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Python: get maximum occurrence in array

I implemented codes to try to get maximum occurrence in numpy array. I was satisfactory using numba, but got limitations. I wonder whether it can be improved to a general case.
numba implementation
import numba as nb
import numpy as np
import collections
#nb.njit("int64(int64[:])")
def max_count_unique_num(x):
"""
Counts maximum number of unique integer in x.
Args:
x (numpy array): Integer array.
Returns:
Int
"""
# get maximum value
m = x[0]
for v in x:
if v > m:
m = v
if m == 0:
return x.size
# count each unique value
num = np.zeros(m + 1, dtype=x.dtype)
for k in x:
num[k] += 1
# maximum count
m = 0
for k in num:
if k > m:
m = k
return m
For comparisons, I also implemented numpy's unique and collections.Counter
def np_unique(x):
""" Counts maximum occurrence using numpy's unique. """
ux, uc = np.unique(x, return_counts=True)
return uc.max()
def counter(x):
""" Counts maximum occurrence using collections.Counter. """
counts = collections.Counter(x)
return max(counts.values())
timeit
Edit: Add np.bincount for additional comparison, as suggested by #MechanicPig.
In [1]: x = np.random.randint(0, 2000, size=30000).astype(np.int64)
In [2]: %timeit max_count_unique_num(x)
30 µs ± 387 ns per loop (mean ± std. dev. of 7 runs, 10,000 loops each)
In [3]: %timeit np_unique(x)
1.14 ms ± 1.65 µs per loop (mean ± std. dev. of 7 runs, 1,000 loops each)
In [4]: %timeit counter(x)
2.68 ms ± 33.9 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
In [5]: x = np.random.randint(0, 200000, size=30000).astype(np.int64)
In [6]: %timeit counter(x)
3.07 ms ± 40.5 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
In [7]: %timeit np_unique(x)
1.3 ms ± 7.35 µs per loop (mean ± std. dev. of 7 runs, 1,000 loops each)
In [8]: %timeit max_count_unique_num(x)
490 µs ± 1.47 µs per loop (mean ± std. dev. of 7 runs, 1,000 loops each)
In [9]: x = np.random.randint(0, 2000, size=30000).astype(np.int64)
In [10]: %timeit np.bincount(x).max()
32.3 µs ± 250 ns per loop (mean ± std. dev. of 7 runs, 10,000 loops each)
In [11]: x = np.random.randint(0, 200000, size=30000).astype(np.int64)
In [12]: %timeit np.bincount(x).max()
830 µs ± 6.09 µs per loop (mean ± std. dev. of 7 runs, 1,000 loops each)
The limitations of numba implementation are quite obvious: efficiency only when all values in x are small positive int and will be significantly reduced for very large int; not applicable to float and negative values.
Any way I can generalize the implementation and keep the speed?
Update
After checking the source code of np.unique, an implementation for general cases can be:
#nb.njit(["int64(int64[:])", "int64(float64[:])"])
def max_count_unique_num_2(x):
x.sort()
n = 0
k = 0
x0 = x[0]
for v in x:
if x0 == v:
k += 1
else:
if k > n:
n = k
k = 1
x0 = v
# for last item in x if it equals to previous one
if k > n:
n = k
return n
timeit
In [154]: x = np.random.randint(0, 200000, size=30000).astype(np.int64)
In [155]: %timeit max_count_unique_num(x)
519 µs ± 5.33 µs per loop (mean ± std. dev. of 7 runs, 1,000 loops each)
In [156]: %timeit np_unique(x)
1.3 ms ± 9.88 µs per loop (mean ± std. dev. of 7 runs, 1,000 loops each)
In [157]: %timeit max_count_unique_num_2(x)
240 µs ± 1.92 µs per loop (mean ± std. dev. of 7 runs, 1,000 loops each)
In [158]: x = np.random.randint(0, 200000, size=300000).astype(np.int64)
In [159]: %timeit max_count_unique_num(x)
1.01 ms ± 7.2 µs per loop (mean ± std. dev. of 7 runs, 1,000 loops each)
In [160]: %timeit np_unique(x)
18.1 ms ± 395 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
In [161]: %timeit max_count_unique_num_2(x)
3.58 ms ± 28.7 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
So:
If large integer in x and the size is not large, max_count_unique_num_2 beats max_count_unique_num.
Both max_count_unique_num and max_count_unique_num_2 are significantly faster than np.unique.
Small modification on max_count_unique_num_2 can return the item that has maximum occurrence, even all items having same maximum occurrence.
max_count_unique_num_2 can even be accelerated if x is itself sorted by removing x.sort().

