I'm new to Numba and I'm trying to implement an old Fortran code in Python using Numba (version 0.54.1), but when I add parallel = True the program actually slows down. My program is very simple: I change the positions x and y in a L x L grid and for each position in the grid I perform a summation
import numpy as np
import numba as nb
#nb.njit(parallel=True)
def lyapunov_grid(x_grid, y_grid, k, N):
L = len(x_grid)
lypnv = np.zeros((L, L))
for ii in nb.prange(L):
for jj in range(L):
x = x_grid[ii]
y = y_grid[jj]
beta0 = 0
sumT11 = 0
for j in range(N):
y = (y - k*np.sin(x)) % (2*np.pi)
x = (x + y) % (2*np.pi)
J = np.array([[1.0, -k*np.cos(x)], [1.0, 1.0 - k*np.cos(x)]])
beta = np.arctan((-J[1,0]*np.cos(beta0) + J[1,1]*np.sin(beta0))/(J[0,0]*np.cos(beta0) - J[0,1]*np.sin(beta0)))
T11 = np.cos(beta0)*(J[0,0]*np.cos(beta) - J[1,0]*np.sin(beta)) - np.sin(beta0)*(J[0,1]*np.cos(beta) - J[1,1]*np.sin(beta))
sumT11 += np.log(abs(T11))/np.log(2)
beta0 = beta
lypnv[ii, jj] = sumT11/N
return lypnv
# Compile
_ = lyapunov_grid(np.linspace(0, 1, 10), np.linspace(0, 1, 10), 1, 10)
# Parameters
N = int(1e3)
L = 128
pi = np.pi
k = 1.5
# Limits of the phase space
x0 = -pi
xf = pi
y0 = -pi
yf = pi
# Grid positions
x = np.linspace(x0, xf, L, endpoint=True)
y = np.linspace(y0, yf, L, endpoint=True)
lypnv = lyapunov_grid(x, y, k, N)
With parallel=False it takes about 8s to run, however with parallel=True it takes about 14s. I also tested with another code from https://github.com/animator/mandelbrot-numba and in this case the parallelization works.
import math
import numpy as np
import numba as nb
WIDTH = 1000
MAX_ITER = 1000
#nb.njit(parallel=True)
def mandelbrot(width, max_iter):
pixels = np.zeros((width, width, 3), dtype=np.uint8)
for y in nb.prange(width):
for x in range(width):
c0 = complex(3.0*x/width - 2, 3.0*y/width - 1.5)
c = 0
for i in range(1, max_iter):
if abs(c) > 2:
log_iter = math.log(i)
pixels[y, x, :] = np.array([int(255*(1+math.cos(3.32*log_iter))/2),
int(255*(1+math.cos(0.774*log_iter))/2),
int(255*(1+math.cos(0.412*log_iter))/2)],
dtype=np.uint8)
break
c = c * c + c0
return pixels
# compile
_ = mandelbrot(WIDTH, 10)
calcpixels = mandelbrot(WIDTH, MAX_ITER)
One main issue is that the second function call compile the function again. Indeed, the types of the provided arguments change: in the first call the third argument is an integer (int transformed to a np.int_) while in the second call the third argument (k) is a floating point number (float transformed to a np.float64). Numba recompiles the function for different parameter types because they are deduced from the type of the arguments and it does not know you want to use a np.float64 type for the third argument (since the first time the function is compiled with for a np.int_ type). One simple solution to fix the problem is to change the first call to:
_ = lyapunov_grid(np.linspace(0, 1, 10), np.linspace(0, 1, 10), 1.0, 10)
However, this is not a robust way to fix the problem. You can specify the parameter types to Numba so it will compile the function at declaration time. This also remove the need to artificially call the function (with useless parameters).
#nb.njit('float64[:,:](float64[::1], float64[::1], float64, float64)', parallel=True)
Note that (J[0,0]*np.cos(beta0) - J[0,1]*np.sin(beta0)) is zero the first time resulting in a division by 0.
