I want to vectorise the triple sum
\sum_{i=1}^I\sum_{j=1}^J\sum_{m=1}^J a_{ijm}
such that I end up with a matrix
A \in \mathbb{R}^{I \times J}
where A_{kl} = \sum_{i=1}^k\sum_{j=1}^l\sum_{m=1}^l a_{ijm} for k = 1,...,I and l = 1, ...,J
carrying forward the sums to avoid pointless recomputation.
I currently use this code:
np.cumsum(np.cumsum(np.cumsum(a, axis = 0), axis = 1), axis = 2).diagonal(axis1 = 1, axis2 = 2)
but it is inefficient as it does lots of extra work and extracts the correct matrix at the end with the diagonal method. I can't think of how to make this faster.
The main challenge here is to compute the inner two sums, i.e. the sum of the square slices of a matrix originating from the top left. The final sum is just a cumsum on top of that along the 0th axis.
Setup:
import numpy as np
I, J = 100, 100
arr = np.random.rand(I, J, J)
Your implementation:
%%timeit
out = np.cumsum(np.cumsum(np.cumsum(arr, axis = 0), axis = 1), axis = 2).diagonal(axis1 = 1, axis2 = 2)
# 10.9 ms ± 162 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
Your implementation improved by taking the diagonal before cumsumming over the 0th axis:
%%timeit
out = arr.cumsum(axis=1).cumsum(axis=2).diagonal(axis1=1, axis2=2).cumsum(axis=0)
# 6.25 ms ± 34.4 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
Finally, some tril/triu trickery:
%%timeit
out = np.cumsum(np.cumsum(np.tril(arr, k=-1).sum(axis=2) + np.triu(arr).sum(axis=1), axis=1), axis=0)
# 3.15 ms ± 71.4 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
which is already better, but admittedly still not ideal. I don't see a better way to compute the inner two sums noted above with pure numpy.
You can use Numba so to produce a very fast implementation. Here is the code:
import numba as nb
import numpy as np
#nb.njit('(float64[:,:,::1],)', parallel=True)
def compute(arr):
ni, nj, nk = arr.shape
assert nj == nk
result = np.empty((ni, nj))
# Parallel cumsum along the axis 1 and 2 + extraction of the diagonal
for i in nb.prange(ni):
tmp = np.zeros(nk)
for j in range(nj):
for k in range(nk):
tmp[k] += arr[i, j, k]
result[i, j] = np.sum(tmp[:j+1])
# Cumsum along the axis 0
for i in range(1, ni):
for k in range(nk):
result[i, k] += result[i-1, k]
return result
result = compute(a)
Here are performance results on my 6-core i5-9600KF with a 100x100x100 float64 input array:
Initial code: 12.7 ms
Chryophylaxs v1: 7.1 ms
Chryophylaxs v2: 5.5 ms
Numba: 0.2 ms
This implementation is significantly faster than all others. It is about 64 times faster than the initial implementation. It is also actually optimal on my machine since it completely saturate the bandwidth of my RAM only for reading the input array (which is mandatory). Note that it is better not to use multiple threads for very small arrays.
Note that this code also use far less memory as it only need 8 * nk * num_threads bytes of temporary storage as opposed to 16 * ni * nj * nk bytes for the initial solution.
Related
Here's some data I've generated:
import numpy as np
import pandas as pd
import scipy
import scipy.spatial
df = pd.DataFrame(
{
"item_1": np.random.randint(low=0, high=10, size=1000),
"item_2": np.random.randint(low=0, high=10, size=1000),
}
)
embeddings = {item_id: np.random.randn(100) for item_id in range(0, 10)}
def get_distance(item_1, item_2):
arr1 = embeddings[item_1]
arr2 = embeddings[item_2]
return scipy.spatial.distance.cosine(arr1, arr2)
I'd like to apply get_distance to each row. I can do:
df.apply(lambda row: get_distance(row["item_1"], row["item_2"]), axis=1)
But that would be very slow for large datasets.
