[英]Why is numpy's einsum slower than numpy's built-in functions?
我通常從numpy的einsum函數中獲得了很好的表現(我喜歡它的語法)。 @Ophion對這個問題的回答表明 - 對於測試的案例 - einsum始終優於“內置”功能(有時候會有一些,有時會很多)。 但我剛遇到一個einsum慢得多的情況。 考慮以下等效函數:
(M, K) = (1000000, 20)
C = np.random.rand(K, K)
X = np.random.rand(M, K)
def func_dot(C, X):
Y = X.dot(C)
return np.sum(Y * X, axis=1)
def func_einsum(C, X):
return np.einsum('ik,km,im->i', X, C, X)
def func_einsum2(C, X):
# Like func_einsum but break it into two steps.
A = np.einsum('ik,km', X, C)
return np.einsum('ik,ik->i', A, X)
我希望func_einsum
運行得最快,但這不是我遇到的。 在具有超線程,numpy版本1.9.0.dev-7ae0206的四核CPU上運行,以及使用OpenBLAS進行多線程處理,我得到以下結果:
In [2]: %time y1 = func_dot(C, X)
CPU times: user 320 ms, sys: 312 ms, total: 632 ms
Wall time: 209 ms
In [3]: %time y2 = func_einsum(C, X)
CPU times: user 844 ms, sys: 0 ns, total: 844 ms
Wall time: 842 ms
In [4]: %time y3 = func_einsum2(C, X)
CPU times: user 292 ms, sys: 44 ms, total: 336 ms
Wall time: 334 ms
當我將K
增加到200時,差異更加極端:
In [2]: %time y1= func_dot(C, X)
CPU times: user 4.5 s, sys: 1.02 s, total: 5.52 s
Wall time: 2.3 s
In [3]: %time y2= func_einsum(C, X)
CPU times: user 1min 16s, sys: 44 ms, total: 1min 16s
Wall time: 1min 16s
In [4]: %time y3 = func_einsum2(C, X)
CPU times: user 15.3 s, sys: 312 ms, total: 15.6 s
Wall time: 15.6 s
有人能解釋為什么einsum這么慢嗎?
如果重要,這是我的numpy配置:
In [6]: np.show_config()
lapack_info:
libraries = ['openblas']
library_dirs = ['/usr/local/lib']
language = f77
atlas_threads_info:
libraries = ['openblas']
library_dirs = ['/usr/local/lib']
define_macros = [('ATLAS_WITHOUT_LAPACK', None)]
language = c
include_dirs = ['/usr/local/include']
blas_opt_info:
libraries = ['openblas']
library_dirs = ['/usr/local/lib']
define_macros = [('ATLAS_INFO', '"\\"None\\""')]
language = c
include_dirs = ['/usr/local/include']
atlas_blas_threads_info:
libraries = ['openblas']
library_dirs = ['/usr/local/lib']
define_macros = [('ATLAS_INFO', '"\\"None\\""')]
language = c
include_dirs = ['/usr/local/include']
lapack_opt_info:
libraries = ['openblas', 'openblas']
library_dirs = ['/usr/local/lib']
define_macros = [('ATLAS_WITHOUT_LAPACK', None)]
language = f77
include_dirs = ['/usr/local/include']
lapack_mkl_info:
NOT AVAILABLE
blas_mkl_info:
NOT AVAILABLE
mkl_info:
NOT AVAILABLE
你可以充分利用這兩個方面:
def func_dot_einsum(C, X):
Y = X.dot(C)
return np.einsum('ij,ij->i', Y, X)
在我的系統上:
In [7]: %timeit func_dot(C, X)
10 loops, best of 3: 31.1 ms per loop
In [8]: %timeit func_einsum(C, X)
10 loops, best of 3: 105 ms per loop
In [9]: %timeit func_einsum2(C, X)
10 loops, best of 3: 43.5 ms per loop
In [10]: %timeit func_dot_einsum(C, X)
10 loops, best of 3: 21 ms per loop
如果可用, np.dot
使用BLAS,MKL或您擁有的任何庫。 所以對np.dot
的調用幾乎肯定是多線程的。 np.einsum
有自己的循環,因此不使用任何這些優化,除了它自己使用SIMD來加速通過vanilla C實現。
然后是多輸入einsum調用運行得慢得多...... einsum的numpy源非常復雜,我不完全理解它。 所以請注意以下內容充其量只是推測,但這是我認為正在發生的事情......
