sign() much slower in python than matlab? - python

I have a function in python that basically takes the sign of an array (75,150), for example.
I'm coming from Matlab and the time execution looks more or less the same less this function.
I'm wondering if sign() works very slowly and you know an alternative to do the same.
Thx,

I can't tell you if this is faster or slower than Matlab, since I have no idea what numbers you're seeing there (you provided no quantitative data at all). However, as far as alternatives go:
import numpy as np
a = np.random.randn(75, 150)
aSign = np.sign(a)
Testing using %timeit in IPython:
In [15]: %timeit np.sign(a)
10000 loops, best of 3: 180 µs per loop
Because the loop over the array (and what happens inside it) is implemented in optimized C code rather than generic Python code, it tends to be about an order of magnitude faster—in the same ballpark as Matlab.
Comparing the exact same code as a numpy vectorized operation vs. a Python loop:
In [276]: %timeit [np.sign(x) for x in a]
1000 loops, best of 3: 276 us per loop
In [277]: %timeit np.sign(a)
10000 loops, best of 3: 63.1 us per loop
So, only 4x as fast here. (But then a is pretty small here.)

Related

numba #jit slower that pure python?

so i need to improve the execution time for a script that i have been working on. I started working with numba jit decorator to try parallel computing however it throws me
KeyError: "Does not support option: 'parallel'"
so i decided to test the nogil if it unlocks the whole capabilities from my cpu but it was slower than pure python i dont understand why this happened, and if someone can help me or guide me i will be very grateful
import numpy as np
from numba import *
#jit(['float64[:,:],float64[:,:]'],'(n,m),(n,m)->(n,m)',nogil=True)
def asd(x,y):
return x+y
u=np.random.random(100)
w=np.random.random(100)
%timeit asd(u,w)
%timeit u+w
10000 loops, best of 3: 137 µs per loop
The slowest run took 7.13 times longer than the fastest. This could mean that an intermediate result is being cached
1000000 loops, best of 3: 1.75 µs per loop
You cannot expect numba to outperform numpy on such a simple vectorized operation. Also your comparison isn't exactly fair since the numba function includes the cost of the outside function call. If you sum a larger array, you'll see that the performance of the two converge and what you are seeing is just overhead on a very fast operation:
import numpy as np
import numba as nb
#nb.njit
def asd(x,y):
return x+y
def asd2(x, y):
return x + y
u=np.random.random(10000)
w=np.random.random(10000)
%timeit asd(u,w)
%timeit asd2(u,w)
The slowest run took 17796.43 times longer than the fastest. This could mean
that an intermediate result is being cached.
100000 loops, best of 3: 6.06 µs per loop
The slowest run took 29.94 times longer than the fastest. This could mean that
an intermediate result is being cached.
100000 loops, best of 3: 5.11 µs per loop
As far as parallel functionality, for this simple operation, you can use nb.vectorize:
#nb.vectorize([nb.float64(nb.float64, nb.float64)], target='parallel')
def asd3(x, y):
return x + y
u=np.random.random((100000, 10))
w=np.random.random((100000, 10))
%timeit asd(u,w)
%timeit asd2(u,w)
%timeit asd3(u,w)
But again, if you operate on small arrays, you are going to be seeing the overhead of thread dispatch. For the array sizes above, I see the parallel giving me a 2x speedup.
Where numba really shines is doing operations that are difficult to do in numpy using broadcasting, or when operations would result in a lot of temporary intermediate array allocations.

Numpy: get the column and row index of the minimum value of a 2D array

For example,
x = array([[1,2,3],[3,2,5],[9,0,2]])
some_func(x) gives (2,1)
I know one can do it by a custom function:
def find_min_idx(x):
k = x.argmin()
ncol = x.shape[1]
return k/ncol, k%ncol
However, I am wondering if there's a numpy built-in function that does this faster.
Thanks.
EDIT: thanks for the answers. I tested their speeds as follows:
%timeit np.unravel_index(x.argmin(), x.shape)
#100000 loops, best of 3: 4.67 µs per loop
%timeit np.where(x==x.min())
#100000 loops, best of 3: 12.7 µs per loop
%timeit find_min_idx(x) # this is using the custom function above
#100000 loops, best of 3: 2.44 µs per loop
Seems the custom function is actually faster than unravel_index() and where(). unravel_index() does similar things as the custom function plus the overhead of checking extra arguments. where() is capable of returning multiple indices but is significantly slower for my purpose. Perhaps pure python code is not that slow for doing just two simple arithmetic and the custom function approach is as fast as one can get.
You may use np.where:
In [9]: np.where(x == np.min(x))
Out[9]: (array([2]), array([1]))
Also as #senderle mentioned in comment, to get values in an array, you can use np.argwhere:
In [21]: np.argwhere(x == np.min(x))
Out[21]: array([[2, 1]])
Updated:
As OP's times show, and much clearer that argmin is desired (no duplicated mins etc.), one way I think may slightly improve OP's original approach is to use divmod:
divmod(x.argmin(), x.shape[1])
Timed them and you will find that extra bits of speed, not much but still an improvement.
%timeit find_min_idx(x)
1000000 loops, best of 3: 1.1 µs per loop
%timeit divmod(x.argmin(), x.shape[1])
1000000 loops, best of 3: 1.04 µs per loop
If you are really concerned about performance, you may take a look at cython.
You can use np.unravel_index
print(np.unravel_index(x.argmin(), x.shape))
(2, 1)