What if shortening your code:
#nb.njit("int64(int64[:])", fastmath=True)
def shortened(x):
num = np.zeros(x.max() + 1, dtype=x.dtype)
for k in x:
num[k] += 1
return num.max()
or paralleled:
#nb.njit("int64(int64[:])", parallel=True, fastmath=True)
def shortened_paralleled(x):
num = np.zeros(x.max() + 1, dtype=x.dtype)
for k in nb.prange(x.size):
num[x[k]] += 1
return num.max()
Parallelizing will beat for larger data sizes. Note that parallel will get different result in some runs and need to be cured if be possible.
For handling the floats (or negative values) using Numba:
#nb.njit("int8(float64[:])", fastmath=True)
def shortened_float(x):
num = np.zeros(x.size, dtype=np.int8)
for k in x:
for j in range(x.shape[0]):
if k == x[j]:
num[j] += 1
return num.max()
IMO, np.unique(x, return_counts=True)[1].max() is the best choice which handle both integers and floats in a very fast implementation. Numba can be faster for integers (it depends on the data sizes as larger data sizes weaker performance; AIK, it is due to looping instinct than arrays), but for floats the code must be optimized in terms of performance if it could; But I don't think that Numba can beat NumPy unique, particularly when we faced to large data.
Notes: np.bincount can handle just integers.

You can do that without using numpy too.
arr = [1,1,2,2,3,3,4,5,6,1,3,5,7,1]
counts = list(map(list(arr).count, set(arr)))
list(set(arr))[counts.index(max(counts))]
If you want to use numpy then try this,
arr = np.array([1,1,2,2,3,3,4,5,6,1,3,5,7,1])
uniques, counts = np.unique(arr, return_counts = True)
uniques[np.where(counts == counts.max())]
Both do the exact same job. To check which method is more efficient just do this,
time_i = time.time()
<arr declaration> # Creating a new array each iteration can cause the total time to increase which would be biased against the numpy method.
for i in range(10**5):
<method you want>
time_f = time.time()
When I ran this I got 0.39 seconds for the first method and 2.69 for the second one. So it's pretty safe to say that the first method is more efficient.

What I want to say is that your implementation is almost the same as numpy.bincount. If you want to make it universal, you can consider encoding the original data:
def encode(ar):
# Equivalent to numpy.unique(ar, return_inverse=True)[1] when ar.ndim == 1
flatten = ar.ravel()
perm = flatten.argsort()
sort = flatten[perm]
mask = np.concatenate(([False], sort[1:] != sort[:-1]))
encoded = np.empty(sort.shape, np.int64)
encoded[perm] = mask.cumsum()
encoded.shape = ar.shape
return encoded
def count_max(ar):
return max_count_unique_num(encode(ar))

python: fastest way to compute euclidean distance of a vector to every row of a matrix?

Consider this python code, where I try to compute the eucliean distance of a vector to every row of a matrix. It's very slow compared to the best Julia version I can find using Tullio.jl.
The python version takes 30s but the Julia version only takes 75ms.
I am sure I am not doing the best in Python. Are there faster solutions? Numba and numpy solutions welcome.
import numpy as np
# generate
a = np.random.rand(4000000, 128)
b = np.random.rand(128)
print(a.shape)
print(b.shape)
def lin_norm_ever(a, b):
return np.apply_along_axis(lambda x: np.linalg.norm(x - b), 1, a)
import time
t = time.time()
res = lin_norm_ever(a, b)
print(res.shape)
elapsed = time.time() - t
print(elapsed)
The Julia verions
using Tullio
function comp_tullio(a, c)
dist = zeros(Float32, size(a, 2))
#tullio dist[i] = (c[j] - a[j,i])^2
dist
end
#time comp_tullio(a, c)
#benchmark comp_tullio(a, c) # 75ms on my computer

