While I was trying to understand numpy.tensordot(), I tried out the examples from the documentation and was convinced that we can get exactly same tensordoted result by different permutation of axes argument. For example, the below two permutations of the axes are equivalent (i.e. they both yield same result):
In [28]: a = np.arange(60.).reshape(3,4,5)
In [29]: b = np.arange(24.).reshape(4,3,2)
In [30]: perm1 = np.tensordot(a, b, axes=[(1, 0), (0, 1)])
In [31]: perm2 = np.tensordot(a, b, axes=[(0, 1), (1, 0)])
In [32]: np.all(perm1 == perm2)
Out[32]: True
However, while measuring the performance, I found that one permutation is little over 2x faster than the other and that puzzles me..
# setting up input arrays
In [19]: a = np.arange(30*40*50).reshape(30,40,50)
In [20]: b = np.arange(40*30*20).reshape(40,30,20)
# contracting the first two axes from the input tensors
In [21]: %timeit np.tensordot(a, b, axes=[(0, 1), (1, 0)])
3.23 ms ± 166 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
# an equivalent way of contraction of the first two
# axes from the input tensors as in the above case
In [22]: %timeit np.tensordot(a, b, axes=[(1, 0), (0, 1)])
1.62 ms ± 16.9 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
So, what is the reason for the 2x speedup in the latter case? Does it have to do with how NumPy ndarrays are structured internally in memory? Or something else? Thanks in advance for your insights!
Without going into the details, these two calculations recreate the actions taken by tensordot, and produce the same perm values.
They show the same sort of 2x speed difference:
In [24]: timeit np.dot(a.transpose(2,0,1).reshape(50,-1), b.transpose(1,0,2).reshape(-1,20))
4.39 ms ± 103 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
In [25]: timeit np.dot(a.transpose(2,1,0).reshape(50,-1), b.reshape(-1,20))
2.99 ms ± 97.4 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
My guess is that the 2nd is faster because the b.reshape(-1,20) does not require a copy, whereas the transpose followed by reshape in the 1st does.
And timing the different reshapes:
In [28]: timeit a.transpose(2,1,0).reshape(50,-1)
128 µs ± 978 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
In [29]: timeit a.transpose(2,0,1).reshape(50,-1)
1.04 µs ± 21.1 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
In [30]: timeit b.reshape(-1,20)
501 ns ± 14.6 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
In [31]: timeit b.transpose(1,0,2).reshape(-1,20)
27.5 µs ± 1.34 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
There are significant differences in speed. [30] is just a view, so that explains why it is so fast. I'm guessing [28] is so much slower because it involves a full reversal of elements, where as [29] copy (40,50) blocks.
Related
I have to do a lot of dot products in my data processing pipeline. So, I was experimenting with the following two pieces of code where one is 3 times efficient (in terms of runtime) when compared to its slowest counterpart.
slowest method (with arrays created on the fly)
In [33]: %timeit np.dot(np.arange(200000), np.arange(200000, 400000))
352 µs ± 958 ns per loop (mean ± std. dev. of 7 runs, 1000 loops each)
fastest method (with static arrays)
In [34]: vec1_arr = np.arange(200000)
In [35]: vec2_arr = np.arange(200000, 400000)
In [36]: %timeit np.dot(vec1_arr, vec2_arr)
121 µs ± 90.3 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
why is the first method of dynamically generating arrays 3x slower when compared to second method? Is it because in the first method much of these extra time is spent in allocating memory for the elements? Or some other factors contributing to this degradation?
To gain little more understanding, I also replicated the setup in pure Python. And surprisingly there is no performance difference between doing it one way or the other, although it is slower than the numpy implementation, which is obvious and expected.
In [42]: %timeit sum(map(operator.mul, range(200000), range(200000, 400000)))
12.5 ms ± 71.7 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
In [38]: vec1 = range(200000)
In [39]: vec2 = range(200000, 400000)
In [40]: %timeit sum(map(operator.mul, vec1, vec2))
12.5 ms ± 27.6 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
The behaviour in the case of pure Python is clear because range function doesn't actually create all those elements. It does lazy evaluation (i.e. it is generated on the fly).
Note: The pure Python impl. is just to make myself convinced that the array allocation might be the factor that is causing the drag. It's not meant to compare it with NumPy implementation.
