I have an array of shape (N, 3) and I'd like to randomly shuffle the rows. N is on the order of 100,000.
I discovered that np.random.shuffle was bottlenecking my application. I tried replacing the shuffle with a call to np.random.choice and experienced a 10x speed-up. What's going on here? Why is it so much faster to call np.random.choice? Does the np.random.choice version generate a uniformly distributed shuffle?
import timeit
task_choice = '''
N = 100000
x = np.zeros((N, 3))
inds = np.random.choice(N, N, replace=False)
x[np.arange(N), :] = x[inds, :]
'''
task_shuffle = '''
N = 100000
x = np.zeros((N, 3))
np.random.shuffle(x)
'''
task_permute = '''
N = 100000
x = np.zeros((N, 3))
x = np.random.permutation(x)
'''
setup = 'import numpy as np'
timeit.timeit(task_choice, setup=setup, number=10)
>>> 0.11108078400138766
timeit.timeit(task_shuffle, setup=setup, number=10)
>>> 1.0411593900062144
timeit.timeit(task_permute, setup=setup, number=10)
>>> 1.1140159380011028
Edit: For anyone curious, I decided to go with the following solution since it is readable and outperformed all other methods in my benchmarks:
task_ind_permute = '''
N = 100000
x = np.zeros((N, 3))
inds = np.random.permutation(N)
x[np.arange(N), :] = x[inds, :]
'''
You're comparing very different sized arrays here. In your first example, although you create an array of zeros, you simply use random.choice(100000, 100000), which pulls 100000 random values between 1-100000. In your second example your are shuffling a (100000, 3) shape array.
>>> x.shape
(100000, 3)
>>> np.random.choice(N, N, replace=False).shape
(100000,)
Timings on more equivalent samples:
In [979]: %timeit np.random.choice(N, N, replace=False)
2.6 ms ± 201 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
In [980]: x = np.arange(100000)
In [981]: %timeit np.random.shuffle(x)
2.29 ms ± 67.3 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
In [982]: x.shape == np.random.choice(N, N, replace=False).shape
Out[982]: True
permutation and shuffle are linked, in fact permutation calls shuffle under the hood!!
The reason why shuffle is slower than permutation for multidimensional array is that permutation only need to shuffle the index along the first axis. Thus becomes a special case of shuffle of 1d array (the 1st if-else block).
This special case is also explained in the source as well:
# We trick gcc into providing a specialized implementation for
# the most common case, yielding a ~33% performance improvement.
# Note that apparently, only one branch can ever be specialized.
For shuffle on the otherhand, multidimensional ndarray operation requires a bounce buffer, creating that buffer, especially when the dimension is relative big, becomes expensive. Additionally, we can no longer use the trick mentioned above that helps the 1d case.
With replace=False and using choice to generate a new array of the same size, choice and permutation is the same, see here. The extra time would have to come from the time spend in creating intermediate index arrays.
Related
I want to calculate a large distance matrix, based on a higher dimensional vector. For instance, I have 1000 instances each represented by 20 vectors of length 10. The distance between each two instances is given by the mean distance between each of the 20 vectors associated to each vector. So I want to go from a 1000 by 20 by 10 matrix to a 1000 by 1000 (lower-triangular) matrix. Because these calculations can get slow, I want to use Dask distributed to block the algorithm and spread it over several CPU's. Below is how far I've gotten:
Preamble
import itertools
import random
import numpy as np
import dask.array
from dask.distributed import Client
The distance function is defined by
def distance(u, v):
result = np.empty([int((len(u)*(len(u)+1))/2)], dtype=float)
for i, j in itertools.product(range(len(u)),range(len(v))):
if j <= i:
differences = []
k = int(((i*(i+1))/2 +j-1)+1)
for x,y in itertools.product(u[i], v[j]):
difference = np.abs(np.array(x) - np.array(y)).sum(axis=1)
differences.apply(difference)
result[k] = np.mean(differences)
return result
and returns an array of length n*(n+1)/2 to describe the lower triangular matrix for this block of the distance matrix.
