I have a numpy array with a shape of:
(11L, 5L, 5L)
I want to calculate the mean over the 25 elements of each 'slice' of the array [0, :, :], [1, :, :] etc, returning 11 values.
It seems silly, but I can't work out how to do this. I've thought the mean(axis=x) function would do this, but I've tried all possible combinations of axis and none of them give me the result I want.
I can obviously do this using a for loop and slicing, but surely there is a better way?
Use a tuple for axis :
>>> a = np.arange(11*5*5).reshape(11,5,5)
>>> a.mean(axis=(1,2))
array([ 12., 37., 62., 87., 112., 137., 162., 187., 212.,
237., 262.])
Edit: This works only with numpy version 1.7+.
You can reshape(11, 25) and then call mean only once (faster):
a.reshape(11, 25).mean(axis=1)
Alternatively, you can call np.mean twice (about 2X slower on my computer):
a.mean(axis=2).mean(axis=1)
Can always use np.einsum:
>>> a = np.arange(11*5*5).reshape(11,5,5)
>>> np.einsum('...ijk->...i',a)/(a.shape[-1]*a.shape[-2])
array([ 12, 37, 62, 87, 112, 137, 162, 187, 212, 237, 262])
Works on higher dimensional arrays (all of these methods would if the axis labels are changed):
>>> a = np.arange(10*11*5*5).reshape(10,11,5,5)
>>> (np.einsum('...ijk->...i',a)/(a.shape[-1]*a.shape[-2])).shape
(10, 11)
Faster to boot:
a = np.arange(11*5*5).reshape(11,5,5)
%timeit a.reshape(11, 25).mean(axis=1)
10000 loops, best of 3: 21.4 us per loop
%timeit a.mean(axis=(1,2))
10000 loops, best of 3: 19.4 us per loop
%timeit np.einsum('...ijk->...i',a)/(a.shape[-1]*a.shape[-2])
100000 loops, best of 3: 8.26 us per loop
Scales slightly better then the other methods as array size increases.
Using dtype=np.float64 does not change the above timings appreciably, so just to double check:
a = np.arange(110*50*50,dtype=np.float64).reshape(110,50,50)
%timeit a.reshape(110,2500).mean(axis=1)
1000 loops, best of 3: 307 us per loop
%timeit a.mean(axis=(1,2))
1000 loops, best of 3: 308 us per loop
%timeit np.einsum('...ijk->...i',a)/(a.shape[-1]*a.shape[-2])
10000 loops, best of 3: 145 us per loop
Also something that is interesting:
%timeit np.sum(a) #37812362500.0
100000 loops, best of 3: 293 us per loop
%timeit np.einsum('ijk->',a) #37812362500.0
100000 loops, best of 3: 144 us per loop
Related
For a numpy array
a = np.array([0, 1, 2, 3, 4, 5, 6, 7, 8])
You can get a slice using something like a[3:6]
But what about getting the rest of the slice? What is the most computationally efficient method for this? So something like a[:3, 6:].
The best I can come up with is to use a concatenate.
np.concatenate([a[:3], a[6:]], axis=0)
I am wondering if this is the best method, as I will be doing millions of these operations for a data processing pipeline.
Your solution seems to be the most efficient one since it is more than 2x faster than the next best thing.
a = np.array([0, 1, 2, 3, 4, 5, 6, 7, 8])
%timeit -n 100000 np.concatenate([a[:3], a[6:]], axis=0)
%timeit -n 100000 np.delete(a, slice(3, 6))
%timeit -n 100000 a[np.r_[:3,6:]]
>2.03 µs ± 75.5 ns per loop (mean ± std. dev. of 7 runs, 100,000 loops each)
>4.61 µs ± 146 ns per loop (mean ± std. dev. of 7 runs, 100,000 loops each)
>11 µs ± 350 ns per loop (mean ± std. dev. of 7 runs, 100,000 loops each)
However, the real question is if these operations (complement set of slice/deletion) need to be applied consecutively. Otherwise, you could aggregate the indices via set operations and slice the compliment a single time in the end to obtain the proper NumPy array.