The pure Python test is not fair. Because np.arange(200000) really returns an array while range(200000) only returns a generator. So these two methods both create arrays on the fly.
import operator
%timeit sum(map(operator.mul, range(200000), range(200000, 400000)))
# 15.1 ms ± 45.7 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
vec1 = range(200000)
vec2 = range(200000, 400000)
%timeit sum(map(operator.mul, vec1, vec2))
# 15.2 ms ± 117 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
vec1 = list(range(200000))
vec2 = list(range(200000, 400000))
%timeit sum(map(operator.mul, vec1, vec2))
# 12.4 ms ± 716 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
And we can see the time cost on allocation:
import numpy as np
%timeit np.arange(200000), np.arange(200000, 400000)
# 632 µs ± 9.45 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%timeit np.dot(np.arange(200000), np.arange(200000, 400000))
# 703 µs ± 5.27 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
vec1_arr = np.arange(200000)
vec2_arr = np.arange(200000, 400000)
%timeit np.dot(vec1_arr, vec2_arr)
# 77.7 µs ± 427 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
Make sense.
The difference in speed is due to allocating the arrays in the slower case. I pasted the output of %timeit that takes into account the allocation of arrays in the two cases. The OP's timeit commands only took into account allocation in the slower case but not in the faster case.
%timeit np.dot(np.arange(200000), np.arange(200000, 400000))
# 524 µs ± 11.1 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%timeit vec1_arr = np.arange(200000); vec2_arr = np.arange(200000, 400000); np.dot(vec1_arr, vec2_arr)
# 523 µs ± 17.8 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
The allocation of both arrays takes about 360 microseconds on my machine, and the np.dot operation takes 169 microseconds. The sum of those two durations is 529 microseconds, which is equivalent to the output of %timeit output above.
%timeit vec1_arr = np.arange(200000); vec2_arr = np.arange(200000, 400000)
# 360 µs ± 16.9 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
vec1_arr = np.arange(200000)
vec2_arr = np.arange(200000, 400000)
%timeit np.dot(vec1_arr, vec2_arr)
# 169 µs ± 5.24 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
I'm trying to find the best way to compute the minimum element wise products between two sets of vectors. The usual matrix multiplication C=A#B computes Cij as the sum of the pairwise products of the elements of the vectors Ai and B^Tj. I would like to perform instead the minimum of the pairwise products. I can't find an efficient way to do this between two matrices with numpy.
One way to achieve this would be to generate the 3D matrix of the pairwise products between A and B (before the sum) and then take the minimum over the third dimension. But this would lead to a huge memory footprint (and I actually dn't know how to do this).
Do you have any idea how I could achieve this operation ?
Example:
A = [[1,1],[1,1]]
B = [[0,2],[2,1]]
matrix matmul:
C = [[1*0+1*2,1*2+1*1][1*0+1*2,1*2+1*1]] = [[2,3],[2,3]]
minimum matmul:
C = [[min(1*0,1*2),min(1*2,1*1)][min(1*0,1*2),min(1*2,1*1)]] = [[0,1],[0,1]]
Use broadcasting after extending A to 3D -
A = np.asarray(A)
B = np.asarray(B)
C_out = np.min(A[:,None]*B,axis=2)
If you care about memory footprint, use numexpr module to be efficient about it -
import numexpr as ne
C_out = ne.evaluate('min(A3D*B,2)',{'A3D':A[:,None]})
Timings on large arrays -
In [12]: A = np.random.rand(200,200)
In [13]: B = np.random.rand(200,200)
In [14]: %timeit np.min(A[:,None]*B,axis=2)
34.4 ms ± 614 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
In [15]: %timeit ne.evaluate('min(A3D*B,2)',{'A3D':A[:,None]})
29.3 ms ± 316 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
In [16]: A = np.random.rand(300,300)
In [17]: B = np.random.rand(300,300)
In [18]: %timeit np.min(A[:,None]*B,axis=2)
113 ms ± 2.27 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
In [19]: %timeit ne.evaluate('min(A3D*B,2)',{'A3D':A[:,None]})
102 ms ± 691 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
So, there's some improvement with numexpr, but maybe not as much I was expecting it to be.