Is there a way to calculate the cosine similarity of the embeddings corresponding to each row, without using DataFrame.apply?
For scipy version
%%timeit
df.apply(lambda row: get_distance(row["item_1"], row["item_2"]), axis=1)
# 38.3 ms ± 84 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
For what its worth I added numba with extra complication
Thinking about memory and numpy broadcast use tmp allocation, I used for loops
Also it is worth considering passing arguments, maybe you can pass vectors instead of dictionary.
Also first run is slow due to compilation
Also you can make it parallel with numba
#nb.njit((nb.float64[:, ::100], nb.float64[:, ::100]))
def cos(a, b):
norm_a = np.empty((a.shape[0],), dtype=np.float64)
norm_b = np.empty((b.shape[0],), dtype=np.float64)
cos_ab = np.empty((a.shape[0],), dtype=np.float64)
for i in nb.prange(a.shape[0]):
sq_norm = 0.0
for j in range(100):
sq_norm += a[i][j] ** 2
norm_a[i] = sq_norm ** 0.5
for i in nb.prange(b.shape[0]):
sq_norm = 0.0
for j in range(100):
sq_norm += b[i][j] ** 2
norm_b[i] = sq_norm ** 0.5
for i in nb.prange(a.shape[0]):
dot = 0.0
for j in range(100):
dot += a[i][j] * b[i][j]
cos_ab[i] = 1 - dot / (norm_a[i] * norm_b[i])
return cos_ab
%%timeit
cos(item_1_embedded, item_2_embedded)
# 218 µs ± 1.23 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
Using vectorized numpy operations directly is much faster:
item_1_embedded = np.array([embeddings[x]for x in df.item_1])
item_2_embedded = np.array([embeddings[x]for x in df.item_2])
cos_dist = 1 - np.sum(item_1_embedded*item_2_embedded, axis=1)/(np.linalg.norm(item_1_embedded, axis=1)*np.linalg.norm(item_2_embedded, axis=1))
(This version runs in 771 µs on average on my pc, vs 37.4 ms for the DataFrame.apply, which makes the pure numpy version about 50 times faster).
You can vectorize the call to cosine with numpy.vectorize. There is a slight gain in speed (34 ms vs 53 ms)
vec_cosine = np.vectorize(scipy.spatial.distance.cosine)
vec_cosine(df['item_1'].map(embeddings),
df['item_2'].map(embeddings))
output:
array([0.90680875, 0.90999454, 0.99212814, 1.12455852, 1.06354469,
0.95542037, 1.07133003, 1.07133003, 0. , 1.00837058,
0. , 0.93961103, 0.8943738 , 1.04872436, 1.21171375,
1.04621226, 0.90392229, 1.0365102 , 0. , 0.90180297,
0.90180297, 1.04516879, 0.94877277, 0.90180297, 0.93713404,
...
1.17548653, 1.11700641, 0.97926805, 0.8943738 , 0.93961103,
1.21171375, 0.91817959, 0.91817959, 1.04674315, 0.88210679,
1.11806218, 1.07816675, 1.00837058, 1.12455852, 1.04516879,
0.93713404, 0.93713404, 0.95542037, 0.93876964, 0.91817959])
This question already has answers here:
Restart cumsum and get index if cumsum more than value
(3 answers)
Closed 2 years ago.
I want to bin the data every time the threshold 10000 is exceeded.
I have tried this with no luck:
# data which is an array of floats
diff = np.diff(np.cumsum(data)//10000, prepend=0)
indices = (np.argwhere(diff > 0)).flatten()
The problem is that all the bins does not contain 10000, which was my goal.
Expected output
input_data = [4000, 5000, 6000, 2000, 8000, 3000]
# (4000+5000+6000 >= 10000. Index 2)
# (2000+8000 >= 10000. Index 4)
Output: [2, 4]
I wonder if there is any alternative to a for loop?
