Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
236 views
in Technique[技术] by (71.8m points)

python - Is 2-dimensional numpy.take fast?

numpy.take can be applied in 2 dimensions with

np.take(np.take(T,ix,axis=0), iy,axis=1 )

I tested the stencil of the discret 2-dimensional Laplacian

ΔT = T[ix-1,iy] + T[ix+1, iy] + T[ix,iy-1] + T[ix,iy+1] - 4 * T[ix,iy]

with 2 take-schemes and the usual numpy.array scheme. The functions p and q are introduced for a leaner code writing and adress the axis 0 and 1 in different order. This is the code:

nx = 300; ny= 300
T  = np.arange(nx*ny).reshape(nx, ny)
ix = np.linspace(1,nx-2,nx-2,dtype=int) 
iy = np.linspace(1,ny-2,ny-2,dtype=int)
#------------------------------------------------------------
def p(Φ,kx,ky):
    return np.take(np.take(Φ,ky,axis=1), kx,axis=0 )
#------------------------------------------------------------
def q(Φ,kx,ky):
    return np.take(np.take(Φ,kx,axis=0), ky,axis=1 )
#------------------------------------------------------------
%timeit ΔT_n = T[0:nx-2,1:ny-1] + T[2:nx,1:ny-1] + T[1:nx-1,0:ny-2]  + T[1:nx-1,2:ny] - 4.0 * T[1:nx-1,1:ny-1] 
%timeit ΔT_t = p(T,ix-1,iy)  + p(T,ix+1,iy)  + p(T,ix,iy-1)  + p(T,ix,iy+1)  - 4.0 * p(T,ix,iy)
%timeit ΔT_t = q(T,ix-1,iy)  + q(T,ix+1,iy)  + q(T,ix,iy-1)  + q(T,ix,iy+1)  - 4.0 * q(T,ix,iy)
.
1000 loops, best of 3: 944 μs per loop
100 loops, best of 3: 3.11 ms per loop
100 loops, best of 3: 2.02 ms per loop

The results seem to be obvious:

  1. usual numpy index arithmeitk is fastest
  2. take-scheme q takes 100% longer (= C-ordering ?)
  3. take-scheme p takes 200% longer (= Fortran-ordering ?)

Not even the 1-dimensional example of the scipy manual indicates that numpy.take is fast:

a = np.array([4, 3, 5, 7, 6, 8])
indices = [0, 1, 4]
%timeit np.take(a, indices)
%timeit a[indices]
.
The slowest run took 6.58 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 4.32 μs per loop
The slowest run took 7.34 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 3.87 μs per loop

Does anybody has experiences how to make numpy.take fast ? It would be an flexible and attractive way for lean code writing that is fast in coding and
is told to be fast in execution as well. Thank your for some hints to improve my approach !

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

The indexed version might be cleaned up with slice objects like this:

T[0:nx-2,1:ny-1] + T[2:nx,1:ny-1] + T[1:nx-1,0:ny-2]  + T[1:nx-1,2:ny] - 4.0 * T[1:nx-1,1:ny-1]

sy1 = slice(1,ny-1)
sx1 = slice(1,nx-1)
sy2 = slice(2,ny)
sy_2 = slice(0,ny-2)
T[0:nx-2,sy1] + T[2:nx,sy1] + T[sx1,xy_2]  + T[sx1,sy2] - 4.0 * T[sx1,sy1]

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...