當你運行像np.einsum('ij,ij->i', a, b)
,做np.sum(a*b, axis=1)
的好處來自於避免必須用所有實例化中間數組產品,並在其上循環兩次。 所以在低級別發生的事情是這樣的:
for i in range(I):
out[i] = 0
for j in range(J):
out[i] += a[i, j] * b[i, j]
現在說你正在追求類似的東西:
np.einsum('ij,jk,ik->i', a, b, c)
你可以做同樣的操作
np.sum(a[:, :, None] * b[None, :, :] * c[:, None, :], axis=(1, 2))
而我認為einsum所做的就是運行這個最后的代碼,而不必實例化巨大的中間數組,這肯定會讓事情變得更快:
In [29]: a, b, c = np.random.rand(3, 100, 100)
In [30]: %timeit np.einsum('ij,jk,ik->i', a, b, c)
100 loops, best of 3: 2.41 ms per loop
In [31]: %timeit np.sum(a[:, :, None] * b[None, :, :] * c[:, None, :], axis=(1, 2))
100 loops, best of 3: 12.3 ms per loop
但是如果你仔細看一下,擺脫中間存儲可能是一件可怕的事情。 這就是我認為einsum在低級別做的事情:
for i in range(I):
out[i] = 0
for j in range(J):
for k in range(K):
out[i] += a[i, j] * b[j, k] * c[i, k]
但是你正在重復大量的操作! 如果您改為:
for i in range(I):
out[i] = 0
for j in range(J):
temp = 0
for k in range(K):
temp += b[j, k] * c[i, k]
out[i] += a[i, j] * temp
你會做I * J * (K-1)
減少乘法(和I * J
額外增加),並節省你自己很多時間。 我的猜測是,einsum不夠智能,不能在這個級別上優化事物。 在源代碼中有一個提示,它只用1或2個操作數優化操作,而不是3.在任何情況下,為一般輸入自動執行此操作似乎只是簡單...
einsum
有一個'2操作數,ndim = 2'的專門案例。 在這種情況下,有3個操作數,總共3個維度。 所以它必須使用一般的nditer
。
在嘗試理解如何解析字符串輸入時,我編寫了一個純Python einsum模擬器, https://github.com/hpaulj/numpy-einsum/blob/master/einsum_py.py
(剝離的)einsum和產品總和函數是:
def myeinsum(subscripts, *ops, **kwargs):
# dropin preplacement for np.einsum (more or less)
<parse subscript strings>
<prepare op_axes>
x = sum_of_prod(ops, op_axes, **kwargs)
return x
def sum_of_prod(ops, op_axes,...):
...
it = np.nditer(ops, flags, op_flags, op_axes)
it.operands[nop][...] = 0
it.reset()
for (x,y,z,w) in it:
w[...] += x*y*z
return it.operands[nop]
使用(M,K)=(10,5)
調試myeinsum('ik,km,im->i',X,C,X,debug=True)
輸出myeinsum('ik,km,im->i',X,C,X,debug=True)
(M,K)=(10,5)
{'max_label': 109,
'min_label': 105,
'nop': 3,
'shapes': [(10, 5), (5, 5), (10, 5)],
....}}
...
iter labels: [105, 107, 109],'ikm'
op_axes [[0, 1, -1], [-1, 0, 1], [0, -1, 1], [0, -1, -1]]
如果你在cython
編寫這樣的sum-of-prod
函數,你應該得到一些接近廣義einsum
。
使用滿(M,K)
,這個模擬的einsum慢6-7倍。
一些時間建立在其他答案上:
In [84]: timeit np.dot(X,C)
1 loops, best of 3: 781 ms per loop
In [85]: timeit np.einsum('ik,km->im',X,C)
1 loops, best of 3: 1.28 s per loop
In [86]: timeit np.einsum('im,im->i',A,X)
10 loops, best of 3: 163 ms per loop
這個'im,im->i' step is substantially faster than the other. The sum dimension,
'im,im->i' step is substantially faster than the other. The sum dimension,
m is only 20. I suspect
einsum`將此視為特殊情況。
In [87]: timeit np.einsum('im,im->i',np.dot(X,C),X)
1 loops, best of 3: 950 ms per loop
In [88]: timeit np.einsum('im,im->i',np.einsum('ik,km->im',X,C),X)
1 loops, best of 3: 1.45 s per loop
這些復合計算的時間只是相應部分的總和。
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