in [(100, 5000, 4, np.int32, np.int32),
(100, 5000, 8, np.int32, np.int32),
(100, 5000, 4, float, np.int32),
(100, 5000, 8, float, np.int32)]:
k = params[0]
dat = data(*params)
ref = f_vander(dat[0][0], dat[1][0])
print('rep={} n_b={} n_e={} b_tp={} e_tp={}'.format(*params))
for name, func in list(globals().items()):
if not name.startswith('f_') or not isinstance(func, types.FunctionType):
continue
try:
assert np.allclose(ref, func(dat[0][0], dat[1][0]))
if not func(dat[0][0], dat[1][0]).flags.c_contiguous:
print('cheat', end=' ')
print("{:16s}{:16.8f} ms".format(name[2:], np.min(repeat(
'f(b.pop(), e.pop())', setup='b, e = data(*p)', globals={'f':func, 'data':data, 'p':params}, number=k)) * 1000 / k))
except:
print("{:16s} apparently failed".format(name[2:]))
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