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numpy polyfit alternative for speed

I am utilizing numpy's polyfit a numerous amount of times to run calculations and get the slope between two datasets. However, the speed at which it performs these calculations are not fast enough for what would be desired.
Two things to note about the calculations:
The value for x in the call numpy.polyfit(x,y,n) will always be the same slope value, and
The value for n = 1. So it is a linear regression.
I know there are many different alternatives, including numpy.polynomial.polynomial.polyfit(x,y,n), but they seem to provide the same slow speed performance. I have had little luck getting np.linalg to work properly. Therefore, I am wondering what might be an alternative to speed up calculations?

As others have commented, this can be done using linear least squares.
Using numpy.linalg.lstsq, this could look like:
import numpy as np
def lstsq(x, y):
a = np.stack((x, np.ones_like(x)), axis=1)
return np.linalg.lstsq(a, y)[0]
This offers a slight speed improvement over polyfit. To obtain a significant speed increase (at the expense of numerical stability - for a summary of methods see Numerical methods for linear least squares) you can instead solve the normal equations:
def normal(x, y):
a = np.stack((x, np.ones_like(x)), axis=1)
aT = a.T
return np.linalg.solve(aT#a, aT#y)
As you say that x is constant, you can precompute a.T#a providing a further speed increase:
def normal2(aT, aTa, y):
return np.linalg.solve(aTa, aT#y)
Make up some test data and time:
rng = np.random.default_rng()
N = 1000
x = rng.random(N)
y = rng.random(N)
a = np.stack((x, np.ones_like(x)), axis=1)
aT = a.T
aTa = aT#a
assert np.allclose(lstsq(x, y), np.polyfit(x, y, 1))
assert np.allclose(normal(x, y), np.polyfit(x, y, 1))
assert np.allclose(normal2(aT, aTa, y), np.polyfit(x, y, 1))
%timeit np.polyfit(x, y, 1)
%timeit lstsq(x, y)
%timeit normal(x, y)
%timeit normal2(aT, aTa, y)
Output:
256 µs ± 270 ns per loop (mean ± std. dev. of 7 runs, 1,000 loops each)
220 µs ± 1.87 µs per loop (mean ± std. dev. of 7 runs, 1,000 loops each)
20.2 µs ± 32.3 ns per loop (mean ± std. dev. of 7 runs, 10,000 loops each)
6.54 µs ± 13.5 ns per loop (mean ± std. dev. of 7 runs, 100,000 loops each)

Cumsum with restarts [duplicate]

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

Why numpy.var is O(N) space?

I have an array of ~13GB. I call numpy.var on it to compute the variance. However, it allocates another ~13GB to do this. Why does it need O(N) space? Or am I calling numpy.var in a wrong way?
import numpy as np
# data = ...
print('Variance: ', np.var(data))

NumPy will create an intermediate array to compute abs(data - data.mean()) ** 2 in order to compute the variance. You can write your own variance function with a loop and make it fast with Numba:
import numpy as np
import numba as nb
#nb.njit(parallel=True)
def var_nb(a, ddof=0):
n = len(a)
s = a.sum()
m = s / (n - ddof)
v = 0
for i in nb.prange(n):
v += abs(a[i] - m) ** 2
return v / (n - ddof)
np.random.seed(100)
a = np.random.rand(100_000)
print(np.var(a))
# 0.08349747560941487
print(var_nb(a))
# 0.08349747560941487
%timeit np.var(a)
# 143 µs ± 414 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
%timeit var_nb(a)
# 40.2 µs ± 530 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)

This is faster whitout parallelization:
import numpy as np
def var(a: np.ndarray, axis: int = 0):
return np.sum(abs(a - (a.sum(axis=axis) / len(a))) ** 2, axis=axis) / len(a)

A tedious loop looking for improvements

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)]

Numpy efficient matrix self-multiplication (gram matrix)

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