Numba can be also an option
I was a bit surprised of the not particularly good Numexpr Timings, so I tried a Numba Version. For large Arrays this can be optimized further. (Quite the same principles like for a dgemm can be applied)
import numpy as np
import numba as nb
import numexpr as ne
#nb.njit(fastmath=True,parallel=True)
def min_pairwise_prod(A,B):
assert A.shape[1]==B.shape[1]
res=np.empty((A.shape[0],B.shape[0]))
for i in nb.prange(A.shape[0]):
for j in range(B.shape[0]):
min_prod=A[i,0]*B[j,0]
for k in range(B.shape[1]):
prod=A[i,k]*B[j,k]
if prod<min_prod:
min_prod=prod
res[i,j]=min_prod
return res
Timings
A=np.random.rand(300,300)
B=np.random.rand(300,300)
%timeit res_1=min_pairwise_prod(A,B) #parallel=True
5.56 ms ± 1.6 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
%timeit res_1=min_pairwise_prod(A,B) #parallel=False
26 ms ± 163 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
%timeit res_2 = ne.evaluate('min(A3D*B,2)',{'A3D':A[:,None]})
87.7 ms ± 265 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
%timeit res_3=np.min(A[:,None]*B,axis=2)
110 ms ± 214 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
A=np.random.rand(1000,300)
B=np.random.rand(1000,300)
%timeit res_1=min_pairwise_prod(A,B) #parallel=True
50.6 ms ± 401 µs per loop (mean ± std. dev. of 7 runs, 1 loop each)
%timeit res_1=min_pairwise_prod(A,B) #parallel=False
296 ms ± 5.02 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
%timeit res_2 = ne.evaluate('min(A3D*B,2)',{'A3D':A[:,None]})
992 ms ± 7.59 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
%timeit res_3=np.min(A[:,None]*B,axis=2)
1.27 s ± 15.4 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
I was comparing the performance of counting how many letters 'C' are in a very long string, using a numpy array of characters and the string method count.
genome is a very long string.
g1 = genome
g2 = np.array([i for i in genome])
%timeit np.sum(g2=='C')
4.43 s ± 230 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
%timeit g1.count('C')
955 ms ± 6.42 ms per loop (mean ± std. dev. of 7 runs, 1 loop each).
I expected that a numpy array would compute it faster but I am wrong.
Can someone explain me how the count method works and what is it faster than using a numpy array?
Thank you!
Let's explore some variations on the problem. I won't try to make as large a string as yours.
In [393]: astr = 'ABCDEF'*10000
First the string count:
In [394]: astr.count('C')
Out[394]: 10000
In [395]: timeit astr.count('C')
70.2 µs ± 115 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
Now try a 1 element array with that string:
In [396]: arr = np.array(astr)
In [397]: arr.shape
Out[397]: ()
In [398]: np.char.count(arr, 'C')
Out[398]: array(10000)
In [399]: timeit np.char.count(arr, 'C')
200 µs ± 2.97 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
In [400]: arr.dtype
Out[400]: dtype('<U60000')
My experience with other uses of char is that it iterates on the array elements and applies the string method. So it can't be faster than applying the string method directly. I suppose the rest of the time is some sort of numpy overhead.
Make a list from the string - one character per list element:
In [402]: alist = list(astr)
In [403]: alist.count('C')
Out[403]: 10000
In [404]: timeit alist.count('C')
955 µs ± 18.8 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
The list count has to loop through the elements, and do the test against C each time. Still it is faster than sum(i=='C' for i in alist) (and variants).
Now make an array from that list - single character elements:
In [405]: arr1 = np.array(alist)
In [406]: arr1.shape
Out[406]: (60000,)
In [407]: timeit arr1=='C'
634 µs ± 12.8 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
In [408]: timeit np.sum(arr1=='C')
740 µs ± 23.3 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
The np.sum is relatively fast. It's the check against 'C' that takes the most time.
If I construct a numeric array of the same size, the count time is quite a bit faster. The equality test against a number is faster than the equivalent string test.
In [431]: arr2 = np.resize(np.array([1,2,3,4,5,6]),arr1.shape[0])
In [432]: np.sum(arr2==3)
Out[432]: 10000
In [433]: timeit np.sum(arr2==3)
155 µs ± 1.66 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
numpy does not promise to be faster for all Python operations. For the most part when working string elements, it is heavily dependent on Python's own string code.