Not sure how this could be vectorized, if it even can be, since by taking the cumulative sum you'll be propagating the remainders each time the threshold is surpassed. So probably this is a good case for numba, which will compile the code down to C level, allowing for a loopy but performant approach:
from numba import njit, int32
#njit('int32[:](int32[:], uintc)')
def windowed_cumsum(a, thr):
indices = np.zeros(len(a), int32)
window = 0
ix = 0
for i in range(len(a)):
window += a[i]
if window >= thr:
indices[ix] = i
ix += 1
window = 0
return indices[:ix]
The explicit signature implies ahead of time compilation, though this enforces specific dtypes on the input array. The inferred dtype for the example array is of int32, though if this might not always be the case or for a more flexible solution you can always ignore the dtypes in the signature, which will only imply that the function will be compiled on the first execution.
input_data = np.array([4000, 5000, 6000, 2000, 8000, 3000])
windowed_cumsum(input_data, 10000)
# array([2, 4])
Also #jdehesa raises an interesting point, which is that for very long arrays compared to the number of bins, a better option might be to just append the indices to a list. So here is an alternative approach using lists (also in no-python mode), along with timings under different scenarios:
from numba import njit, int32
#njit
def windowed_cumsum_list(a, thr):
indices = []
window = 0
for i in range(len(a)):
window += a[i]
if window >= thr:
indices.append(i)
window = 0
return indices
a = np.random.randint(0,10,10_000)
%timeit windowed_cumsum(a, 20)
# 16.1 µs ± 232 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
%timeit windowed_cumsum_list(a, 20)
# 65.5 µs ± 623 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
%timeit windowed_cumsum(a, 2000)
# 7.38 µs ± 167 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
%timeit windowed_cumsum_list(a, 2000)
# 7.1 µs ± 103 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
So it seems that under most scenarios using numpy will be a faster option, since even in the second case, with a length 10000 array and a resulting array of 20 indices of bins, both perform similarly, though for memory efficiency reasons the latter might be more convenient in some cases.
Here is how you can do it fairly efficiently with a loop, using np.searchsorted to find bin boundaries fast:
import numpy as np
np.random.seed(0)
bin_size = 10_000
data = np.random.randint(100, size=20_000)
# Naive solution (incorrect, for comparison)
data_f = np.floor(np.cumsum(data) / bin_size).astype(int)
bin_starts = np.r_[0, np.where(np.diff(data_f) > 0)[0] + 1]
# Check bin sizes
bin_sums = np.add.reduceat(data, bin_starts)
# We go over the limit!
print(bin_sums.max())
# 10080
# Better solution with loop
data_c = np.cumsum(data)
ref_val = 0
bin_starts = [0]
while True:
# Search next split point
ref_idx = bin_starts[-1]
# Binary search through remaining cumsum
next_idx = np.searchsorted(data_c[ref_idx:], ref_val + bin_size, side='right')
next_idx += ref_idx
# If we finished the array stop
if next_idx >= len(data_c):
break
# Add new bin boundary
bin_starts.append(next_idx)
ref_val = data_c[next_idx - 1]
# Convert bin boundaries to array
bin_starts = np.array(bin_starts)
# Check bin sizes
bin_sums = np.add.reduceat(data, bin_starts)
# Does not go over limit
print(bin_sums.max())
# 10000
in my code I need to calculate the values of a vector many times which are the mean values from different patches of another array.
Here is an example of my code showing how I do it but I found that it is too less-efficient in running...
import numpy as np
vector_a = np.zeros(10)
array_a = np.random.random((100,100))
for i in range(len(vector_a)):
vector_a[i] = np.mean(array_a[:,i+20:i+40]
Is there any way to make it more efficient? Any comments or suggestions are very welcome! Many thanks!
-yes, the 20 and 40 are fixed.
