Develop Reference

Comparing scipy.stats.t.sf vs GSL using Cython

I would like to calculate the p values of a large 2D numpy t values array. However, this takes long time and I would like to improve its speed. I tried using GSL.
Although a single gsl_cdf_tdist_P is much much faster than scipy.stats.t.sf, when iterating over the ndarray, the process is very slow. I would like help to improve this.
See the code below.
GSL_Test.pyx
import cython
cimport cython
import numpy
cimport numpy
from cython_gsl cimport *
DTYPE = numpy.float32
ctypedef numpy.float32_t DTYPE_t
cdef get_gsl_p(double t, double nu):
return (1 - gsl_cdf_tdist_P(t, nu)) * 2
#cython.boundscheck(False)
#cython.wraparound(False)
#cython.nonecheck(False)
cdef get_gsl_p_for_2D_matrix(numpy.ndarray[DTYPE_t, ndim=2] t_matrix, int n):
cdef unsigned int rows = t_matrix.shape[0]
cdef numpy.ndarray[DTYPE_t, ndim=2] out = numpy.zeros((rows, rows), dtype='float32')
cdef unsigned int row, col
for row in range(rows):
for col in range(rows):
out[row, col] = get_gsl_p(t_matrix[row, col], n-2)
return out
def get_gsl_p_for_2D_matrix_def(numpy.ndarray[DTYPE_t, ndim=2] t_matrix, int n):
return get_gsl_p_for_2D_matrix(t_matrix, n)
ipython
import GSL_Test
import numpy
import scipy.stats
a = numpy.random.rand(3544, 3544).astype('float32')
%timeit -n 1 GSL_Test.get_gsl_p_for_2D_matrix(a, 25)
1 loop, best of 3: 7.87 s per loop
%timeit -n 1 scipy.stats.t.sf(a, 25)*2
1 loop, best of 3: 4.66 s per loop
UPDATE: Adding cdef declarations I was able to reduce the computational time but not lower than scipy still. I modified the code to have the cdef declarations.
%timeit -n 1 GSL_Test.get_gsl_p_for_2D_matrix_def(a, 25)
1 loop, best of 3: 6.73 s per loop

You can get some small gain in raw performance by using a raw special function instead of stats.t.sf. Looking at the source, you find (https://github.com/scipy/scipy/blob/master/scipy/stats/_continuous_distns.py#L3849)
def _sf(self, x, df):
return sc.stdtr(df, -x)
So that you can use stdtr directly:
np.random.seed(1234)
x = np.random.random((3740, 374))
t1 = stats.t.sf(x, 25)
t2 = stdtr(25, -x)
1 loop, best of 3: 653 ms per loop
1 loop, best of 3: 562 ms per loop
If you do reach out for cython, the typed memoryview syntax often gives you faster code than the old ndarray syntax:
from scipy.special.cython_special cimport stdtr
from numpy cimport npy_intp
import numpy as np
def tsf(double [:, ::1] x, int df=25):
cdef double[:, ::1] out = np.empty_like(x)
cdef npy_intp i, j
cdef double tmp, xx
for i in range(x.shape[0]):
for j in range(x.shape[1]):
xx = x[i, j]
out[i, j] = stdtr(df, -xx)
return np.asarray(out)
Here I'm also using the cython_special interface, which is only avaialble in the dev version of scipy (http://scipy.github.io/devdocs/special.cython_special.html#module-scipy.special.cython_special), but you can use GSL if you want.
Finally, if you suspect a bottleneck in iterations, don't forget to inspect the output of cython -a to see if there's some python overhead in the hot loops.