I would use Numba in this example for best performance. I also added 2 approaches from Divakars linked answer for comparison.
Code
import numpy as np
import numba as nb
from scipy.spatial.distance import cdist
#nb.njit(fastmath=True,parallel=True,cache=True)
def dist_1(mat,vec):
res=np.empty(mat.shape[0],dtype=mat.dtype)
for i in nb.prange(mat.shape[0]):
acc=0
for j in range(mat.shape[1]):
acc+=(mat[i,j]-vec[j])**2
res[i]=np.sqrt(acc)
return res
#from https://stackoverflow.com/a/52364284/4045774
def dist_2(mat,vec):
return cdist(mat, np.atleast_2d(vec)).ravel()
#from https://stackoverflow.com/a/52364284/4045774
def dist_3(mat,vec):
M = mat.dot(vec)
d = np.einsum('ij,ij->i',mat,mat) + np.inner(vec,vec) -2*M
return np.sqrt(d)
Timings
#Float64
a = np.random.rand(4000000, 128)
b = np.random.rand(128)
%timeit dist_1(a,b)
#122 ms ± 3.86 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
%timeit dist_2(a,b)
#484 ms ± 3.02 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
%timeit dist_3(a,b)
#432 ms ± 14.4 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
#Float32
a = np.random.rand(4000000, 128).astype(np.float32)
b = np.random.rand(128).astype(np.float32)
%timeit dist_1(a,b)
#68.6 ms ± 414 µs per loop (mean ± std. dev. of 7 runs, 1 loop each)
%timeit dist_2(a,b)
#2.2 s ± 32.9 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
#looks like there is a costly type-casting to float64
%timeit dist_3(a,b)
#228 ms ± 8.13 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

Python function with list

learn python by myself. I made a function for filling list. But I have 2 variants, and I want to discover which one is better and why. Or they both awful anyway I want to know truth.
def foo (x):
l = [0] * x
for i in range(x):
l[i] = i
return l
def foo1 (x):
l = []
for i in range(x):
l.append(i)
return l

from a performance perspective the first version foo is better:
%timeit foo(1000000)
# 52.4 ms ± 1.99 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
%timeit foo1(1000000)
# 67.2 ms ± 916 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
but the pythonic way to unpack an iterator in a list will be:
list(range(x))
also is faster:
%timeit list(range(1000000))
# 26.7 ms ± 661 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

Fastest Python log-sum-exp in a 'reduceat'

As part of a statistical programming package, I need to add log-transformed values together with the LogSumExp Function. This is significantly less efficient than adding unlogged values together.
Furthermore, I need to add values together using the numpy.ufunc.reduecat functionality.
There are various options I've considered, with code below:
(for comparison in non-log-space) use numpy.add.reduceat
Numpy's ufunc for adding logged values together: np.logaddexp.reduceat
Handwritten reduceat function with the following logsumexp functions:
scipy's implemention of logsumexp
logsumexp function in Python (with numba)
Streaming logsumexp function in Python (with numba)
def logsumexp_reduceat(arr, indices, logsum_exp_func):
res = list()
i_start = indices[0]
for cur_index, i in enumerate(indices[1:]):
res.append(logsum_exp_func(arr[i_start:i]))
i_start = i
res.append(logsum_exp_func(arr[i:]))
return res
#numba.jit(nopython=True)
def logsumexp(X):
r = 0.0
for x in X:
r += np.exp(x)
return np.log(r)
#numba.jit(nopython=True)
def logsumexp_stream(X):
alpha = -np.Inf
r = 0.0
for x in X:
if x != -np.Inf:
if x <= alpha:
r += np.exp(x - alpha)
else:
r *= np.exp(alpha - x)
r += 1.0
alpha = x
return np.log(r) + alpha
arr = np.random.uniform(0,0.1, 10000)
log_arr = np.log(arr)
indices = sorted(np.random.randint(0, 10000, 100))
# approach 1
%timeit np.add.reduceat(arr, indices)
12.7 µs ± 503 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
# approach 2
%timeit np.logaddexp.reduceat(log_arr, indices)
462 µs ± 17.1 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
# approach 3, scipy function
%timeit logsum_exp_reduceat(arr, indices, scipy.special.logsumexp)
3.69 ms ± 273 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
# approach 3 handwritten logsumexp
%timeit logsumexp_reduceat(log_arr, indices, logsumexp)
139 µs ± 7.1 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
# approach 3 streaming logsumexp
%timeit logsumexp_reduceat(log_arr, indices, logsumexp_stream)
164 µs ± 10.9 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
The timeit results show that handwritten logsumexp functions with numba are the fastest options, but are still 10x slower than numpy.add.reduceat.
A few questions:
Are there any other approaches (or tweaks to the options I've presented) which are faster? For instance, is there a way to use a lookup table to compute the logsumexp function?
Why is Sebastian Nowozin's "streaming logsumexp" function not faster than the naive approach?

