numpy 使用@jit实现自动并行化

hk8txs48  于 2023-03-30  发布在  其他
关注(0)|答案(2)|浏览(186)

我想利用numba自动并行化与@jit，以帮助我加快运行我的函数，因为我使用非常大的输入，我没有parralisation之前的经验，我尝试了几个解决方案使用多处理，我有很多问题，我张贴它作为问题在堆栈溢出，但几乎没有响应，所以我移动的东西，可以帮助我自动和看起来更简单，
下面是一个实现numba @jit的简单例子：

@jit(nopython=True, parallel=True)
def f(x, y):
    return x + y

所以我试着在代码中简单地使用它：

from numba import njit, prange
@njit(parallel=True)
def defineMatrices(C0x, C0y, C0xy, C0yx, Cxx_err, Cyy_err, Cxy_err, Cyx_err, dCx, dCy, dCxy,dCyx):
    Nk = len(dCx)  # number of free parameters
    Nm = len(C0x)  # number of measurements
    #print('NK:', Nk)
    #print('Nm:', Nm)
    Ax = np.zeros([Nk, Nk])
    Ay = np.zeros([Nk, Nk])
    Axy = np.zeros([Nk, Nk])
    Ayx = np.zeros([Nk, Nk])
    A = np.zeros([4 * Nk, Nk])
    ##
    Bx = np.zeros([Nk, 1])
    By = np.zeros([Nk, 1])
    Bxy = np.zeros([Nk, 1])
    Byx = np.zeros([Nk, 1])
    B = np.zeros([4 * Nk, 1])
    ##
    Dx = (Cxx_err[0:Nm, :] - C0x[0:Nm, :])  ### dk ?
    Dy = (Cyy_err[0:Nm, :] - C0y[0:Nm, :])
    Dxy = (Cxy_err[0:Nm, :] - C0xy[0:Nm, :])
    Dyx = (Cyx_err[0:Nm, :] - C0yx[0:Nm, :])
    for i in range(Nk):  ## i represents each quad
        # print('done A:', 100.* i ,'%')
        for j in range(Nk):
            Ax[i, j] = np.sum(np.dot(dCx[i], dCx[j].T))
            Ay[i, j] = np.sum(np.dot(dCy[i], dCy[j].T))
            Axy[i, j] = np.sum(np.dot(dCxy[i], dCxy[j].T))
            Ayx[i, j] = np.sum(np.dot(dCyx[i], dCyx[j].T))
        A[i, :] = Ax[i, :]
        A[i + Nk, :] = Ay[i, :]
        A[i + 2 * Nk, :] = Axy[i, :]
        A[i + 3 * Nk, :] = Ayx[i, :]
    ##
    for i in range(Nk):
        Bx[i] = np.sum(np.dot(dCx[i], Dx.T))
        By[i] = np.sum(np.dot(dCy[i], Dy.T))
        Bxy[i] = np.sum(np.dot(dCxy[i], Dxy.T))
        Byx[i] = np.sum(np.dot(dCyx[i], Dyx.T))
        B[i] = Bx[i]
        B[i + Nk] = By[i]
        B[i + 2 * Nk] = Bxy[i]
        B[i + 3 * Nk] = Byx[i]
    
    return A, B

我的意见是：
所有C：是两个dim矩阵所有dC：三维矩阵
输出：
A：两个调光B：一维
但我看到了错误：

TypingError                               Traceback (most recent call last)
~\AppData\Local\Temp/ipykernel_17908/1125053646.py in <module>
      1 start_time = time.time()
----> 2 A, B = defineMatrices(C0x, C0y, C0xy, C0yx, Cxx1, Cyy1, Cxy1, Cyx1, dCx, dCy, dCxy,dCyx)
      3 end_time = time.time()
      4 
      5 # Calculate the time taken and print it

~\Anaconda3\lib\site-packages\numba\core\dispatcher.py in _compile_for_args(self, *args, **kws)
    418                 e.patch_message(msg)
    419 
--> 420             error_rewrite(e, 'typing')
    421         except errors.UnsupportedError as e:
    422             # Something unsupported is present in the user code, add help info

~\Anaconda3\lib\site-packages\numba\core\dispatcher.py in error_rewrite(e, issue_type)
    359                 raise e
    360             else:
--> 361                 raise e.with_traceback(None)
    362 
    363         argtypes = []

TypingError: Failed in nopython mode pipeline (step: nopython frontend)
No implementation of function Function(<function fill_diagonal at 0x0000019FF09F3940>) found for signature:
 
 >>> fill_diagonal(array(float64, 1d, C), float64)
 