Optimize python for Connected Component Labeling Area of Subsets

I have a binary map on which I do Connected Component Labeling and get something like this for a 64x64 grid - http://pastebin.com/bauas0NJ
Now I want to group them by label, so that I can find their area and their center of mass. This is what I do:
#ccl_np is the computed array from the previous step (see pastebin)
#I discard the label '1' as its the background
unique, count = np.unique(ccl_np, return_counts = True)
xcm_array = []
ycm_array = []
for i in range(1,len(unique)):
subarray = np.where(ccl_np == unique[i])
xcm_array.append("{0:.5f}".format((sum(subarray[0]))/(count[i]*1.)))
ycm_array.append("{0:.5f}".format((sum(subarray[1]))/(count[i]*1.)))
final_array = zip(xcm_array,ycm_array,count[1:])
I want a fast code (as I will be doing this for grids of size 4096x4096) and was told to check out numba. Here's my naive attempt :
unique, inverse, count = np.unique(ccl_np, return_counts = True, return_inverse = True)
xcm_array = np.zeros(len(count),dtype=np.float32)
ycm_array = np.zeros(len(count),dtype=np.float32)
inverse = inverse.reshape(64,64)
#numba.autojit
def mysolver(xcm_array, ycm_array, inverse, count):
for i in range(64):
for j in range(64):
pos = inverse[i][j]
local_count = count[pos]
xcm_array[pos] += i/(local_count*1.)
ycm_array[pos] += j/(local_count*1.)
mysolver(xcm_array, ycm_array, inverse, count)
final_array = zip(xcm_array,ycm_array,count)
To my surprise, using numba was slower or at best equal to the speed of the previous way. What am I doing wrong ?
Also, can this be done in Cython and will that be faster ?
I am using the included packages in the latest Anaconda python 2.7 distribution.

I believe the issue might be that you are timing jit'd code incorrectly. The first time you run the code, your timing includes the time it takes numba to compile the code. This is called warming up the jit. If you call it again, that cost is gone.
import numpy as np
import numba as nb
unique, inverse, count = np.unique(ccl_np, return_counts = True, return_inverse = True)
xcm_array = np.zeros(len(count),dtype=np.float32)
ycm_array = np.zeros(len(count),dtype=np.float32)
inverse = inverse.reshape(64,64)
def mysolver(xcm_array, ycm_array, inverse, count):
for i in range(64):
for j in range(64):
pos = inverse[i][j]
local_count = count[pos]
xcm_array[pos] += i/(local_count*1.)
ycm_array[pos] += j/(local_count*1.)
#nb.jit(nopython=True)
def mysolver_nb(xcm_array, ycm_array, inverse, count):
for i in range(64):
for j in range(64):
pos = inverse[i,j]
local_count = count[pos]
xcm_array[pos] += i/(local_count*1.)
ycm_array[pos] += j/(local_count*1.)
Then the timings with timeit which runs the code multiple times. First the plain python version:
In [4]:%timeit mysolver(xcm_array, ycm_array, inverse, count)
10 loops, best of 3: 25.8 ms per loop
and then with numba:
In [5]: %timeit mysolver_nb(xcm_array, ycm_array, inverse, count)
The slowest run took 3630.44 times longer than the fastest. This could mean that an intermediate result is being cached
10000 loops, best of 3: 33.1 µs per loop
The numba code is ~1000 times faster.

Speed up Python/Cython loops.