There is some room for improvement
But never expect logsumexp to be as fast as a standard summation, because exp is quite a expensive operation.
Example
import numpy as np
#from version 0.43 until 0.47 this has to be set before importing numba
#Bug: https://github.com/numba/numba/issues/4689
from llvmlite import binding
binding.set_option('SVML', '-vector-library=SVML')
import numba as nb
#nb.njit(fastmath=True,parallel=False)
def logsum_exp_reduceat(arr, indices):
res = np.empty(indices.shape[0],dtype=arr.dtype)
for i in nb.prange(indices.shape[0]-1):
r = 0.
for j in range(indices[i],indices[i+1]):
r += np.exp(arr[j])
res[i]=np.log(r)
r = 0.
for j in range(indices[-1],arr.shape[0]):
r += np.exp(arr[j])
res[-1]=np.log(r)
return res
Timings
#small example where parallelization doesn't make sense
arr = np.random.uniform(0,0.1, 10_000)
log_arr = np.log(arr)
#use arrays if possible
indices = np.sort(np.random.randint(0, 10_000, 100))
%timeit logsum_exp_reduceat(arr, indices)
#without parallelzation 22 µs ± 173 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
#with parallelization 84.7 µs ± 32.2 µs per loop (mean ± std. dev. of 7 runs, 1 loop each)
%timeit np.add.reduceat(arr, indices)
#4.46 µs ± 61.1 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
#large example where parallelization makes sense
arr = np.random.uniform(0,0.1, 1000_000)
log_arr = np.log(arr)
indices = np.sort(np.random.randint(0, 1000_000, 100))
%timeit logsum_exp_reduceat(arr, indices)
#without parallelzation 1.57 ms ± 14.1 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
#with parallelization 409 µs ± 14.7 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%timeit np.add.reduceat(arr, indices)
#340 µs ± 11.1 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

Creating an array of numbers that add up to 1 with given length

I'm trying to use different weights for my model and I need those weights add up to 1 like this;
def func(length):
return ['a list of numbers add up to 1 with given length']
func(4) returns [0.1, 0.2, 0.3, 0.4]
The numbers should be linearly spaced and they should not start from 0. Is there any way to achieve this with numpy or scipy?

This can be done quite simply using numpy arrays:
def func(length):
linArr = np.arange(1, length+1)
return linArr/sum(x)
First we create an array of length length ranging from 1 to length. Then we normalize the sum.
Thanks to Paul Panzer for pointing out that the efficiency of this function can be improved by using Gauss's formula for the sum of the first n integers:
def func(length):
linArr = np.arange(1, length+1)
arrSum = length * (length+1) // 2
return linArr/arrSum

For large inputs, you might find that using np.linspace is faster than the accepted answer
def f1(length):
linArr = np.arange(1, length+1)
arrSum = length * (length+1) // 2
return linArr/arrSum
def f2(l):
delta = 2/(l*(l+1))
return np.linspace(delta, l*delta, l)
Ensure that the two things produce the same result:
In [39]: np.allclose(f1(1000000), f2(1000000))
Out[39]: True
Check timing of both:
In [68]: %timeit f1(10000000)
515 ms ± 28.2 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
In [69]: %timeit f2(10000000)
247 ms ± 4.57 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
It's tempting to just use np.arange(delta, l*delta, delta) which should be even faster, but this does present the risk of rounding errors causing the array to have lengths different from l (as will happen e.g. for l = 10000000).
If speed is more important than code style, it might also possible to squeeze out a bit more by using Numba:
from numba import jit
#jit
def f3(l):
a = np.empty(l, dtype=np.float64)
delta = 2/(l*(l+1))
for n in range(l):
a[n] = (n+1)*delta
return a
In [96]: %timeit f3(10000000)
216 ms ± 16.4 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
While we're at it, let's note that it's possible to parallelize this loop. Doing so naively with Numba doesn't appear to give much, but helping it out a bit and pre-splitting the array into num_parallel parts does give further improvement on a quad core system:
from numba import njit, prange
#njit(parallel=True)
def f4(l, num_parallel=4):
a = np.empty(l, dtype=np.float64)
delta = 2/(l*(l+1))
for j in prange(num_parallel):
# The last iteration gets whatever's left from rounding
offset = 0 if j != num_parallel - 1 else l % num_parallel
for n in range(l//num_parallel + offset):
i = j*(l//num_parallel) + n
a[i] = (i+1)*delta
return a
In [171]: %timeit f4(10000000, 4)
163 ms ± 13.2 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
In [172]: %timeit f4(10000000, 8)
158 ms ± 5.58 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
In [173]: %timeit f4(10000000, 12)
157 ms ± 8.77 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)