EDIT:
Actually you can do this much faster. The previous function can be improved by operating on summed columns like this:
def rolling_means_faster1(array_a, n, first, size):
# Sum each relevant columns
sum_a = np.sum(array_a[:, first:(first + size + n - 1)], axis=0)
# Reshape as before
strides_b = (sum_a.strides[0], sum_a.strides[0])
array_b = np.lib.stride_tricks.as_strided(sum_a, (n, size), (strides_b))
# Average
v = np.sum(array_b, axis=1)
v /= (len(array_a) * size)
return v
Another way is to work with accumulated sums, adding and removing as necessary for each output element.
def rolling_means_faster2(array_a, n, first, size):
# Sum each relevant columns
sum_a = np.sum(array_a[:, first:(first + size + n - 1)], axis=0)
# Add a zero a the beginning so the next operation works fine
sum_a = np.insert(sum_a, 0, 0)
# Sum the initial `size` elements and add and remove partial sums as necessary
v = np.sum(sum_a[:size]) - np.cumsum(sum_a[:n]) + np.cumsum(sum_a[-n:])
# Average
v /= (size * len(array_a))
return v
Benchmarking with the previous solution from before:
import numpy as np
np.random.seed(100)
array_a = np.random.random((1000, 1000))
n = 100
first = 100
size = 200
%timeit rolling_means_orig(array_a, n, first, size)
# 12.7 ms ± 55.8 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit rolling_means(array_a, n, first, size)
# 5.49 ms ± 43.9 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit rolling_means_faster1(array_a, n, first, size)
# 166 µs ± 874 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
%timeit rolling_means_faster2(array_a, n, first, size)
# 182 µs ± 2.04 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
So these last two seem to be very close in performance. It may depend on the relative sizes of the inputs.
This is a possible vectorized solution:
import numpy as np
# Data
np.random.seed(100)
array_a = np.random.random((100, 100))
# Take all the relevant columns
slice_a = array_a[:, 20:40 + 10]
# Make a "rolling window" with stride tricks
strides_b = (slice_a.strides[1], slice_a.strides[0], slice_a.strides[1])
array_b = np.lib.stride_tricks.as_strided(slice_a, (10, 100, 20), (strides_b))
# Take mean
result = np.mean(array_b, axis=(1, 2))
# Original method for testing correctness
vector_a = np.zeros(10)
idv1 = np.arange(10) + 20
idv2 = np.arange(10) + 40
for i in range(len(vector_a)):
vector_a[i] = np.mean(array_a[:,idv1[i]:idv2[i]])
print(np.allclose(vector_a, result))
# True
Here is a quick benchmark in IPython (sizes increased for appreciation):
import numpy as np
def rolling_means(array_a, n, first, size):
slice_a = array_a[:, first:(first + size + n)]
strides_b = (slice_a.strides[1], slice_a.strides[0], slice_a.strides[1])
array_b = np.lib.stride_tricks.as_strided(slice_a, (n, len(array_a), size), (strides_b))
return np.mean(array_b, axis=(1, 2))
def rolling_means_orig(array_a, n, first, size):
vector_a = np.zeros(n)
idv1 = np.arange(n) + first
idv2 = np.arange(n) + (first + size)
for i in range(len(vector_a)):
vector_a[i] = np.mean(array_a[:,idv1[i]:idv2[i]])
return vector_a
np.random.seed(100)
array_a = np.random.random((1000, 1000))
n = 100
first = 100
size = 200
%timeit rolling_means(array_a, n, first, size)
# 5.48 ms ± 26.6 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit rolling_means_orig(array_a, n, first, size)
# 32.8 ms ± 762 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
This solution works on the assumption that you are trying to compute rolling average of a subset of window of columns.
As an example and ignoring rows, given [0, 1, 2, 3, 4] and a window of 2 the averages are [0.5, 1.5, 2.5, 3.5], and that you might only want the second and third averages.