I've trying to get a loop in python to run as fast as possible. So I've dived into NumPy and Cython.
Here's the original Python code:
def calculate_bsf_u_loop(uvel,dy,dz):
"""
Calculate barotropic stream function from zonal velocity
uvel (t,z,y,x)
dy (y,x)
dz (t,z,y,x)
bsf (t,y,x)
"""
nt = uvel.shape[0]
nz = uvel.shape[1]
ny = uvel.shape[2]
nx = uvel.shape[3]
bsf = np.zeros((nt,ny,nx))
for jn in range(0,nt):
for jk in range(0,nz):
for jj in range(0,ny):
for ji in range(0,nx):
bsf[jn,jj,ji] = bsf[jn,jj,ji] + uvel[jn,jk,jj,ji] * dz[jn,jk,jj,ji] * dy[jj,ji]
return bsf
It's just a sum over k indices. Array sizes are nt=12, nz=75, ny=559, nx=1442, so ~725 million elements.
That took 68 seconds. Now, I've done it in cython as
import numpy as np
cimport numpy as np
cimport cython
#cython.boundscheck(False) # turn off bounds-checking for entire function
#cython.wraparound(False) # turn off negative index wrapping for entire function
## Use cpdef instead of def
## Define types for arrays
cpdef calculate_bsf_u_loop(np.ndarray[np.float64_t, ndim=4] uvel, np.ndarray[np.float64_t, ndim=2] dy, np.ndarray[np.float64_t, ndim=4] dz):
"""
Calculate barotropic stream function from zonal velocity
uvel (t,z,y,x)
dy (y,x)
dz (t,z,y,x)
bsf (t,y,x)
"""
## cdef the constants
cdef int nt = uvel.shape[0]
cdef int nz = uvel.shape[1]
cdef int ny = uvel.shape[2]
cdef int nx = uvel.shape[3]
## cdef loop indices
cdef ji,jj,jk,jn
## cdef. Note that the cdef is followed by cython type
## but the np.zeros function as python (numpy) type
cdef np.ndarray[np.float64_t, ndim=3] bsf = np.zeros([nt,ny,nx], dtype=np.float64)
for jn in xrange(0,nt):
for jk in xrange(0,nz):
for jj in xrange(0,ny):
for ji in xrange(0,nx):
bsf[jn,jj,ji] += uvel[jn,jk,jj,ji] * dz[jn,jk,jj,ji] * dy[jj,ji]
return bsf
and that took 49 seconds.
However, swapping the loop for
for jn in range(0,nt):
for jk in range(0,nz):
bsf[jn,:,:] = bsf[jn,:,:] + uvel[jn,jk,:,:] * dz[jn,jk,:,:] * dy[:,:]
only takes 0.29 seconds! Unfortunately, I can't do this in my full code.
Why is NumPy slicing so much faster than the Cython loop?
I thought NumPy was fast because it is Cython under the hood. So shouldn't they be of similar speed?
As you can see, I've disabled boundary checks in cython, and I've also compiled using "fast math". However, this only gives a tiny speedup.
Is there anyway to get a loop to be of similar speed as NumPy slicing, or is looping always slower than slicing?
Any help is greatly appreciated!
/Joakim

That code is screaming for numpy.einsum's's intervention, given that you are doing elementwise-multiplication and then sum-reduction on the second axis of the 4D product array, which essenti
ally numpy.einsum does in a highly efficient manner. To solve your case, you can use numpy.einsum in two ways -
bsf = np.einsum('ijkl,ijkl,kl->ikl',uvel,dz,dy)
bsf = np.einsum('ijkl,ijkl->ikl',uvel,dz)*dy
Runtime tests & Verify outputs -
In [100]: # Take a (1/5)th of original input shapes
...: original_shape = [12,75, 559,1442]
...: m,n,p,q = (np.array(original_shape)/5).astype(int)
...:
...: # Generate random arrays with given shapes
...: uvel = np.random.rand(m,n,p,q)
...: dy = np.random.rand(p,q)
...: dz = np.random.rand(m,n,p,q)
...:
In [101]: bsf = calculate_bsf_u_loop(uvel,dy,dz)
In [102]: print(np.allclose(bsf,np.einsum('ijkl,ijkl,kl->ikl',uvel,dz,dy)))
True
In [103]: print(np.allclose(bsf,np.einsum('ijkl,ijkl->ikl',uvel,dz)*dy))
True
In [104]: %timeit calculate_bsf_u_loop(uvel,dy,dz)
1 loops, best of 3: 2.16 s per loop
In [105]: %timeit np.einsum('ijkl,ijkl,kl->ikl',uvel,dz,dy)
100 loops, best of 3: 3.94 ms per loop
In [106]: %timeit np.einsum('ijkl,ijkl->ikl',uvel,dz)*dy
100 loops, best of 3: 3.96 ms per loo