def distance_matrix(X):
X = np.asarray(X, dtype=object)
X = dask.array.from_array(X, (100, 20, 10)).astype(float)
print("chunksize: ", X.chunksize)
resulting_length = [int((X.chunksize[0]*(X.chunksize[0])+1)/2)]
result = dask.array.map_blocks(distance, X, X, chunks=(resulting_length), drop_axis=[1,2], dtype=float)
return result.compute()
I split up the input array in chunks and use dask.array.map_blocks to apply the distance calculation to all the blocks.
if __name__ == '__main__':
workers = 6
X = np.array([[[random.random() for _ in range(10)] for _ in range(20)] for _ in range(1000)])
client = Client(n_workers=workers)
results = similarity_matrix(X)
client.close()
print(results)
Unfortunately, this approach returns the wrong length of array at the end of the process. Would somebody to help me out here? I don't have much experience in distributed computing.
I'm a big fan of dask, but this problem is way too small to need it. The runtime issue you're seeing is because you are looping through each element in python rather than using vectorized operations in numpy.
As with many packages in python, numpy relies on highly efficient compiled code written in other, faster languages such as C to carry out array operations. When you do something like an array operation A + B numpy calls these fast routines once, and the array operations are carried out within a highly optimized C routine. There is overhead involved with making calls to other languages, but this is overwhelmed by the performance gain due to the single call to a very fast routine. If instead you loop over every element, adding cell-wise, you have a (slow) python process, and on each element, this calls the C code, which adds overhead for each element of the array. Because of this, you actually would be better off not using numpy if you're going to do this once for each element.
To implement this in a vectorized manner, you can exploit numpy's broadcasting rules to ensure the first dimensions of your two arrays expand to a new dimension. I don't totally understand what's going on in your distance function, but you could extend this simple version to do whatever you want:
In [1]: import numpy as np
In [2]: A = np.random.random((1000, 20))
...: B = np.random.random((1000, 20))
In [3]: distance = np.abs(A[:, np.newaxis, :] - B[np.newaxis, :, :]).sum(axis=-1)
In [4]: distance
Out[4]:
array([[7.22985776, 7.76185666, 5.61824886, ..., 7.62092039, 6.35189562,
7.06365986],
[5.73359499, 5.8422105 , 7.2644021 , ..., 5.72230353, 6.79390303,
5.03074007],
[7.27871151, 8.6856818 , 5.97489449, ..., 8.86620029, 7.49875638,
6.57389575],
...,
[7.67783107, 7.24419076, 4.17941596, ..., 8.68674754, 6.65078093,
5.67279811],
[7.1550136 , 6.10590227, 5.75417987, ..., 7.05953998, 5.8306628 ,
6.55112672],
[5.81748615, 6.79246838, 6.95053088, ..., 7.63994705, 6.77720511,
7.5663236 ]])
In [5]: distance.shape
Out[5]: (1000, 1000)
The performance difference can be seen clearly against a looped implementation:
In [6]: %%timeit
...: np.abs(A[:, np.newaxis, :] - B[np.newaxis, :, :]).sum(axis=-1)
...:
...:
45 ms ± 326 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
In [7]: %%timeit
...: distances = np.empty((1000, 1000))
...: for i in range(1000):
...: for j in range(1000):
...: distances[i, j] = np.abs(A[i, :] - B[j, :]).sum()
...:
2.42 s ± 7.6 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
The looped version takes more than 50x as long!
I am looking for suggestions on the most efficient way to solve the following problem:
I have two arrays called A and B. They are both of shape NxNx3. They represent two 2D matrix of positions, where each position is a vector of x, y, and z coordinates.
I want to create a new array, called C, of shape NxN, where C[i, j] is the dot product of the vectors A[i, j] and B[i, j].
Here are the solutions I've come up with so far. The first uses the numpy's einsum function (which is beautifully described here). The second uses numpy's broadcasting rules along with its sum function.