Another main issue comes from the allocations of many small arrays in the loop causing a contention of the standard allocator (see this post for more information). While Numba could theoretically optimize it (ie. replace the array with local variables), it actually does not, resulting in a huge slowdown and a contention. Hopefully, in your case, you do not need to actually create the array. At last, you can create it only in the encompassing loop and modify it in the innermost loop. Here is the optimized code:
#nb.njit('float64[:,:](float64[::1], float64[::1], float64, float64)', parallel=True)
def lyapunov_grid(x_grid, y_grid, k, N):
L = len(x_grid)
lypnv = np.zeros((L, L))
for ii in nb.prange(L):
J = np.ones((2, 2), dtype=np.float64)
for jj in range(L):
x = x_grid[ii]
y = y_grid[jj]
beta0 = 0
sumT11 = 0
for j in range(N):
y = (y - k*np.sin(x)) % (2*np.pi)
x = (x + y) % (2*np.pi)
J[0, 1] = -k*np.cos(x)
J[1, 1] = 1.0 - k*np.cos(x)
beta = np.arctan((-J[1,0]*np.cos(beta0) + J[1,1]*np.sin(beta0))/(J[0,0]*np.cos(beta0) - J[0,1]*np.sin(beta0)))
T11 = np.cos(beta0)*(J[0,0]*np.cos(beta) - J[1,0]*np.sin(beta)) - np.sin(beta0)*(J[0,1]*np.cos(beta) - J[1,1]*np.sin(beta))
sumT11 += np.log(abs(T11))/np.log(2)
beta0 = beta
lypnv[ii, jj] = sumT11/N
return lypnv
Here is the results on a old 2-core machine (with 4 hardware threads):
Original sequential: 15.9 s
Original parallel: 11.9 s
Fix-build sequential: 15.7 s
Fix-build parallel: 10.1 s
Optimized sequential: 2.73 s
Optimized parallel: 0.94 s
The optimized implementation is much faster than the others. The parallel optimized version scale very well compared than the original one (2.9 times faster than the sequential one). Finally, the best version is about 12 times faster than the original parallel version. I expect a much faster computation on a recent machine with many more cores.
I want to find (efficiently) all pairs of points that are closer than some distance max_d. My current method, using cdist, is:
import numpy as np
from scipy.spatial.distance import cdist
def close_pairs(X,max_d):
d = cdist(X,X)
I,J = (d<max_d).nonzero()
IJ = np.sort(np.vstack((I,J)), axis=0)
# remove diagonal element
IJ = IJ[:,np.diff(IJ,axis=0).ravel()<>0]
# remove duplicate
dt = np.dtype([('i',int),('j',int)])
pairs = np.unique(IJ.T.view(dtype=dt)).view(int).reshape(-1,2)
return pairs
def test():
X = np.random.rand(100,2)*20
p = close_pairs(X,2)
from matplotlib import pyplot as plt
plt.clf()
plt.plot(X[:,0],X[:,1],'.r')
plt.plot(X[p,0].T,X[p,1].T,'-b')
But I think this is overkill (and not very readable), because most of the work is done only to remove distance-to-self and duplicates.
My main question is: is there a better way to do it?
(Note: the type of outputs (array, set, ...) is not important at this point)
My current thinking is on using pdist which returns a condensed distance array which contains only the right pairs. However, once I found the suitable coordinates k's from the condensed distance array, how do I compute which i,j pairs it is equivalent to?
So the alternative question is: is there an easy way to get the list of coordinate pairs relative to the entries of pdist outputs:
a function f(k)->i,j
such that cdist(X,X)[i,j] = pdist(X)[k]
In my experience, there are two fastest ways to find neighbor lists in 3D. One is to use a most naive double-for-loop code written in C++ or Cython (in my case, both). It runs in N^2, but is very fast for small systems. The other way is to use a linear time algorithm. Scipy ckdtree is a good choice, but has limitations. Neighbor list finders from molecular dynamics software are most powerful, but are very hard to wrap, and likely have slow initialization time.