cython times 10 slower

I am trying to boost up some calculations in python by using cython ...
In my calculations I will be doing double loops or more plus I can't always use numpy vectorization so I need to boost up the python loops with cython.
here I benchmark some simple calculation and it shows so far that cython is 10 times slower than using numpy. I am sure that numpy is optimized to the max and I doubt I could be able to beat its performance but still factor of 10 slower means I am doing something wrong. Suggestions ?
test.py
import numpy as np
from histogram import distances
import time
REPEAT = 10
def printTime(message, t):
print "%s total: %.7f(s) --> average: %.7f(s) %.7f(Ms)"%(message, t, t/REPEAT, 1000000*t/REPEAT)
DATA = np.array( np.random.random((10000, 3)), dtype=np.float32)
POINT = np.array( np.random.random((1,3)), dtype=np.float32)
# numpy histogram
r = REPEAT
startTime = time.clock()
while r:
diff = (DATA-POINT)%1
diffNumpy = np.where(diff<0, diff+1, diff)
distNumpy = np.sqrt( np.add.reduce(diff**2,1) )
r-=1
printTime("numpy", time.clock()-startTime)
# cython test
r = REPEAT
startTime = time.clock()
while r:
distCython = distances(POINT, DATA)
r-=1
printTime("cython", time.clock()-startTime)
histogram.pyx
import numpy as np
import cython
cimport cython
cimport numpy as np
DTYPE=np.float32
ctypedef np.float32_t DTYPE_C
#cython.nonecheck(False)
#cython.boundscheck(False)
#cython.wraparound(False)
def distances(np.ndarray[DTYPE_C, ndim=2] point, np.ndarray[DTYPE_C, ndim=2] data):
# declare variables
cdef int i
cdef float x,y,z
cdef np.ndarray[DTYPE_C, mode="c", ndim=1] dist = np.empty((data.shape[0]), dtype=DTYPE)
# loop
for i from 0 <= i < data.shape[0]:
# calculate distance
x = (data[i,0]-point[0,0])%1
y = (data[i,1]-point[0,1])%1
z = (data[i,2]-point[0,2])%1
# fold between 0 and 1
if x<0: x+=1
if y<0: y+=1
if z<0: z+=1
# assign to array
dist[i] = np.sqrt(x**2+y**2+z**2)
return dist
setup.py
from distutils.core import setup
from Cython.Build import cythonize
import numpy as np
setup(
ext_modules = cythonize("histogram.pyx"),
include_dirs=[np.get_include()]
)
to compile do the following
python setup.py build_ext --inplace
to launch benchmarch
python test.py
My results are
numpy total: 0.0153390(s) --> average: 0.0015339(s) 1533.9000000(Ms)
cython total: 0.1509920(s) --> average: 0.0150992(s) 15099.2000000(Ms)

Your problem is almost definitely
np.sqrt(x**2+y**2+z**2)
You should use the C sqrt function. It will look something like
from libc.math cimport sqrt
sqrt(x*x + y*y + z*z)