>>> import numpy as np
>>> A = np.random.randint(0, 10, (100, 100, 3))
>>> B = np.random.randint(0, 10, (100, 100, 3))
>>> C = np.einsum("ijk,ijk->ij", A, B)
>>> D = np.sum(A * B, axis=2)
>>> np.allclose(C, D)
True
Is there a faster way? I've heard murmurs that numpy's tensordot function can be blazing fast but I've always struggled to understand it. What about using numpy's dot, or inner functions?
For some context, the A and B arrays will typically have between 100 and 1000 elements.
Any guidance is much appreciated!
With a bit of reshaping, we can use matmul. The idea is to treat the first 2 dimensions as the 'batch' dimensions, and to the dot on the last:
In [278]: E = A[...,None,:]#B[...,:,None]
In [279]: E.shape
Out[279]: (100, 100, 1, 1)
In [280]: E = np.squeeze(A[...,None,:]#B[...,:,None])
In [281]: np.allclose(C,E)
Out[281]: True
In [282]: timeit E = np.squeeze(A[...,None,:]#B[...,:,None])
130 µs ± 2.01 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
In [283]: timeit C = np.einsum("ijk,ijk->ij", A, B)
90.2 µs ± 1.53 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
Comparative timings can be a bit tricky. In the current versions, einsum can take different routes depending on the dimensions. In some cases it appears to delegate the task to matmul (or at least the same underlying BLAS-like code). While it's nice that einsum is faster in this test, I wouldn't generalize that.
tensordot just reshapes (and if needed transposes) the arrays so it can apply the ordinary 2d np.dot. Actually it doesn't work here because you are treating the first 2 axes as a 'batch', where as it does an outer product on them.
If you don't care about the details of what I'm trying to implement, just skip past the lower horizontal line
I am trying to do a bootstrap error estimation on some statistic with NumPy. I have an array x, and wish to compute the error on the statistic f(x) for which usual gaussian assumptions in error analysis do not hold. x is very large.
To do this, I resample x using numpy.random.choice(), where the size of my resample is the size of the original array, with replacement:
resample = np.random.choice(x, size=len(x), replace=True)
This gives me a new realization of x. This operation must now be repeated ~1,000 times to give an accurate error estimate. If I generate 1,000 resamples of this nature;
resamples = [np.random.choice(x, size=len(x), replace=True) for i in range(1000)]
and then compute the statistic f(x) on each realization;
results = [f(arr) for arr in resamples]
then I have inferred the error of f(x) to be something like
np.std(results)
the idea being that even though f(x) itself cannot be described using gaussian error analysis, a distribution of f(x) measures subject to random error can be.
Okay, so that's a bootstrap. Now, my problem is that the line
resamples = [np.random.choice(x, size=len(x), replace=True) for i in range(1000)]
is very slow for large arrays. Is there a smarter way to do this without a list comprehension? The second list comprehension
results = [f(arr) for arr in resamples]
can be pretty slow too, depending on the details of the function f(x).
Since we are allowing repetitions, we could generate all the indices in one go with np.random.randint and then simply index to get resamples equivalent, like so -
num_samples = 1000
idx = np.random.randint(0,len(x),size=(num_samples,len(x)))
resamples_arr = x[idx]
One more approach would be to generate random number from uniform distribution with numpy.random.rand and scale to length of array, like so -
resamples_arr = x[(np.random.rand(num_samples,len(x))*len(x)).astype(int)]
Runtime test with x of 5000 elems -
In [221]: x = np.random.randint(0,10000,(5000))
# Original soln
In [222]: %timeit [np.random.choice(x, size=len(x), replace=True) for i in range(1000)]
10 loops, best of 3: 84 ms per loop
# Proposed soln-1
In [223]: %timeit x[np.random.randint(0,len(x),size=(1000,len(x)))]
10 loops, best of 3: 76.2 ms per loop
# Proposed soln-2
In [224]: %timeit x[(np.random.rand(1000,len(x))*len(x)).astype(int)]
10 loops, best of 3: 59.7 ms per loop
For very large x
With a very large array x of 600,000 elements, you might not want to create all those indices for 1000 samples. In that case, per sample solution would have their timings something like this -
In [234]: x = np.random.randint(0,10000,(600000))
# Original soln
In [235]: %timeit np.random.choice(x, size=len(x), replace=True)
100 loops, best of 3: 13 ms per loop
# Proposed soln-1
In [238]: %timeit x[np.random.randint(0,len(x),len(x))]
100 loops, best of 3: 12.5 ms per loop
# Proposed soln-2
In [239]: %timeit x[(np.random.rand(len(x))*len(x)).astype(int)]
100 loops, best of 3: 9.81 ms per loop
As alluded to by #Divakar you can pass a tuple to size to get a 2d array of resamples rather than using list comprehension.