I have a numpy array of shape(64,64,3). How can we get three arrays of dimension (64,64) each.
You can use moveaxis to move the to split axis all the way to the left and then use sequence unpacking:
x,y,z = np.moveaxis(arr, -1, 0)
This is another case where iteration, for a few steps, is not bad.
In [145]: arr = np.ones((64,64,3))
Unpacking a list comprehension:
In [146]: a,b,c = [arr[:,:,i] for i in range(3)]
In [147]: a.shape
Out[147]: (64, 64)
Unpacking a transposed array:
In [148]: a,b,c = np.moveaxis(arr,-1,0)
In [149]: a.shape
Out[149]: (64, 64)
Timings for such a small example aren't ground breaking, but they hint at the relative advantages of the two approaches:
In [150]: timeit a,b,c = [arr[:,:,i] for i in range(3)]
3.02 µs ± 10.6 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
In [151]: timeit a,b,c = np.moveaxis(arr,-1,0)
18.4 µs ± 946 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
The array unpacking (on the 1st dimension) requires converting the array into a list of 3 subarrays, as can be seen by these similar timings:
In [154]: timeit a,b,c = list(np.moveaxis(arr,-1,0))
17.8 µs ± 15.1 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
In [155]: timeit a,b,c = [i for i in np.moveaxis(arr,-1,0)]
17.9 µs ± 14.6 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
It's not the unpacking or iteration that taking time. It's the moveaxis:
In [156]: timeit np.moveaxis(arr,-1,0)
14 µs ± 4.41 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
Looking at its code, we see that it uses transpose, after first constructing an order from the parameters:
Calling transpose directly is fast (since it just involves changing shape and strides):
In [157]: timeit arr.transpose(2,0,1)
688 ns ± 1.87 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
And unpacking that transpose is a bit faster than my original list comprehension.
In [158]: timeit a,b,c = arr.transpose(2,0,1)
2.78 µs ± 9.3 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
So moveaxis has a significant overhead relative to the rest of the task. That said, it probably does not increase with the side of the array. It's a fixed overhead.
Since for my program fast indexing of Numpy arrays is quite necessary and fancy indexing doesn't have a good reputation considering performance, I decided to make a few tests. Especially since Numba is developing quite fast, I tried which methods work well with numba.
As inputs I've been using the following arrays for my small-arrays-test:
import numpy as np
import numba as nb
x = np.arange(0, 100, dtype=np.float64) # array to be indexed
idx = np.array((0, 4, 55, -1), dtype=np.int32) # fancy indexing array
bool_mask = np.zeros(x.shape, dtype=np.bool) # boolean indexing mask
bool_mask[idx] = True # set same elements as in idx True
y = np.zeros(idx.shape, dtype=np.float64) # output array
y_bool = np.zeros(bool_mask[bool_mask == True].shape, dtype=np.float64) #bool output array (only for convenience)
And the following arrays for my large-arrays-test (y_bool needed here to cope with dupe numbers from randint):
x = np.arange(0, 1000000, dtype=np.float64)
idx = np.random.randint(0, 1000000, size=int(1000000/50))
bool_mask = np.zeros(x.shape, dtype=np.bool)
bool_mask[idx] = True
y = np.zeros(idx.shape, dtype=np.float64)
y_bool = np.zeros(bool_mask[bool_mask == True].shape, dtype=np.float64)
This yields the following timings without using numba:
%timeit x[idx]
#1.08 µs ± 21 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
#large arrays: 129 µs ± 3.45 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
%timeit x[bool_mask]
#482 ns ± 18.1 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
#large arrays: 621 µs ± 15.9 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%timeit np.take(x, idx)
#2.27 µs ± 104 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
# large arrays: 112 µs ± 5.76 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
%timeit np.take(x, idx, out=y)
#2.65 µs ± 134 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
# large arrays: 134 µs ± 4.47 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
%timeit x.take(idx)
#919 ns ± 21.3 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
# large arrays: 108 µs ± 1.71 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
%timeit x.take(idx, out=y)
#1.79 µs ± 40.7 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
# larg arrays: 131 µs ± 2.92 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
%timeit np.compress(bool_mask, x)
#1.93 µs ± 95.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
# large arrays: 618 µs ± 15.8 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%timeit np.compress(bool_mask, x, out=y_bool)
#2.58 µs ± 167 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