Numba code slower than pure python

I've been working on speeding up a resampling calculation for a particle filter. As python has many ways to speed it up, I though I'd try them all. Unfortunately, the numba version is incredibly slow. As Numba should result in a speed up, I assume this is an error on my part.
I tried 4 different versions:
Numba
Python
Numpy
Cython
The code for each is below:
import numpy as np
import scipy as sp
import numba as nb
from cython_resample import cython_resample
#nb.autojit
def numba_resample(qs, xs, rands):
n = qs.shape[0]
lookup = np.cumsum(qs)
results = np.empty(n)
for j in range(n):
for i in range(n):
if rands[j] < lookup[i]:
results[j] = xs[i]
break
return results
def python_resample(qs, xs, rands):
n = qs.shape[0]
lookup = np.cumsum(qs)
results = np.empty(n)
for j in range(n):
for i in range(n):
if rands[j] < lookup[i]:
results[j] = xs[i]
break
return results
def numpy_resample(qs, xs, rands):
results = np.empty_like(qs)
lookup = sp.cumsum(qs)
for j, key in enumerate(rands):
i = sp.argmax(lookup>key)
results[j] = xs[i]
return results
#The following is the code for the cython module. It was compiled in a
#separate file, but is included here to aid in the question.
"""
import numpy as np
cimport numpy as np
cimport cython
DTYPE = np.float64
ctypedef np.float64_t DTYPE_t
#cython.boundscheck(False)
def cython_resample(np.ndarray[DTYPE_t, ndim=1] qs,
np.ndarray[DTYPE_t, ndim=1] xs,
np.ndarray[DTYPE_t, ndim=1] rands):
if qs.shape[0] != xs.shape[0] or qs.shape[0] != rands.shape[0]:
raise ValueError("Arrays must have same shape")
assert qs.dtype == xs.dtype == rands.dtype == DTYPE
cdef unsigned int n = qs.shape[0]
cdef unsigned int i, j
cdef np.ndarray[DTYPE_t, ndim=1] lookup = np.cumsum(qs)
cdef np.ndarray[DTYPE_t, ndim=1] results = np.zeros(n, dtype=DTYPE)
for j in range(n):
for i in range(n):
if rands[j] < lookup[i]:
results[j] = xs[i]
break
return results
"""
if __name__ == '__main__':
n = 100
xs = np.arange(n, dtype=np.float64)
qs = np.array([1.0/n,]*n)
rands = np.random.rand(n)
print "Timing Numba Function:"
%timeit numba_resample(qs, xs, rands)
print "Timing Python Function:"
%timeit python_resample(qs, xs, rands)
print "Timing Numpy Function:"
%timeit numpy_resample(qs, xs, rands)
print "Timing Cython Function:"
%timeit cython_resample(qs, xs, rands)
This results in the following output:
Timing Numba Function:
1 loops, best of 3: 8.23 ms per loop
Timing Python Function:
100 loops, best of 3: 2.48 ms per loop
Timing Numpy Function:
1000 loops, best of 3: 793 µs per loop
Timing Cython Function:
10000 loops, best of 3: 25 µs per loop
Any idea why the numba code is so slow? I assumed it would be at least comparable to Numpy.
Note: if anyone has any ideas on how to speed up either the Numpy or Cython code samples, that would be nice too:) My main question is about Numba though.

The problem is that numba can't intuit the type of lookup. If you put a print nb.typeof(lookup) in your method, you'll see that numba is treating it as an object, which is slow. Normally I would just define the type of lookup in a locals dict, but I was getting a strange error. Instead I just created a little wrapper, so that I could explicitly define the input and output types.
#nb.jit(nb.f8[:](nb.f8[:]))
def numba_cumsum(x):
return np.cumsum(x)
#nb.autojit
def numba_resample2(qs, xs, rands):
n = qs.shape[0]
#lookup = np.cumsum(qs)
lookup = numba_cumsum(qs)
results = np.empty(n)
for j in range(n):
for i in range(n):
if rands[j] < lookup[i]:
results[j] = xs[i]
break
return results
Then my timings are:
print "Timing Numba Function:"
%timeit numba_resample(qs, xs, rands)
print "Timing Revised Numba Function:"
%timeit numba_resample2(qs, xs, rands)
Timing Numba Function:
100 loops, best of 3: 8.1 ms per loop
Timing Revised Numba Function:
100000 loops, best of 3: 15.3 µs per loop
You can go even a little faster still if you use jit instead of autojit:
#nb.jit(nb.f8[:](nb.f8[:], nb.f8[:], nb.f8[:]))
For me that lowers it from 15.3 microseconds to 12.5 microseconds, but it's still impressive how well autojit does.

Faster numpy version (10x speedup compared to numpy_resample)
def numpy_faster(qs, xs, rands):
lookup = np.cumsum(qs)
mm = lookup[None,:]>rands[:,None]
I = np.argmax(mm,1)
return xs[I]