Here assume for a second that f is just sum rather than some other function. Then:
x = np.random.randn(100000)
resamples = np.random.choice(x, size=(1000, x.shape[0]), replace=True)
# resamples.shape = (1000, 1000000)
results = np.apply_along_axis(f, axis=1, arr=resamples)
print(results.shape)
# (1000,)
Here np.apply_along_axis is admittedly just a glorified for-loop equivalent to [f(arr) for arr in resamples]. But I am not exactly sure if you need to index x here based on your question.
I ran into a memory problem when trying to use .reshape on a numpy array and figured if I could somehow reshape the array in place that would be great.
I realised that I could reshape arrays by simply changing the .shape value.
Unfortunately when I tried using .shape I again got a memory error which has me thinking that it doesn't reshape in place.
I was wondering when do I use one when do I use the other?
Any help is appreciated.
If you want additional information please let me know.
EDIT:
I added my code and how the matrix I want to reshape is created in case that is important.
Change the N value depending on your memory.
import numpy as np
N = 100
a = np.random.rand(N, N)
b = np.random.rand(N, N)
c = a[:, np.newaxis, :, np.newaxis] * b[np.newaxis, :, np.newaxis, :]
c = c.reshape([N*N, N*N])
c.shape = ([N, N, N, N])
EDIT2:
This is a better representation. Apparently the transpose seems to be important as it changes the arrays from C-contiguous to F-contiguous, and the resulting multiplication in above case is contiguous while in the one below it is not.
import numpy as np
N = 100
a = np.random.rand(N, N).T
b = np.random.rand(N, N).T
c = a[:, np.newaxis, :, np.newaxis] * b[np.newaxis, :, np.newaxis, :]
c = c.reshape([N*N, N*N])
c.shape = ([N, N, N, N])
numpy.reshape will copy the data if it can't make a proper view, whereas setting the shape will raise an error instead of copying the data.
It is not always possible to change the shape of an array without copying the data. If you want an error to be raise if the data is copied, you should assign the new shape to the shape attribute of the array.
I would like to revisit this question focusing on OOP paradigm, despite memory issues presented as the problem.
When to use .shape and when to use .reshape?
OOP principle of Encapsulation
Following OOP paradigms, since shape is a property of the object numpy.array it is always advisable to call an object.method to change properties. This adheres to OOP principle of encapsulation.
Performance Issues
As for performance, there seems to be no difference.
import numpy as np
# creates an array of 1,000,000 random floats
a = np.array(np.random.rand(1_000_000))
# (1000000,)
a.shape
# using IPython to time both operations resulted in
# 201 ns ± 4.85 ns per loop (mean ± std. dev. of 7 runs, 10000000 loops each)
%timeit a.shape = (5_000, 200)
# 217 ns ± 0.957 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
%timeit a.reshape (5_000, 200)
Running hardware
OS : Linux 4.15.0-142-generic #146~16.04.1-Ubuntu
CPU: Intel(R) Core(TM) i3-4170 CPU # 3.70GHz 4 cores
RAM: 16BG
Normally I would invert an array of 3x3 matrices in a for loop like in the example below. Unfortunately for loops are slow. Is there a faster, more efficient way to do this?