Below I compare four methods:
Naive cython code
Wrapper around OpenMM (is very hard to install, see below)
Scipy.spatial.ckdtree
scipy.spatial.distance.pdist
Test setup: n points scattered in a rectangular box at volume density 0.2. System size ranging from 10 to a 1000000 (a million) particles. Contact radius is taken from 0.5, 1, 2, 4, 7, 10. Note that because density is 0.2, at contact radius 0.5 we'll have on average about 0.1 contacts per particle, at 1 = 0.8, at 2 = 6.4, and at 10 - about 800! Contact finding was repeated several times for small systems, done once for systems >30k particles. If time per call exceeded 5 seconds, the run was aborted.
Setup: dual xeon 2687Wv3, 128GB RAM, Ubuntu 14.04, python 2.7.11, scipy 0.16.0, numpy 1.10.1. None of the code was using parallel optimizations (except for OpenMM, though parallel part went so quick that it was not even noticeable on a CPU graph, most of the time was spend piping data to-from OpenMM).
Results: Note that plots below are logscale, and spread over 6 orders of magnitude. Even small visual difference may be actually 10-fold.
For systems less than 1000 particles, Cython code was always faster. However, after 1000 particles results are dependent on the contact radius. pdist implementation was always slower than cython, and takes much more memory, because it explicitly creates a distance matrix, which is slow because of sqrt.
At small contact radius (<1 contact per particle), ckdtree is a good choice for all system sizes.
At medium contact radius, (5-50 contacts per particle) naive cython implementation is the best up to 10000 particles, then OpenMM starts to win by about several orders of magnitude, but ckdtree performs just 3-10 times worse
At high contact radius (>200 contacts per particle) naive methods work up to 100k or 1M particles, then OpenMM may win.
Installing OpenMM is very tricky; you can read more in http://bitbucket.org/mirnylab/openmm-polymer file "contactmaps.py" or in the readme. However, the results below show that it is only advantageous for 5-50 contacts per particle, for N>100k particles.
Cython code below:
import numpy as np
cimport numpy as np
cimport cython
cdef extern from "<vector>" namespace "std":
cdef cppclass vector[T]:
cppclass iterator:
T operator*()
iterator operator++()
bint operator==(iterator)
bint operator!=(iterator)
vector()
void push_back(T&)
T& operator[](int)
T& at(int)
iterator begin()
iterator end()
np.import_array() # initialize C API to call PyArray_SimpleNewFromData
cdef public api tonumpyarray(int* data, long long size) with gil:
if not (data and size >= 0): raise ValueError
cdef np.npy_intp dims = size
#NOTE: it doesn't take ownership of `data`. You must free `data` yourself
return np.PyArray_SimpleNewFromData(1, &dims, np.NPY_INT, <void*>data)
#cython.boundscheck(False)
#cython.wraparound(False)
def contactsCython(inArray, cutoff):
inArray = np.asarray(inArray, dtype = np.float64, order = "C")
cdef int N = len(inArray)
cdef np.ndarray[np.double_t, ndim = 2] data = inArray
cdef int j,i
cdef double curdist
cdef double cutoff2 = cutoff * cutoff # IMPORTANT to avoid slow sqrt calculation
cdef vector[int] contacts1
cdef vector[int] contacts2
for i in range(N):
for j in range(i+1, N):
curdist = (data[i,0] - data[j,0]) **2 +(data[i,1] - data[j,1]) **2 + (data[i,2] - data[j,2]) **2
if curdist < cutoff2:
contacts1.push_back(i)
contacts2.push_back(j)
cdef int M = len(contacts1)
cdef np.ndarray[np.int32_t, ndim = 2] contacts = np.zeros((M,2), dtype = np.int32)
for i in range(M):
contacts[i,0] = contacts1[i]
contacts[i,1] = contacts2[i]
return contacts
Compilation (or makefile) for Cython code:
cython --cplus fastContacts.pyx
g++ -g -march=native -Ofast -fpic -c fastContacts.cpp -o fastContacts.o `python-config --includes`
g++ -g -march=native -Ofast -shared -o fastContacts.so fastContacts.o `python-config --libs`
Testing code:
from __future__ import print_function, division
import signal
import time
from contextlib import contextmanager
import matplotlib
import matplotlib.pyplot as plt
import numpy as np
from scipy.spatial import ckdtree
from scipy.spatial.distance import pdist
from contactmaps import giveContactsOpenMM # remove this unless you have OpenMM and openmm-polymer libraries installed
from fastContacts import contactsCython
class TimeoutException(Exception): pass
#contextmanager
def time_limit(seconds):
def signal_handler(signum, frame):
raise TimeoutException("Timed out!")