Your current solution is inefficient as it is recomputes the mean for a column for each output in vector_a. Given that (a / n) + (b / n) == (a + b) / n we can get away with computing the mean of each column only once, and then combine the column means as needed to produce the final output.
window_first_start = idv1.min() # or idv1[0]
window_last_end = idv2.max() # or idv2[-1]
window_size = idv2[0] - idv1[0]
assert ((idv2 - idv1) == window_size).all(), "sanity check, not needed if assumption holds true"
# a view of the columns we are interested in, no copying is done here
view = array_a[:,window_first_start:window_last_end]
# calculate the means for each column
col_means = view.mean(axis=0)
# cumsum is used to find the rolling sum of means and so the rolling average
# We use an out variable to make sure we have a 0 in the first element of cum_sum.
# This makes like a little easier in the next step.
cum_sum = np.empty(len(col_means) + 1, dtype=col_means.dtype)
cum_sum[0] = 0
np.cumsum(col_means, out=cum_sum[1:])
result = (cum_sum[window_size:] - cum_sum[:-window_size]) / window_size
Having tested this against your own code, the above is significantly faster (increasing with the size of the input array), and slightly faster than the solution provided by jdehesa. With an input array of 1000x1000, it is two orders of magnitude faster than your solution and one order of magnitude faster than jdehesa's.
Try this:
import numpy as np
array_a = np.random.random((100,100))
vector_a = [np.mean(array_a[:,i+20:i+40]) for i in range(10)]
I want to multiply B = A # A.T in numpy. Obviously, the answer would be a symmetric matrix (i.e. B[i, j] == B[j, i]).
However, it is not clear to me how to leverage this easily to cut the computation time down in half (by only computing the lower triangle of B and then using that to get the upper triangle for free).
Is there a way to perform this optimally?
As noted in #PaulPanzer's link, dot can detect this case. Here's the timing proof:
In [355]: A = np.random.rand(1000,1000)
In [356]: timeit A.dot(A.T)
57.4 ms ± 960 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
In [357]: B = A.T.copy()
In [358]: timeit A.dot(B)
98.6 ms ± 805 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
Numpy dot too clever about symmetric multiplications
You can always use sklearns's pairwise_distances
Usage:
from sklearn.metrics.pairwise import pairwise_distances
gram = pairwise_distance(x, metric=metric)
Where metric is a callable or a string defining one of their implemented metrics (full list in the link above)
But, I wrote this for myself a while back so I can share what I did:
import numpy as np
def computeGram(elements, dist):
n = len(elements)
gram = np.zeros([n, n])
for i in range(n):
for j in range(i + 1):
gram[i, j] = dist(elements[i], elements[j])
upTriIdxs = np.triu_indices(n)
gram[upTriIdxs] = gram.T[upTriIdxs]
return gram
Where dist is a callable, in your case np.inner
In my project (about clustering algorithms, specifically k-medoids) is crucial to be able to compute pairwise distances efficiently. I have a dataset of ~60,000 objects. The problem is, distances must be computed between inhomogeneous vectors, i.e. vectors which may differ in length (in that case, missing items are treated as if they were 0).
Here is a minimal working example:
# %%
MAX_LEN = 11
N = 100
import random
def manhattan_distance(vec1, vec2):
n1, n2 = len(vec1), len(vec2)
n = min(n1, n2)
dist = 0
for i in range(n):
dist += abs(vec1[i] - vec2[i])
if n1 > n2:
for i in range(n, n1):
dist += abs(vec1[i])
else:
for i in range(n, n2):
dist += abs(vec2[i])
return dist
def compute_distances():
n = len(data)
for i in range(n):
for j in range(n):
manhattan_distance(data[i], data[j])
data = []
for i in range(N):
data.append([])
for k in range(random.randint(5, MAX_LEN)):
data[i].append(random.randint(0, 10))
%timeit compute_distances()
import numpy as np
def manhattan_distance_np(vec1, vec2):
return np.absolute(vec1 - vec2).sum()
def compute_distances_np():
n = len(data)
for i in range(n):
for j in range(n):
manhattan_distance_np(data_np[i], data_np[j])
data_np = [np.append(np.asarray(d), np.zeros(MAX_LEN - len(d))) for d in data]
%timeit compute_distances_np()
I was testing my Python lists implementation versus a numpy implementation.