There are 2 candidate implementations:
     - Of which 2 did not match due to:
     Overload in function 'np_fill_diagonal': File: numba\np\arraymath.py: Line 2928.
       With argument(s): '(array(float64, 1d, C), float64)':
      Rejected as the implementation raised a specific error:
        TypingError: The first argument must be at least 2-D (found 1-D)
  raised from C:\Users\musa\Anaconda3\lib\site-packages\numba\np\arraymath.py:2958

During: resolving callee type: Function(<function fill_diagonal at 0x0000019FF09F3940>)
During: typing of call at C:\Users\musa\AppData\Local\Temp/ipykernel_17908/2593834973.py (30)

File "..\..\..\AppData\Local\Temp\ipykernel_17908\2593834973.py", line 30:
<source missing, REPL/exec in use?>

对于最小可归约示例：

C0x = [[2.969129, 3.065011, 3.294439],
       [3.087682, 1.674345, 2.930011],
       [3.355765, 2.863887, 2.588477]]

C0y = [[6.748940, 9.673919, 7.662236],
       [9.527495, 11.328892, 10.147356],
       [7.738452, 10.280770, 7.684719]]

C0xy = [[0.0, 0.0, 0.0],
        [0.0, 0.0, 0.0],
        [0.0, 0.0, 0.0]]

C0yx = [[0.0, 0.0, 0.0],
        [0.0, 0.0, 0.0],
        [0.0, 0.0, 0.0]]

Cxx1 = [[2.970383, 3.065395, 3.295505],
        [3.088111, 1.674282, 2.930299],
        [3.356823, 2.864131, 2.589346]]

Cyy1 = [[6.733786, 9.656677, 7.646381],
        [9.510379, 11.309615, 10.129516],
        [7.722598, 10.262803, 7.668157]]

Cxy1 = [[0.0, 0.0, 0.0],
        [0.0, 0.0, 0.0],
        [0.0, 0.0, 0.0]]

Cyx1 = [[0.0, 0.0, 0.0],
        [0.0, 0.0, 0.0],
        [0.0, 0.0, 0.0]]

dCx = [array([[-0.88391347, -0.9165879 , -1.00445679],
        [-0.9102315 , -0.94390882, -1.03437531],
        [-0.99567547, -1.03249243, -1.13146452]]),
 array([[-0.93911752, -0.49819029, -0.88115751],
        [-0.51945798, -0.2755809 , -0.48739065],
        [-0.87413297, -0.46370905, -0.82018765]])]
dCy = [array([[4.54319981, 6.52256593, 5.20680836],
        [6.38565572, 9.16779592, 7.31840445],
        [5.19995576, 7.46547485, 5.95950073]]),
 array([[ 9.36071422, 11.09190807,  9.94111537],
        [10.98180048, 13.0128158 , 11.66270877],
        [ 9.93855597, 11.7766108 , 10.55478873]])]
dCxy = [array([[0., 0., 0.],
        [0., 0., 0.],
        [0., 0., 0.]]),
 array([[0., 0., 0.],
        [0., 0., 0.],
        [0., 0., 0.]])]
dCyx = [array([[0., 0., 0.],
        [0., 0., 0.],
        [0., 0., 0.]]),
 array([[0., 0., 0.],
        [0., 0., 0.],
        [0., 0., 0.]])]

A, B = defineMatrices(C0x, C0y, C0xy, C0yx, Cxx1, Cyy1, Cxy1, Cyx1, dCx, dCy, dCxy,dCyx)

输出：

A = [[26.203265, 17.026796],
     [17.026796, 11.763451],
     [1138.015961, 1913.702747],
     [1913.702747, 3238.589801],
     [0.0, 0.0],
     [0.0, 0.0],
     [0.0, 0.0],
     [0.0, 0.0]]

B = [-0.016327,     -0.011957,     -2.966823,     -5.028449,     0.0,     0.0,     0.0,     0.0]
numpy

2条答案

按热度按时间
iszxjhcz

iszxjhcz1#

在你的例子中，dCx是一个2（3，3）数组的列表。让我们做一个这样的列表：

In [113]: dcx = [np.arange(9).reshape(3,3), np.arange(10,19).reshape(3,3)]; dcx
Out[113]: 
[array([[0, 1, 2],
        [3, 4, 5],
        [6, 7, 8]]),
 array([[10, 11, 12],
        [13, 14, 15],
        [16, 17, 18]])]

双循环的Ax部分：

In [114]: Ax = np.zeros((2,2))
     ...: for i in range(2):
     ...:         for j in range(2):
     ...:             Ax[i, j] = np.sum(np.dot(dcx[i], dcx[j].T))

In [115]: Ax
Out[115]: 
array([[ 450., 1530.],
       [1530., 5310.]])