signal.signal(signal.SIGALRM, signal_handler)
signal.alarm(seconds)
try:
yield
finally:
signal.alarm(0)
matplotlib.rcParams.update({'font.size': 8})
def close_pairs_ckdtree(X, max_d):
tree = ckdtree.cKDTree(X)
pairs = tree.query_pairs(max_d)
return np.array(list(pairs))
def condensed_to_pair_indices(n, k):
x = n - (4. * n ** 2 - 4 * n - 8 * k + 1) ** .5 / 2 - .5
i = x.astype(int)
j = k + i * (i + 3 - 2 * n) / 2 + 1
return np.array([i, j]).T
def close_pairs_pdist(X, max_d):
d = pdist(X)
k = (d < max_d).nonzero()[0]
return condensed_to_pair_indices(X.shape[0], k)
a = np.random.random((100, 3)) * 3 # test set
methods = {"cython": contactsCython, "ckdtree": close_pairs_ckdtree, "OpenMM": giveContactsOpenMM,
"pdist": close_pairs_pdist}
# checking that each method gives the same value
allUniqueInds = []
for ind, method in methods.items():
contacts = method(a, 1)
uniqueInds = contacts[:, 0] + 100 * contacts[:, 1] # unique index of each contacts
allUniqueInds.append(np.sort(uniqueInds)) # adding sorted unique conatcts
for j in allUniqueInds:
assert np.allclose(j, allUniqueInds[0])
# now actually doing testing
repeats = [30,30,30, 30, 30, 20, 20, 10, 5, 3, 2 , 1, 1, 1]
sizes = [10,30,100, 200, 300, 500, 1000, 2000, 3000, 10000, 30000, 100000, 300000, 1000000]
systems = [[np.random.random((n, 3)) * ((n / 0.2) ** 0.333333) for k in range(repeat)] for n, repeat in
zip(sizes, repeats)]
for j, radius in enumerate([0.5, 1, 2, 4, 7, 10]):
plt.subplot(2, 3, j + 1)
plt.title("Radius = {0}; {1:.2f} cont per particle".format(radius, 0.2 * (4 / 3 * np.pi * radius ** 3)))
times = {i: [] for i in methods}
for name, method in methods.items():
for n, system, repeat in zip(sizes, systems, repeats):
if name == "pdist" and n > 30000:
break # memory issues
st = time.time()
try:
with time_limit(5 * repeat):
for ind in range(repeat):
k = len(method(system[ind], radius))
except:
print("Run aborted")
break
end = time.time()
mytime = (end - st) / repeat
times[name].append((n, mytime))
print("{0} radius={1} n={2} time={3} repeat={4} contPerParticle={5}".format(name, radius, n, mytime,repeat, 2 * k / n))
for name in sorted(times.keys()):
plt.plot(*zip(*times[name]), label=name)
plt.xscale("log")
plt.yscale("log")
plt.xlabel("System size")
plt.ylabel("Time (seconds)")
plt.legend(loc=0)
plt.show()
Here's how to do it with the cKDTree module. See query_pairs
import numpy as np
from scipy.spatial.distance import cdist
from scipy.spatial import ckdtree
def close_pairs(X,max_d):
d = cdist(X,X)
I,J = (d<max_d).nonzero()
IJ = np.sort(np.vstack((I,J)), axis=0)
# remove diagonal element
IJ = IJ[:,np.diff(IJ,axis=0).ravel()<>0]
# remove duplicate
dt = np.dtype([('i',int),('j',int)])
pairs = np.unique(IJ.T.view(dtype=dt)).view(int).reshape(-1,2)