Fast indexed dot-product for numpy/scipy

I'm using numpy to do linear algebra. I want to do fast subset-indexed dot and other linear operations.
When dealing with big matrices, slicing solution like A[:,subset].dot(x[subset]) may be longer than doing the multiplication on the full matrix.
A = np.random.randn(1000,10000)
x = np.random.randn(10000,1)
subset = np.sort(np.random.randint(0,10000,500))
Timings show that sub-indexing can be faster when columns are in one block.
%timeit A.dot(x)
100 loops, best of 3: 4.19 ms per loop
%timeit A[:,subset].dot(x[subset])
100 loops, best of 3: 7.36 ms per loop
%timeit A[:,:500].dot(x[:500])
1000 loops, best of 3: 1.75 ms per loop
Still the acceleration is not what I would expect (20x faster!).
Does anyone know an idea of a library/module that allow these kind of fast operation through numpy or scipy?
For now on I'm using cython to code a fast column-indexed dot product through the cblas library. But for more complex operation (pseudo-inverse, or subindexed least square solving) I'm not shure to reach good acceleration.
Thanks!
Well, this is faster.
%timeit A.dot(x)
#4.67 ms
%%timeit
y = numpy.zeros_like(x)
y[subset]=x[subset]
d = A.dot(y)
#4.77ms
%timeit c = A[:,subset].dot(x[subset])
#7.21ms
And you have all(d-ravel(c)==0) == True.
Notice that how fast this is depends on the input. With subset = array([1,2,3]) you have that the time of my solution is pretty much the same, while the timing of the last solution is 46micro seconds.
Basically this will be faster if the size ofsubset is not much smaller than the size of x

Are Numpy functions slow?

Numpy is supposed to be fast. However, when comparing Numpy ufuncs with standard Python functions I find that the latter are much faster.
For example,
aa = np.arange(1000000, dtype = float)
%timeit np.mean(aa) # 1000 loops, best of 3: 1.15 ms per loop
%timeit aa.mean # 10000000 loops, best of 3: 69.5 ns per loop
I got similar results with other Numpy functions like max, power. I was under the impression that Numpy has an overhead that makes it slower for small arrays but would be faster for large arrays. In the code above aa is not small: it has 1 million elements. Am I missing something?
Of course, Numpy is fast, only the functions seem to be slow:
bb = range(1000000)
%timeit mean(bb) # 1 loops, best of 3: 551 ms per loop
%timeit mean(list(bb)) # 10 loops, best of 3: 136 ms per loop
Others already pointed out that your comparison is not a real comparison (you are not calling the function + both are numpy).
But to give an answer to the question "Are numpy function slow?": generally speaking, no, numpy function are not slow (or not slower than plain python function). Off course there are some side notes to make:
'Slow' depends off course on what you compare with, and it can always faster. With things like cython, numexpr, numba, calling C-code, ... and others it is in many cases certainly possible to get faster results.
Numpy has a certain overhead, which can be significant in some cases. For example, as you already mentioned, numpy can be slower on small arrays and scalar math. For a comparison on this, see eg Are NumPy's math functions faster than Python's?
To make the comparison you wanted to make:
In [1]: import numpy as np
In [2]: aa = np.arange(1000000)
In [3]: bb = range(1000000)
For the mean (note, there is no mean function in python standard library: Calculating arithmetic mean (average) in Python):
In [4]: %timeit np.mean(aa)
100 loops, best of 3: 2.07 ms per loop
In [5]: %timeit float(sum(bb))/len(bb)
10 loops, best of 3: 69.5 ms per loop
For max, numpy vs plain python:
In [6]: %timeit np.max(aa)
1000 loops, best of 3: 1.52 ms per loop
In [7]: %timeit max(bb)
10 loops, best of 3: 31.2 ms per loop
As a final note, in the above comparison I used a numpy array (aa) for the numpy functions and a list (bb) for the plain python functions. If you would use a list with numpy functions, in this case it would again be slower:
In [10]: %timeit np.max(bb)
10 loops, best of 3: 115 ms per loop
because the list is first converted to an array (which consumes most of the time). So, if you want to rely on numpy in your application, it is important to make use of numpy arrays to store you data (or if you have a list, convert it to an array so this conversion has to be done only once).
You're not calling aa.mean. Put the function call parentheses on the end, to actually call it, and the speed difference will nearly vanish. (Both np.mean(aa) and aa.mean() are NumPy; neither uses Python builtins to do the math.)

fractional part of numpy array

I need to get only the fractional part of an array.
using numpy or simply python modf function is convenient.
In case we big arrays of positive fractional data, that can be as big as (1000000,3) for instance, what is more convenient to do:
numpy.modf(array)[0]
array-numpy.trunc(array)
In my opinion 2 is faster and cheaper in memory usage ... but not sure. What do python and numpy experts think ?
I'm not an expert, so I have to use the timeit module to check speed. I use IPython (which makes timing things really easy) but even without it the timeit module is probably the way to go.
In [21]: a = numpy.random.random((10**6, 3))
In [22]: timeit numpy.modf(a)[0]
10 loops, best of 3: 90.1 ms per loop
In [23]: timeit a-numpy.trunc(a)
10 loops, best of 3: 135 ms per loop
In [24]: timeit numpy.mod(a, 1.0)
10 loops, best of 3: 68.3 ms per loop
In [25]: timeit a % 1.0
10 loops, best of 3: 68.1 ms per loop
The last two are equivalent. I don't know much about memory use, but I'd be surprised if modf(a)[0] and a-numpy.trunc(a) both didn't use more memory than simply taking the mod directly.
[BTW, if your code does what you want it to and you're only interested in improvements, you might be interested in the codereview stackexchange. I still don't have a good handle on where the dividing line is, but this feels a little more like their cup of tea.]

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