I find declaring an empty array and filling it up seems to be very slightly better than using concat . As André mentioned in their comment this will vary based on the shape.
a = np.array([0, 1, 2, 3, 4, 5, 6, 7, 8])
def testing123():
new = np.zeros(6, dtype=int)
new[:3] = a[0:3]
new[3:] = a[6:]
return new
%timeit -n 100000 np.concatenate([a[:3], a[6:]], axis=0)
100000 loops, best of 5: 2.18 µs per loop
%timeit -n 100000 np.delete(a, slice(3, 6))
100000 loops, best of 5: 6.11 µs per loop
%timeit -n 100000 a[np.r_[:3,6:]]
100000 loops, best of 5: 16.4 µs per loop
%timeit -n 100000 testing123()
100000 loops, best of 5: 2.01 µs per loop
a = np.arange(10_000)
def testing123():
new = np.empty(5000, dtype=int)
new[:2500] = a[:2500]
new[2500:] = a[7500:]
return new
%timeit -n 100000 np.concatenate([a[:2500], a[7500:]], axis=0)
100000 loops, best of 5: 3.99 µs per loop
%timeit -n 100000 np.delete(a, slice(2500, 7500))
100000 loops, best of 5: 7.76 µs per loop
%timeit -n 100000 a[np.r_[:2500,7500:]]
100000 loops, best of 5: 47.3 µs per loop
%timeit -n 100000 testing123()
100000 loops, best of 5: 3.61 µs per loop
The following function apply numpy functions to two numpy arrays.
import numpy as np
def my_func(a: np.ndarray, b: np.ndarray) -> float:
return np.nanmin(a, axis=0) + np.nanmin(b, axis=0)
>>> my_func(np.array([1., 2., np.nan]), np.array([1., np.nan]))
2.0
However what is the best way to apply this same function to an np.array of np.array of different shape ?
a = np.array([np.array([1., 2]), np.array([1, 2., 3, np.nan])], dtype=object) # First array shape (2,), second (3,)
b = np.array([np.array([1]), np.array([1.5, 2.5, np.nan])], dtype=object)
np.vectorize does work
>>> np.vectorize(my_func)(a, b)
array([2. , 2.5])
but as specified by the vectorize documentation:
The vectorize function is provided primarily for convenience, not for
performance. The implementation is essentially a for loop.
Is there a more clever solution ?
I could use np.pad to have identifical shape but it seems sub-optimal as it requires to pad up to the maximum length of the inside arrays (here 4 for a and 3 for b).
I looked at numba and this stack exchange about performance but I am not sure of the best pratice for such a case.
Thanks !
Your function and arrays:
In [222]: def my_func(a: np.ndarray, b: np.ndarray) -> float:
...: return np.nanmin(a, axis=0) + np.nanmin(b, axis=0)
...:
In [223]: a = np.array([np.array([1., 2]), np.array([1, 2., 3, np.nan])], dtype=object
...: ) # First array shape (2,), second (3,)
...: b = np.array([np.array([1]), np.array([1.5, 2.5, np.nan])], dtype=object)
In [224]: a
Out[224]: array([array([1., 2.]), array([ 1., 2., 3., nan])], dtype=object)
In [225]: b
Out[225]: array([array([1]), array([1.5, 2.5, nan])], dtype=object)
Compare vectorize with a straightforward list comprehension:
In [226]: np.vectorize(my_func)(a, b)
Out[226]: array([2. , 2.5])
In [227]: [my_func(i,j) for i,j in zip(a,b)]
Out[227]: [2.0, 2.5]
and their times:
In [228]: timeit np.vectorize(my_func)(a, b)
157 µs ± 117 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
In [229]: timeit [my_func(i,j) for i,j in zip(a,b)]
85.9 µs ± 148 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
In [230]: timeit np.array([my_func(i,j) for i,j in zip(a,b)])
89.7 µs ± 1.03 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
If you are going to work with object arrays, frompyfunc is faster than vectorize:
In [231]: np.frompyfunc(my_func,2,1)(a, b)
Out[231]: array([2.0, 2.5], dtype=object)
In [232]: timeit np.frompyfunc(my_func,2,1)(a, b)
83.2 µs ± 50.1 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
I'm a bit surprised that it's even better than the list comprehension.
frompyfunc (and vectorize) are more useful when the inputs need to 'broadcast' against each other:
In [233]: np.frompyfunc(my_func,2,1)(a[:,None], b)
Out[233]:
array([[2.0, 2.5],
[2.0, 2.5]], dtype=object)
I'm not a numba expert, but I suspect it doesn't handle object dtype arrays, or it it does it doesn't improve speed much. Remember, object dtype means the elements are object references, just like in lists.