return pairs
def close_pairs_ckdtree(X, max_d):
tree = ckdtree.cKDTree(X)
pairs = tree.query_pairs(max_d)
return np.array(list(pairs))
def test():
np.random.seed(0)
X = np.random.rand(100,2)*20
p = close_pairs(X,2)
q = close_pairs_ckdtree(X, 2)
from matplotlib import pyplot as plt
plt.plot(X[:,0],X[:,1],'.r')
plt.plot(X[p,0].T,X[p,1].T,'-b')
plt.figure()
plt.plot(X[:,0],X[:,1],'.r')
plt.plot(X[q,0].T,X[q,1].T,'-b')
plt.show()
t
I finally found it myself. The function converting indices k in condensed distance array to equivalent i,j in square distance array is:
def condensed_to_pair_indices(n,k):
x = n-(4.*n**2-4*n-8*k+1)**.5/2-.5
i = x.astype(int)
j = k+i*(i+3-2*n)/2+1
return i,j
I had to play a little with sympy to find it. Now, to compute all point pairs than are less than a given distance apart:
def close_pairs_pdist(X,max_d):
d = pdist(X)
k = (d<max_d).nonzero()[0]
return condensed_to_pair_indices(X.shape[0],k)
As expected, it is more efficient than the other methods (but I did not test ckdtree). I will update the timeit answer.
slightly faster, didnt test the time difference thoroughly, but if i ran it a few times, it gave a time of about 0.0755529403687 for my method, and 0.0928771495819 for yours. I use the triu method to get rid of upper triangle of the array (where duplicates are) including diagonal (which is where the self-distances are), and i dont sort either, since if you plot it, it does not matter if i plot them in order or not. So i guess it speeds up about 15% or so
import numpy as np
from scipy.spatial.distance import cdist
from scipy.misc import comb
def close_pairs(X,max_d):
d = cdist(X,X)
I,J = (d<max_d).nonzero()
IJ = np.sort(np.vstack((I,J)), axis=0)
# remove diagonal element
IJ = IJ[:,np.diff(IJ,axis=0).ravel()<>0]
# remove duplicate
dt = np.dtype([('i',int),('j',int)])
pairs = np.unique(IJ.T.view(dtype=dt)).view(int).reshape(-1,2)
return pairs
def close_pairs1(X,max_d):
d = cdist(X,X)
d1 = np.triu_indices(len(X)) # indices of the upper triangle including the diagonal
d[d1] = max_d+1 # value that will not get selected when doing d<max_d in the next line
I,J = (d<max_d).nonzero()
pairs = np.vstack((I,J)).T
return pairs
def close_pairs3(X, max_d):
d = pdist(X)
n = len(X)
pairs = np.zeros((0,2))
for i in range(n):
for j in range(i+1,n):
# formula from http://docs.scipy.org/doc/scipy/reference/generated/scipy.spatial.distance.squareform.html
a=d[int(comb(n,2)-comb(n-i,2)+j-i-1+0.1)] # the +0.1 is because otherwise i get floating point trouble
if(a<max_d):
pairs = np.r_[pairs, np.array([i,j])[None,:]]
return pairs
def close_pairs4(X, max_d):
d = pdist(X)
n = len(X)
a = np.where(d<max_d)[0]
i = np.arange(n)[:,None]
j = np.arange(n)[None,:]