# large arrays: 637 µs ± 9.88 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%timeit x.compress(bool_mask)
#900 ns ± 82.4 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
# large arrays: 628 µs ± 17.8 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%timeit x.compress(bool_mask, out=y_bool)
#1.78 µs ± 59.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
# large arrays: 628 µs ± 13.8 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%timeit np.extract(bool_mask, x)
#5.29 µs ± 194 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
# large arrays: 641 µs ± 13 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
And with numba, using jitting in nopython-mode, caching and nogil I decorated the ways of indexing, which are supported by numba:
#nb.jit(nopython=True, cache=True, nogil=True)
def fancy(x, idx):
x[idx]
#nb.jit(nopython=True, cache=True, nogil=True)
def fancy_bool(x, bool_mask):
x[bool_mask]
#nb.jit(nopython=True, cache=True, nogil=True)
def taker(x, idx):
np.take(x, idx)
#nb.jit(nopython=True, cache=True, nogil=True)
def ndtaker(x, idx):
x.take(idx)
This yields the following results for small and large arrays:
%timeit fancy(x, idx)
#686 ns ± 25.1 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
# large arrays: 84.7 µs ± 1.82 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
%timeit fancy_bool(x, bool_mask)
#845 ns ± 31 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
# large arrays: 843 µs ± 14.2 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%timeit taker(x, idx)
#814 ns ± 21.1 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
# large arrays: 87 µs ± 1.52 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
%timeit ndtaker(x, idx)
#831 ns ± 24.5 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
# large arrays: 85.4 µs ± 2.69 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
Summary
While for numpy without numba it is clear that small arrays are by far best indexed with boolean masks (about a factor 2 compared to ndarray.take(idx)), for larger arrays ndarray.take(idx) will perform best, in this case around 6 times faster than boolean indexing. The breakeven-point is at an array-size of around 1000 cells with and index-array-size of around 20 cells.
For arrays with 1e5 elements and 5e3 index array size, ndarray.take(idx) will be around 10 times faster than boolean mask indexing. So it seems that boolean indexing seems to slow down considerably with array size, but catches up a little after some array-size-threshold is reached.
For the numba jitted functions there is a small speedup for all indexing functions except for boolean mask indexing. Simple fancy indexing works best here, but is still slower than boolean masking without jitting.
For larger arrays boolean mask indexing is a lot slower than the other methods, and even slower than the non-jitted version. The three other methods all perform quite good and around 15% faster than the non-jitted version.
For my case with many arrays of different sizes, fancy indexing with numba is the best way to go. Perhaps some other people can also find some useful information in this quite lengthy post.
Edit:
I'm sorry that I forgot to ask my question, which I in fact have. I was just rapidly typing this at the end of my workday and completely forgot it...
Well, do you know any better and faster method than those that I tested? Using Cython my timings were between Numba and Python.
As the index array is predefined once and used without alteration in long iterations, any way of pre-defining the indexing process would be great. For this I thought about using strides. But I wasn't able to pre-define a custom set of strides. Is it possible to get a predefined view into the memory using strides?
Edit 2:
I guess I'll move my question about predefined constant index arrays which will be used on the same value array (where only the values change but not the shape) for a few million times in iterations to a new and more specific question. This question was too general and perhaps I also formulated the question a little bit misleading. I'll post the link here as soon as I opened the new question!
Here is the link to the followup question.
Your summary isn't completely correct, you already did tests with differently sized arrays but one thing that you didn't do was to change the number of elements indexed.
I restricted it to pure indexing and omitted take (which effectively is integer array indexing) and compress and extract (because these are effectively boolean array indexing). The only difference for these are the constant factors. The constant factor for the methods take and compress will be less than the overhead for the numpy functions np.take and np.compress but otherwise the effects will be negligible for reasonably sized arrays.