import numpy as np
A = np.random.rand(3,3,100)
Ainv = np.zeros_like(A)
for i in range(100):
Ainv[:,:,i] = np.linalg.inv(A[:,:,i])
It turns out that you're getting burned two levels down in the numpy.linalg code. If you look at numpy.linalg.inv, you can see it's just a call to numpy.linalg.solve(A, inv(A.shape[0]). This has the effect of recreating the identity matrix in each iteration of your for loop. Since all your arrays are the same size, that's a waste of time. Skipping this step by pre-allocating the identity matrix shaves ~20% off the time (fast_inverse). My testing suggests that pre-allocating the array or allocating it from a list of results doesn't make much difference.
Look one level deeper and you find the call to the lapack routine, but it's wrapped in several sanity checks. If you strip all these out and just call lapack in your for loop (since you already know the dimensions of your matrix and maybe know that it's real, not complex), things run MUCH faster (Note that I've made my array larger):
import numpy as np
A = np.random.rand(1000,3,3)
def slow_inverse(A):
Ainv = np.zeros_like(A)
for i in range(A.shape[0]):
Ainv[i] = np.linalg.inv(A[i])
return Ainv
def fast_inverse(A):
identity = np.identity(A.shape[2], dtype=A.dtype)
Ainv = np.zeros_like(A)
for i in range(A.shape[0]):
Ainv[i] = np.linalg.solve(A[i], identity)
return Ainv
def fast_inverse2(A):
identity = np.identity(A.shape[2], dtype=A.dtype)
return array([np.linalg.solve(x, identity) for x in A])
from numpy.linalg import lapack_lite
lapack_routine = lapack_lite.dgesv
# Looking one step deeper, we see that solve performs many sanity checks.
# Stripping these, we have:
def faster_inverse(A):
b = np.identity(A.shape[2], dtype=A.dtype)
n_eq = A.shape[1]
n_rhs = A.shape[2]
pivots = zeros(n_eq, np.intc)
identity = np.eye(n_eq)
def lapack_inverse(a):
b = np.copy(identity)
pivots = zeros(n_eq, np.intc)
results = lapack_lite.dgesv(n_eq, n_rhs, a, n_eq, pivots, b, n_eq, 0)
if results['info'] > 0:
raise LinAlgError('Singular matrix')
return b
return array([lapack_inverse(a) for a in A])
%timeit -n 20 aI11 = slow_inverse(A)
%timeit -n 20 aI12 = fast_inverse(A)
%timeit -n 20 aI13 = fast_inverse2(A)
%timeit -n 20 aI14 = faster_inverse(A)
The results are impressive:
20 loops, best of 3: 45.1 ms per loop
20 loops, best of 3: 38.1 ms per loop
20 loops, best of 3: 38.9 ms per loop
20 loops, best of 3: 13.8 ms per loop
EDIT: I didn't look closely enough at what gets returned in solve. It turns out that the 'b' matrix is overwritten and contains the result in the end. This code now gives consistent results.
A few things have changed since this question was asked and answered, and now numpy.linalg.inv supports multidimensional arrays, handling them as stacks of matrices with matrix indices being last (in other words, arrays of shape (...,M,N,N)). This seems to have been introduced in numpy 1.8.0. Unsurprisingly this is by far the best option in terms of performance:
import numpy as np
A = np.random.rand(3,3,1000)
def slow_inverse(A):
"""Looping solution for comparison"""
Ainv = np.zeros_like(A)
for i in range(A.shape[-1]):
Ainv[...,i] = np.linalg.inv(A[...,i])
return Ainv
def direct_inverse(A):
"""Compute the inverse of matrices in an array of shape (N,N,M)"""
return np.linalg.inv(A.transpose(2,0,1)).transpose(1,2,0)
Note the two transposes in the latter function: the input of shape (N,N,M) has to be transposed to shape (M,N,N) for np.linalg.inv to work, then the result has to be permuted back to shape (M,N,N).