b = np.array(comb(n,2)-comb(n-i,2)+j-i-1+0.1, dtype=int)
d1 = np.tril_indices(n)
b[d1] = -1
pairs = np.zeros((0,2), dtype=int)
# next part is the bottleneck: the np.where each time,
for v in a:
i, j = np.where(v==b)
pairs = np.r_[pairs, np.array([i[0],j[0]])[None,:]]
return pairs
def close_pairs5(X, max_d):
t0=time.time()
d = pdist(X)
n = len(X)
a = np.where(d<max_d)[0]
i = np.arange(n)[:,None]
j = np.arange(n)[None,:]
t1 = time.time()
b = np.array(comb(n,2)-comb(n-i,2)+j-i-1+0.1, dtype=int)
d1 = np.tril_indices(n)
b[d1] = -1
t2 = time.time()
V = b[:,:,None]-a[None,None,:] # takes a little time
t3 = time.time()
p = np.where(V==0) # takes most of the time, thought that removing the for-loop from the previous method might improve it, but it does not do that much. This method contains the formula you wanted though, but apparently it is still faster if you use the cdist methods
t4 = time.time()
pairs = np.vstack((p[0],p[1])).T
print t4-t3,t3-t2, t2-t1, t1-t0
return pairs
def test():
X = np.random.rand(1000,2)*20
import time
t0 = time.time()
p = close_pairs(X,2)
t1 = time.time()
p2 = close_pairs1(X,2)
t2 = time.time()
print t2-t1, t1-t0
from matplotlib import pyplot as plt
plt.figure()
plt.clf()
plt.plot(X[:,0],X[:,1],'.r')
plt.plot(X[p,0].T,X[p,1].T,'-b')
plt.figure()
plt.clf()
plt.plot(X[:,0],X[:,1],'.r')
plt.plot(X[p2,0].T,X[p2,1].T,'-b')
plt.show()
test()
NOTE: plotting laggs if you do it for 1K points, but it needs 1K points to compare speeds, but i checked that it works correctly when plotting it if doing it with 100 points
The speed difference is something like ten to twenty percent, and i think it will not get much better than this, since i got rid of all the sorting and unique elements things, so the part that takes most of the time probably is the d = cdist(X, X) line
Edit: some more testing shows that in those times, that cdist line takes up about 0.065 sec, while the rest for your method is about 0.02 and for me about 0.015 sec or so. Conclusion: the main bottleneck of your code is the d = cdist(X, X) line, and the stuff i changed speeds up the rest of the code you got, but the main bottleneck stays
Edit: added the method close_pairs3, which gives you the formula, but speed blows, (still need to figure out how to invert that formula, and than it will be superfast, will do that tomorrow - will use np.where(pdist(X)
Edit: added method close_pairs4, which is slightly better than 3, and explains what happens, but is veeery slow, and same with method 5, does not have that for-loop, but is still very slow
I made some code to compare the proposed solutions.
Note: I use scipy 0.11 and cannot use the ckdtree solution (only kdtree) which I expect to be slower. Could anyone with scipy v0.12+ run this code?