我想说的是，你可以用一个“向量化”的操作来代替双循环。einsum，matmul，甚至dot都可以在更高的维度上工作。
你的列表可以被转换成一个3d数组：

In [116]: x =np.array(dcx); x
Out[116]: 
array([[[ 0,  1,  2],
        [ 3,  4,  5],
        [ 6,  7,  8]],

       [[10, 11, 12],
        [13, 14, 15],
        [16, 17, 18]]])

下面是一个使用einsum的Ax的等价物：

In [117]: np.einsum('ikl,jml->ij', x,x)
Out[117]: 
array([[ 450, 1530],
       [1530, 5310]])

进一步考虑，我也可以使用np.dot：

In [118]: np.dot(x,x.transpose(0,2,1)).sum(axis=(1,3))
Out[118]: 
array([[ 450, 1530],
       [1530, 5310]])

matmul版本有点混乱（它对“批次”维度有不同的规则）：

In [121]: (x[:,None,:,:]@x.transpose(0,2,1)[None,:,:,:]).sum(axis=(2,3))
Out[121]: 
array([[ 450, 1530],
       [1530, 5310]])

其中einsum是最快的，并且明显优于双嵌套

In [124]: timeit np.einsum('ikl,jml->ij', x,x)
13 µs ± 22.7 ns per loop (mean ± std. dev. of 7 runs, 100,000 loops each)

In [125]: %%timeit
     ...: Ax = np.zeros((2,2))
     ...: for i in range(2):
     ...:         for j in range(2):
     ...:             Ax[i, j] = np.sum(np.dot(dcx[i], dcx[j].T))
91.2 µs ± 696 ns per loop (mean ± std. dev. of 7 runs, 10,000 loops each)

使用示例dCx：

In [127]: dCx = [np.array([[-0.88391347, -0.9165879 , -1.00445679],
     ...:         [-0.9102315 , -0.94390882, -1.03437531],
     ...:         [-0.99567547, -1.03249243, -1.13146452]]),
     ...:  np.array([[-0.93911752, -0.49819029, -0.88115751],
     ...:         [-0.51945798, -0.2755809 , -0.48739065],
     ...:         [-0.87413297, -0.46370905, -0.82018765]])]

In [128]: x=np.array(dCx)

In [129]: np.einsum('ikl,jml->ij', x,x)
Out[129]: 
array([[26.20326497, 17.02679642],
       [17.02679642, 11.7634506 ]])
赞(0）分享  回复(0）举报  2023-03-30
dw1jzc5e

dw1jzc5e2#

我已经实现了np.einsum解决方案：我在这里使用的dCx具有形状（40，40，40），A具有形状（160，40）

lt = time.time()

for i in range(Nk):  ## i represents each quad
    # print('done A:', 100.* i ,'%')
    for j in range(Nk):
        Ax[i, j] = np.sum(np.dot(dCx[i], dCx[j].T))
        Ay[i, j] = np.sum(np.dot(dCy[i], dCy[j].T))
        Axy[i, j] = np.sum(np.dot(dCxy[i], dCxy[j].T))
        Ayx[i, j] = np.sum(np.dot(dCyx[i], dCyx[j].T))
    A[i, :] = Ax[i, :]
    A[i + Nk, :] = Ay[i, :]
    A[i + 2 * Nk, :] = Axy[i, :]
    A[i + 3 * Nk, :] = Ayx[i, :]
le = time.time()
elapsed_time = le - lt
print('Execution time:', elapsed_time, 'seconds')
Execution time: 0.43982982635498047 seconds

我用两个for循环比较时间和我的代码

lt = time.time()

x =np.array(dCx)
Ax = np.einsum('ikl,jml->ij', x,x)
x =np.array(dCy)
Ay = np.einsum('ikl,jml->ij', x,x)
x =np.array(dCxy)
Axy = np.einsum('ikl,jml->ij', x,x)
x =np.array(dCyx)
Ayx = np.einsum('ikl,jml->ij', x,x)
A = np.vstack([Ax, Ay, Axy, Ayx])

le = time.time()
elapsed_time = le - lt
print('Execution time:', elapsed_time, 'seconds')
Execution time: 1.488039493560791 seconds

我没想到第二个解决方案所花费的时间会比for循环更长

赞(0）分享  回复(0）举报  2023-03-30
我来回答

相关问题

    查看更多

    热门标签

    更多
    JavaquerypythonNode开发语言requestUtil数据库Table后端算法LoggerMessageElementParser

    最新问答

    更多
    • 蜀ICP备2022004421号-2 NULL123 https://www.null123.com © All rights reserved
    • 本站内容来源互联网,如果侵犯您的权益请联系我们删除, 联系方式：448109455@qq.com