[英]Numpy: vectorizing a function that integrate 2D array
我需要為2D數組執行以下集成: 也就是說,網格中的每個點得到值RC,這是整數字段與字段U在某一點(x,y)的值之間的差值的積分,乘以標准化內核,即1D版本是:
到目前為止我所做的是對索引的低效迭代:
def normalized_bimodal_kernel_2D(x,y,a,x0=0.0,y0=0.0):
""" Gives a kernel that is zero in x=0, and its integral from -infty to
+infty is 1.0. The parameter a is a length scale where the peaks of the
function are."""
dist = (x-x0)**2 + (y-y0)**2
return (dist*np.exp(-(dist/a)))/(np.pi*a**2)
def RC_2D(U,a,dx):
nx,ny=U.shape
x,y = np.meshgrid(np.arange(0,nx, dx),np.arange(0,ny,dx), sparse=True)
UB = np.zeros_like(U)
for i in xrange(0,nx):
for j in xrange(0,ny):
field=(U-U[i,j])*normalized_bimodal_kernel_2D(x,y,a,x0=i*dx,y0=j*dx)
UB[i,j]=np.sum(field)*dx**2
return UB
def centerlizing_2D(U,a,dx):
nx,ny=U.shape
x,y = np.meshgrid(np.arange(0,nx, dx),np.arange(0,ny,dx), sparse=True)
UB = np.zeros((nx,ny,nx,ny))
for i in xrange(0,nx):
for j in xrange(0,ny):
UB[i,j]=normalized_bimodal_kernel_2D(x,y,a,x0=i*dx,y0=j*dx)
return UB
你可以在這里看到centeralizing
功能的結果:
U=np.eye(20)
plt.imshow(centerlizing(U,10,1)[10,10])
假設dx=1
因為我不確定您要對該離散化做什么:
def normalized_bimodal_kernel_2D(x, y, a):
#generating a 4-d tensor instead of 1d vector
dist = (x[:,None,None,None] - x[None,None,:,None])**2 +\
(y[None,:,None,None] - y[None,None,None,:])**2
return (dist * np.exp(-(dist / a))) / (np.pi * a**2)
def RC_2D(U, a):
nx, ny = U.shape
x, y = np.arange(nx), np.arange(ny)
U4 = U[:, :, None, None] - U[None, None, :, :] #Another 4d
k = normalized_bimodal_kernel_2D(x, y, a)
return np.einsum('ijkl,ijkl->ij', U4, k)
def centerlizing_2D(U, a):
nx, ny = U.shape
x, y = np.arange(nx), np.arange(ny)
return normalized_bimodal_kernel_2D(x, y, a)
基本上, for
numpy
中的循環進行歸化是一個增加更多維度的問題。 你在2D U
矢量上做了兩個循環,所以要向量化只需將它變成4D。
normalized_bimodal_kernel_2D
在兩個嵌套循環中被調用,每個循環只移動它的小步長偏移量。 這復制了許多計算。
centerlizing_2D
的優化是為較大范圍計算一次內核,然后定義UB
以將移位視圖轉換為更大范圍。 這可以使用stride_tricks
,不幸的是它是相當先進的numpy。
def centerlizing_2D_opt(U,a,dx):
nx,ny=U.shape
x,y = np.meshgrid(np.arange(-nx//2, nx+nx//2, dx),
np.arange(-nx//2, ny+ny//2, dx), # note the increased range
sparse=True)
k = normalized_bimodal_kernel_2D(x, y, a, x0=nx//2, y0=ny//2)
sx, sy = k.strides
UB = as_strided(k, shape=(nx, ny, nx*2, ny*2), strides=(sy, sx, sx, sy))
return UB[:, :, nx:0:-1, ny:0:-1]
assert np.allclose(centerlizing_2D(U,10,1), centerlizing_2D_opt(U,10,1)) # verify it's correct
是的,它更快:
%timeit centerlizing_2D(U,10,1) # 100 loops, best of 3: 9.88 ms per loop
%timeit centerlizing_2D_opt(U,10,1) # 10000 loops, best of 3: 85.9 µs per loop
接下來,我們通過優化的centerlizing_2D
例程來表示它來優化RC_2D
:
def RC_2D_opt(U,a,dx):
UB_tmp = centerlizing_2D_opt(U, a, dx)
U_tmp = U[:, :, None, None] - U[None, None, :, :]
UB = np.sum(U_tmp * UB_tmp, axis=(0, 1))
return UB
assert np.allclose(RC_2D(U,10,1), RC_2D_opt(U,10,1))
%timeit RC_2D(U,10, 1)
:
#original: 100 loops, best of 3: 13.8 ms per loop
#@DanielF's: 100 loops, best of 3: 6.98 ms per loop
#mine: 1000 loops, best of 3: 1.83 ms per loop
為了適合你的公式,讓U
成為一個函數。
然后你只需要用np.ix_
將x,y,x',y'
放在四個不同的維度中,然后刻意翻譯你的公式。 Numpy廣播將完成其余的工作。
a=20
x,y,xp,yp=np.ix_(*[np.linspace(0,1,a)]*4)
def U(x,y) : return np.float32(x == y) # function "eye"
def f(x,y,xp,yp,a):
r2=(x-xp)**2+(y-yp)**2
return r2*np.exp(-r2/a)*(U(xp,yp) - U(x,y))/np.pi/a/a
#f(x,y,xp,yp,a).shape is (20, 20, 20, 20)
RC=f(x,y,xp,yp,a).sum(axis=(2,3))
#RC.shape is (20, 20)
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