I get better times by using otypes and taking the function creation out of the timing loop:
In [235]: %%timeit f=np.vectorize(my_func, otypes=[float])
...: f(a, b)
...:
...:
95.5 µs ± 316 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
In [236]: %%timeit f=np.frompyfunc(my_func,2,1)
...: f(a, b)
...:
...:
81.1 µs ± 103 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
If you don't know about otypes, you haven't read the np.vectorize docs well enough.
I have a code where I do a lot of basic arithmetic calculations with a bunch of numerical data that is is multiple arrays. I have realized that in most concievable operations, numpy classes are always slower than the default python ones. Why is this?
For example I have a simple snippet where all I do is just update 1 numpy array element with another one retrieved from another numpy array, or I update it with the mathematical product of 2 other numpy array elements. It should be a basic operation, yet it will always be at least 2-3x slower than if I do it with list.
First I thought that it's because I haven't harmonized the data structures and the compiler has to do a lot of unecessary transformations. So then I recoded the whole thing and replaced every float with numpy.float64 and every list with numpy.ndarray, and the entire data is numpy.float64 all across the code so that it doesn't have to do any unecessary transformations.
The code is still 2-3 times slower than if I just use list and float.
For example:
ALPHA = [[random.uniform(*a_param) for k in range(l2)] for l in range(l1)]
COEFF = [[random.uniform(*c_param) for k in range(l2)] for l in range(l1)]
summa=0.0
for l in range(l1):
for k in range(l2):
summa+=COEFF[l][k] * ALPHA[l][k]
will always be 2-3x faster than:
ALPHA = numpy.random.uniform(*a_param, (l1,l2))
COEFF = numpy.random.uniform(*c_param, (l1,l2))
summa=0.0
for l in range(l1):
for k in range(l2):
summa+=COEFF[l][k] * ALPHA[l][k]
How is this possible, am I doing something wrong , since numpy is supposed to speed up things.
For the record I am using Python 3.5.3 and numpy (1.12.1), should I update?
Modifying a single element of a NumPy array is not expected to be faster than modifying a single element of a Python list. The speedup from using NumPy comes when you perform "vectorized" operations on entire arrays (or subsets of arrays). Try assigning the first 10000 elements of a NumPy array to be equal to the first 10000 elements of another, and compare that with using lists.
If your data and/or operations are very small (one or just a few elements), you are probably better off not using NumPy.
I tried two things:
Running your two blocks of code. For me, they were about the same speed.
Writing a new function that exploits numpy's vectorized math. This is several times faster than the other methods.
Here are my functions:
import numpy as np
def with_lists(l1, l2):
ALPHA = [[random.uniform(0, 1) for k in range(l2)] for l in range(l1)]
COEFF = [[random.uniform(0, 1) for k in range(l2)] for l in range(l1)]
summa=0.0
for l in range(l1):
for k in range(l2):
summa+=COEFF[l][k] * ALPHA[l][k]
return summa
def with_arrays(l1, l2):
ALPHA = np.random.uniform(size=(l1,l2))
COEFF = np.random.uniform(size=(l1,l2))
summa=0.0
for l in range(l1):
for k in range(l2):
summa+=COEFF[l][k] * ALPHA[l][k]
return summa
def with_ufunc(l1, l2):
"""Avoid the loop completely by exploitng numpy's
elementwise math."""
ALPHA = np.random.uniform(size=(l1,l2))
COEFF = np.random.uniform(size=(l1,l2))
return np.sum(COEFF * ALPHA)
When I compare the speed (I'm using the %timeit magic in IPython), I get the following:
>>> %timeit with_lists(10, 10)
107 µs ± 4.7 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
>>> %timeit with_arrays(10, 10)
91.9 µs ± 10.5 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
>>> %timeit with_ufunc(10, 10)
12.6 µs ± 589 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
The third function, without loops, about 10 to 30 times faster on my machine, depending on the values of l1 and l2.