Just let me present it with different numbers:
# ~ every 500th element
x = np.arange(0, 1000000, dtype=np.float64)
idx = np.random.randint(0, 1000000, size=int(1000000/500)) # changed the ratio!
bool_mask = np.zeros(x.shape, dtype=np.bool)
bool_mask[idx] = True
%timeit x[idx]
# 51.6 µs ± 2.02 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
%timeit x[bool_mask]
# 1.03 ms ± 37.1 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
# ~ every 50th element
idx = np.random.randint(0, 1000000, size=int(1000000/50)) # changed the ratio!
bool_mask = np.zeros(x.shape, dtype=np.bool)
bool_mask[idx] = True
%timeit x[idx]
# 1.46 ms ± 55.1 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%timeit x[bool_mask]
# 2.69 ms ± 154 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
# ~ every 5th element
idx = np.random.randint(0, 1000000, size=int(1000000/5)) # changed the ratio!
bool_mask = np.zeros(x.shape, dtype=np.bool)
bool_mask[idx] = True
%timeit x[idx]
# 14.9 ms ± 495 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit x[bool_mask]
# 8.31 ms ± 181 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
So what happened here? It's simple: Integer array indexing only needs to access as many elements as there are values in the index-array. That means if there are few matches it will be quite fast but slow if there are many indices. Boolean array indexing, however, always needs to walk through the whole boolean array and check for "true" values. That means it should be roughly "constant" for the array.
But, wait, it's not really constant for boolean arrays and why does integer array indexing take longer (last case) than boolean array indexing even if it has to process ~5 times less elements?
That's where it gets more complicated. In this case the boolean array had True at random places which means that it will be subject to branch prediction failures. These will be more likely if True and False will have equal occurrences but at random places. That's why the boolean array indexing got slower - because the ratio of True to False got more equal and thus more "random". Also the result array will be larger if there are more Trues which also consumes more time.
As an example for this branch prediction thing use this as example (could differ with different system/compilers):
bool_mask = np.zeros(x.shape, dtype=np.bool)
bool_mask[:1000000//2] = True # first half True, second half False
%timeit x[bool_mask]
# 5.92 ms ± 118 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
bool_mask = np.zeros(x.shape, dtype=np.bool)
bool_mask[::2] = True # True and False alternating
%timeit x[bool_mask]
# 16.6 ms ± 361 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
bool_mask = np.zeros(x.shape, dtype=np.bool)
bool_mask[::2] = True
np.random.shuffle(bool_mask) # shuffled
%timeit x[bool_mask]
# 18.2 ms ± 325 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
So the distribution of True and False will critically affect the runtime with boolean masks even if they contain the same amount of Trues! The same effect will be visible for the compress-functions.
For integer array indexing (and likewise np.take) another effect will be visible: cache locality. The indices in your case are randomly distributed, so your computer has to do a lot of "RAM" to "processor cache" loads because it's very unlikely two indices will be near to each other.
Compare this:
idx = np.random.randint(0, 1000000, size=int(1000000/5))
%timeit x[idx]
# 15.6 ms ± 703 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
idx = np.random.randint(0, 1000000, size=int(1000000/5))
idx = np.sort(idx) # sort them
%timeit x[idx]
# 4.33 ms ± 366 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
By sorting the indices the chances immensely increased that the next value will already be in the cache and this can lead to huge speedups. That's a very important factor if you know that the indices will be sorted (for example if they were created by np.where they are sorted, which makes the result of np.where especially efficient for indexing).
So, it's not like integer array indexing is slower for small arrays and faster for large arrays it depends on much more factors. Both do have their use-cases and depending on the circumstances one might be (considerably) faster than the other.
Let me also talk a bit about the numba functions. First some general statements:
cache won't make a difference, it just avoids recompiling the function. In interactive environments this is essentially useless. It's faster if you would package the functions in a module though.
nogil by itself won't provide any speed boost. It will be faster if it's called in different threads because each function execution can release the GIL and then multiple calls can run in parallel.
Otherwise I don't know how numba effectivly implements these functions, however when you use NumPy features in numba it could be slower or faster - but even if it's faster it won't be much faster (except maybe for small arrays). Because if it could be made faster the NumPy developers would also implement it. My rule of thumb is: If you can do it (vectorized) with NumPy don't bother with numba. Only if you can't do it with vectorized NumPy functions or NumPy would use too many temporary arrays then numba will shine!