import numpy as np
from scipy.spatial.distance import cdist, pdist
from scipy.spatial import ckdtree
from scipy.spatial import kdtree
def close_pairs(X,max_d):
d = cdist(X,X)
I,J = (d<max_d).nonzero()
IJ = np.sort(np.vstack((I,J)), axis=0)
# remove diagonal element
IJ = IJ[:,np.diff(IJ,axis=0).ravel()<>0]
# remove duplicate
dt = np.dtype([('i',int),('j',int)])
pairs = np.unique(IJ.T.view(dtype=dt)).view(int).reshape(-1,2)
return pairs
def condensed_to_pair_indices(n,k):
x = n-(4.*n**2-4*n-8*k+1)**.5/2-.5
i = x.astype(int)
j = k+i*(i+3-2*n)/2+1
return i,j
def close_pairs_pdist(X,max_d):
d = pdist(X)
k = (d<max_d).nonzero()[0]
return condensed_to_pair_indices(X.shape[0],k)
def close_pairs_triu(X,max_d):
d = cdist(X,X)
d1 = np.triu_indices(len(X)) # indices of the upper triangle including the diagonal
d[d1] = max_d+1 # value that will not get selected when doing d<max_d in the next line
I,J = (d<max_d).nonzero()
pairs = np.vstack((I,J)).T
return pairs
def close_pairs_ckdtree(X, max_d):
tree = ckdtree.cKDTree(X)
pairs = tree.query_pairs(max_d)
return pairs # remove the conversion as it is not required
def close_pairs_kdtree(X, max_d):
tree = kdtree.KDTree(X)
pairs = tree.query_pairs(max_d)
return pairs # remove the conversion as it is not required
methods = [close_pairs, close_pairs_pdist, close_pairs_triu, close_pairs_kdtree] #, close_pairs_ckdtree]
def time_test(n=[10,50,100], max_d=[5,10,50], iter_num=100):
import timeit
for method in methods:
print '-- time using ' + method.__name__ + ' ---'
for ni in n:
for d in max_d:
setup = '\n'.join(['import numpy as np','import %s' % __name__,'np.random.seed(0)','X = np.random.rand(%d,2)*100'%ni])
stmt = 'close_pairs.%s(X,%f)' % (method.__name__,d)
time = timeit.timeit(stmt=stmt, setup=setup, number=iter_num)/iter_num
print 'n=%3d, max_d=%2d: \t%.2fms' % (ni, d,time*1000)
Output of time_test(iter_num=10,n=[20,100,500],max_d=[1,5,10]) are:
-- time using close_pairs ---
n= 20, max_d= 1: 0.22ms
n= 20, max_d= 5: 0.16ms
n= 20, max_d=10: 0.21ms
n=100, max_d= 1: 0.41ms
n=100, max_d= 5: 0.53ms
n=100, max_d=10: 0.97ms
n=500, max_d= 1: 7.12ms
n=500, max_d= 5: 12.28ms
n=500, max_d=10: 33.41ms
-- time using close_pairs_pdist ---
n= 20, max_d= 1: 0.11ms
n= 20, max_d= 5: 0.10ms
n= 20, max_d=10: 0.11ms
n=100, max_d= 1: 0.19ms
n=100, max_d= 5: 0.19ms
n=100, max_d=10: 0.19ms
n=500, max_d= 1: 2.31ms
n=500, max_d= 5: 2.82ms
n=500, max_d=10: 2.49ms
-- time using close_pairs_triu ---
n= 20, max_d= 1: 0.17ms
n= 20, max_d= 5: 0.16ms
n= 20, max_d=10: 0.16ms
n=100, max_d= 1: 0.83ms
n=100, max_d= 5: 0.80ms
n=100, max_d=10: 0.80ms
n=500, max_d= 1: 23.64ms
n=500, max_d= 5: 22.87ms
n=500, max_d=10: 22.96ms
-- time using close_pairs_kdtree ---
n= 20, max_d= 1: 1.71ms
n= 20, max_d= 5: 1.69ms
n= 20, max_d=10: 1.96ms
n=100, max_d= 1: 34.99ms
n=100, max_d= 5: 35.47ms
n=100, max_d=10: 34.91ms
n=500, max_d= 1: 253.87ms
n=500, max_d= 5: 255.05ms
n=500, max_d=10: 256.66ms
Conclusion:
The overall fastest method is close_pairs_pdist
The initial method is relatively fast, but sensitive to both the number of samples and the percentage of pairs to return
both the close_pairs_triu and close_pairs_kdtree are sensitive to the number of samples but relatively insensitive to the number of outputs.
the close_pairs_triu method is much faster than close_pairs_kdtree
However, the ckdtree method needs to be tested.