A check and timing results using IPython, on python 3.6 and numpy 1.14.0:
In [5]: np.allclose(slow_inverse(A),direct_inverse(A))
Out[5]: True
In [6]: %timeit slow_inverse(A)
19 ms ± 138 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
In [7]: %timeit direct_inverse(A)
1.3 ms ± 6.39 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
Numpy-Blas calls are not always the fastest possibility
On problems where you have to calculate lots of inverses, eigenvalues, dot-products of small 3x3 matrices or similar cases, numpy-MKL which I use can often be outperformed by quite a margin.
This external Blas routines are usually made for problems with larger matrices, for smaller ones you can write out a standard algorithm or take a look at eg. Intel IPP.
Please keep also in mind that Numpy uses C-ordered arrays by default (last dimension changes fastest).
For this example I took the code from Matrix inversion (3,3) python - hard coded vs numpy.linalg.inv and modified it a bit.
import numpy as np
import numba as nb
import time
#nb.njit(fastmath=True)
def inversion(m):
minv=np.empty(m.shape,dtype=m.dtype)
for i in range(m.shape[0]):
determinant_inv = 1./(m[i,0]*m[i,4]*m[i,8] + m[i,3]*m[i,7]*m[i,2] + m[i,6]*m[i,1]*m[i,5] - m[i,0]*m[i,5]*m[i,7] - m[i,2]*m[i,4]*m[i,6] - m[i,1]*m[i,3]*m[i,8])
minv[i,0]=(m[i,4]*m[i,8]-m[i,5]*m[i,7])*determinant_inv
minv[i,1]=(m[i,2]*m[i,7]-m[i,1]*m[i,8])*determinant_inv
minv[i,2]=(m[i,1]*m[i,5]-m[i,2]*m[i,4])*determinant_inv
minv[i,3]=(m[i,5]*m[i,6]-m[i,3]*m[i,8])*determinant_inv
minv[i,4]=(m[i,0]*m[i,8]-m[i,2]*m[i,6])*determinant_inv
minv[i,5]=(m[i,2]*m[i,3]-m[i,0]*m[i,5])*determinant_inv
minv[i,6]=(m[i,3]*m[i,7]-m[i,4]*m[i,6])*determinant_inv
minv[i,7]=(m[i,1]*m[i,6]-m[i,0]*m[i,7])*determinant_inv
minv[i,8]=(m[i,0]*m[i,4]-m[i,1]*m[i,3])*determinant_inv
return minv
#I was to lazy to modify the code from the link above more thoroughly
def inversion_3x3(m):
m_TMP=m.reshape(m.shape[0],9)
minv=inversion(m_TMP)
return minv.reshape(minv.shape[0],3,3)
#Testing
A = np.random.rand(1000000,3,3)
#Warmup to not measure compilation overhead on the first call
#You may also use #nb.njit(fastmath=True,cache=True) but this has also about 0.2s
#overhead on fist call
Ainv = inversion_3x3(A)
t1=time.time()
Ainv = inversion_3x3(A)
print(time.time()-t1)
t1=time.time()
Ainv2 = np.linalg.inv(A)
print(time.time()-t1)
print(np.allclose(Ainv2,Ainv))
Performance
np.linalg.inv: 0.36 s
inversion_3x3: 0.031 s
For loops are indeed not necessarily much slower than the alternatives and also in this case, it will not help you much. But here is a suggestion:
import numpy as np
A = np.random.rand(100,3,3) #this is to makes it
#possible to index
#the matrices as A[i]
Ainv = np.array(map(np.linalg.inv, A))
Timing this solution vs. your solution yields a small but noticeable difference:
# The for loop:
100 loops, best of 3: 6.38 ms per loop
# The map:
100 loops, best of 3: 5.81 ms per loop
I tried to use the numpy routine 'vectorize' with the hope of creating an even cleaner solution, but I'll have to take a second look into that. The change of ordering in the array A is probably the most significant change, since it utilises the fact that numpy arrays are ordered column-wise and therefor a linear readout of the data is ever so slightly faster this way.