Say I have the following numpy array
n = 50
a = np.array(range(1, 1000)) / 1000.
I would like to execute this line of code
%timeit v = [a ** k for k in range(0, n)]
1000 loops, best of 3: 2.01 ms per loop
However, this line of code will ultimately be executed in a loop, therefore I have performance issues.
Is there a way to optimize the loop? For example, the result of a specific calculation i in the list comprehension is simply the result of the previous calculation result in the loop, multiplied by a again.
I don't mind storing the results in a 2d-array instead of arrays in a list. That would probably be cleaner. By the way, I also tried the following, but it yields similar performance results:
k = np.array(range(0, n))
ones = np.ones(n)
temp = np.outer(a, ones)
And then performed the following calculation
%timeit temp ** k
1000 loops, best of 3: 1.96 ms per loop
or
%timeit np.power(temp, k)
1000 loops, best of 3: 1.92 ms per loop
But both yields similar results to the list comprehension above. By the way, n will always be an integer in my case.
In quick tests cumprod seems to be faster.
In [225]: timeit v = np.array([a ** k for k in range(0, n)])
2.76 ms ± 1.62 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
In [228]: %%timeit
...: A=np.broadcast_to(a[:,None],(len(a),50))
...: v1=np.cumprod(A,axis=1)
...:
208 µs ± 42.3 ns per loop (mean ± std. dev. of 7 runs, 1000 loops each)
To compare values I have to tweak ranges, since v includes a 0 power, while v1 starts with a 1 power:
In [224]: np.allclose(np.array(v)[1:], v1.T[:-1])
Out[224]: True
But the timings suggest that cumprod is worth refining.
The proposed duplicate was Efficient way to compute the Vandermonde matrix. That still has good ideas.
Say I have a 500000x1 array called A. I want to divide this array into 1000 equal sections, and then calculate the mean of that section. So I will end up with a 1000x1 array called B, in which B[1] is the mean of A[1:500], B[2] is the mean of B[501:1000]`, and so on. Since I will be doing this many many times, I want to do it efficiently. What's the most effective way of doing this in Matlab/Python?
NumPy/Python
We could reshape to have 500 columns and then compute average along the second axis -
A.reshape(-1,500).mean(axis=1)
Sample run -
In [89]: A = np.arange(50)+1;
In [90]: A.reshape(-1,5).mean(1)
Out[90]: array([ 3., 8., 13., 18., 23., 28., 33., 38., 43., 48.])
Runtime test :
An alternative method to get those average values would be with the old-fashioned way of computing the sum and then dividing by the number of elements involved in the summation. Let's time these two methods -
In [107]: A = np.arange(500000)+1;
In [108]: %timeit A.reshape(-1,500).mean(1)
1000 loops, best of 3: 1.19 ms per loop
In [109]: %timeit A.reshape(-1,500).sum(1)/500.0
1000 loops, best of 3: 583 µs per loop
Seems, like quite an improvement there with the alternative method! But wait, it's because with mean method NumPy is converting to float type by default and that conversion overhead showed up here.
So, if we use float type input arrays, we would have a different and a fair scenario -
In [144]: A = np.arange(500000).astype(float)+1;
In [145]: %timeit A.reshape(-1,500).mean(1)
1000 loops, best of 3: 534 µs per loop
In [146]: %timeit A.reshape(-1,500).sum(1)/500.0
1000 loops, best of 3: 516 µs per loop
MATLAB
With column-major ordering, we would reshape to have 500 rows and then average along the first dimension -
mean(reshape(A,500,[]),1)
Sample run -
>> A = 1:50;
>> mean(reshape(A,5,[]),1)
ans =
3 8 13 18 23 28 33 38 43 48
Runtime test :
Let's try out the old-fashioned way here too -
>> A = 1:500000;
>> func1 = #() mean(reshape(A,500,[]),1);
>> timeit(func1)
ans =
0.0013021
>> func2 = #() sum(reshape(A,500,[]),1)/500.0;
>> timeit(func2)
ans =
0.0012291