Here are the results (computation times):
Python lists: 79.6 ms ± 3.78 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
numpy arrays: 226 ms ± 7.18 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
Why is there such a huge difference? I supposed numpy arrays were really fast.
Is there a way to improve my code? Am I misunderstanding the inner workings of numpy?
Edit: I may need, in the future, to be able to use a custom distance function for pairwise distances computations. The method should work also for data sets of length 60'000 without running out of memory.
I believe you can just make your arrays dense and set the unused last elements to 0s.
import numpy as np
from scipy.spatial.distance import cdist, pdist, squareform
def batch_pdist(x, metric, batchsize=1000):
dists = np.zeros((len(x), len(x)))
for i in range(0, len(x), batchsize):
for j in range(0, len(x), batchsize):
dist_batch = cdist(x[i:i+batchsize], x[j:j+batchsize], metric=metric)
dists[i:i+batchsize, j:j+batchsize] = dist_batch
return dists
MIN_LEN = 5
MAX_LEN = 11
N = 10000
M = 10
data = []
data = np.zeros((N,MAX_LEN))
for i in range(N):
num_nonzero = np.random.randint(MIN_LEN, MAX_LEN)
data[i, :num_nonzero] = np.random.randint(0, M, num_nonzero)
dists = squareform(pdist(data, metric='cityblock'))
dists2 = batch_pdist(data, metric='cityblock', batchsize=500)
print((dists == dists2).all())
Timing Output:
%timeit squareform(pdist(data, metric='cityblock'))
43.8 µs ± 134 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
Edit:
For a custom distance function see the very bottom of this documentation.
I finally found probably the most straightforward way to solve this problem without changing too much the code and rely solely on computations and not on memory (since that could be unfeasible for very large datasets).
Based on juanpa.arrivillaga suggestion, I tried numba, that is a library that speeds up array-oriented and math-heavy Python code and is targeted mainly at numpy. You can read a good guide on optimizing Python code here: https://jakevdp.github.io/blog/2015/02/24/optimizing-python-with-numpy-and-numba/.
MAX_LEN = 11
N = 100
# Pure Python lists implementation.
import random
def manhattan_distance(vec1, vec2):
n1, n2 = len(vec1), len(vec2)
n = min(n1, n2)
dist = 0
for i in range(n):
dist += abs(vec1[i] - vec2[i])
if n1 > n2:
for i in range(n, n1):
dist += abs(vec1[i])
else:
for i in range(n, n2):
dist += abs(vec2[i])
return dist
def compute_distances():
n = len(data)
for i in range(n):
for j in range(n):
manhattan_distance(data[i], data[j])
data = []
for i in range(N):
data.append([])
for k in range(random.randint(5, MAX_LEN)):
data[i].append(random.randint(0, 10))
%timeit compute_distances()
# numpy+numba implementation.
import numpy as np
from numba import jit
#jit
def manhattan_distance_np(vec1, vec2):
return np.absolute(vec1 - vec2).sum()
#jit
def compute_distances_np():
n = len(data)
for i in range(n):
for j in range(n):
manhattan_distance_np(data_np[i], data_np[j])
data_np = np.array([np.append(np.asarray(d), np.zeros(MAX_LEN - len(d))) for d in data])
%timeit compute_distances_np()
Timing output:
%timeit compute_distances()
78.4 ms ± 3.44 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
%timeit compute_distances_np()
4.1 ms ± 14.7 µs per loop (mean ± std. dev. of 7 runs, 1 loop each)
As you can see, the numpy with numba optimizations is about 19 times faster (with no other code optimization involved).