I was just revisiting some of my code to improve the performance and stumpled over something strange:
a = np.linspace(10,1000,1000000).reshape(1000,1000)
%timeit np.square(a)
100 loops, best of 3: 8.07 ms per loop
%timeit a*a
100 loops, best of 3: 8.18 ms per loop
%timeit a**2
100 loops, best of 3: 8.32 ms per loop
Ok it seems to have some overhead when using the power-operator (**) but otherwise they seem identical (I guess NumPy is doing that) but then it got strange:
In [46]: %timeit np.power(a, 2)
10 loops, best of 3: 121 ms per loop
So there is no problem but it seems a bit inconsistent to have a fallback for the magic pow but not for the UFUNC. But then I got interested since I am using third powers a lot:
%timeit a*a*a
100 loops, best of 3: 18.1 ms per loop
%timeit a**3
10 loops, best of 3: 121 ms per loop
%timeit np.power(a, 3)
10 loops, best of 3: 121 ms per loop
There seems to be no "shortcut" in the third power and UFUNC and 'magic-pow' work the same (at least in regard to performance).
But that's not that good since I want a consistent method of using powers in my code and I'm not quite sure how to wrap the __pow__ of numpy.
So to get to the point, my question is :
Is there a way I can wrap the numpys __pow__ method? Because I want a consistent way of writing powers in my script not writing a**2 and at another place power(a, 3). Simply writing a**3, and redirecting this to my power function, would be preferred (but for that I would need to somehow wrap the ndarrays __pow__ or?).
Currently I am using a shortcut but that's not that beautiful (I even have to declare the exponent==2 case since np.power performs not optimal there):
def power(array, exponent):
if exponent == 2: #catch this, or it calls the slow np.power(array, exponent)
return np.square(array)
if exponent == 3:
return array * array * array
#As soon as np.cbrt is avaiable catch the exponent 4/3 here too
return np.power(array, exponent)
%timeit power(a, 3)
100 loops, best of 3: 17.8 ms per loop
%timeit a**3
10 loops, best of 3: 121 ms per loop
I am using NumPy v1.9.3 and I do not want to subclass np.ndarray just for wrapping the __pow__ method. :-)
EDIT: I rewrote the part where I get to my question. To clarify it: I am not asking about why NumPy does it the way it does - that is just to explain why I ask the question.
This is a good catch. I too wonder why is that behavior. But to be short and concise answering the question, I would just do:
def mypower(array, exponent):
return reduce(lambda x,y: x*y, [array for _ in range(exponent)])
%timeit mypower(a,2)
100 loops, best of 3: 3.68 ms per loop
%timeit mypower(a,3)
100 loops, best of 3: 8.09 ms per loop
%timeit mypower(a,4)
100 loops, best of 3: 12.6 ms per loop
Obsviouly the overhead increases with the exponent but for low ones is better than 10x the time.
Note that this is different from the original numpy implementation which is not specific for a numeric exponent and supports an array of exponents as the second argument (check it out here).
Overloading the operator
The way to do what you want is to subclass ndarray and use views. See the following example:
import numexpr
import numpy as np
class MyArray(np.ndarray):
def __pow__(self, other):
return reduce(lambda x,y: x*y, [self for _ in range(other)])
class NumExprArray(np.ndarray):
def __pow__(self, other):
return numexpr.evaluate("self**%f" % other)
#This implies extra overhead, is as much as 4x slower:
#return numexpr.evaluate("self**other")
a = np.linspace(10,1000,1000000).reshape(1000,1000).view(MyArray)
na = np.linspace(10,1000,1000000).reshape(1000,1000).view(NumExprArray)
%timeit a**2
1000 loops, best of 3: 1.2 ms per loop
%timeit na**2
1000 loops, best of 3: 1.14 ms per loop
%timeit a**3
100 loops, best of 3: 4.69 ms per loop
%timeit na**3
100 loops, best of 3: 2.36 ms per loop
%timeit a**4
100 loops, best of 3: 6.59 ms per loop
%timeit na**4
100 loops, best of 3: 2.4 ms per loop
For more information on this method please follow this link. Another way would be to use a custom infix operator but for readability purposes is not so good. As one can see, numexpr should be the way to go.
If I read the source correctly, when numpy performs power, it checks whether the numerical value of the exponent is one of the special cases (-0.5, 0, 0.5, 1, and 2). If so, the operation is done using special routines. All other numerical values of the exponent are considered "general", and will be fed into the generic power function, which may be slow (especially if the exponent is promoted to floating-point type, but I'm not sure if this is the case with a ** 3).
Related
I want to compute the output error for a neural network for each input by compare output signal and its true output value so I need two matrix to compute this task.
I have output matrix in shape of (n*1) but in the label I just have the index of neuron that should be activated, so I need a matrix in the same shape with all element equal to zero except the one which it's index is equal to the label. I could do that with a function but I wonder is there a built in method in numpy python that can do that for me?
You can do that multiple ways using numpy or standard libraries, one way is to create an array of zeros, and set the value corresponding to index as 1.
n = len(result)
a = np.zeros((n,));
a[id] = 1
It probably is going to be the fastest one as well:
>> %timeit a = np.zeros((n,)); a[id] = 1
1000000 loops, best of 3: 634 ns per loop
Alternatively you can use numpy.pad to pad [ 1 ] array with zeros. But this will almost definitely will be slower due to padding logic.
np.lib.pad([1],(id,n-id),'constant', constant_values=(0))
As expected order of magnitude slower:
>> %timeit np.lib.pad([1],(id,n-id),'constant', constant_values=(0))
10000 loops, best of 3: 47.4 µs per loop
And you can try list comprehension as suggested by the comments:
results = [7]
np.matrix([1 if x == id else 0 for x in results])
But it is much slower than the first method as well:
>> %timeit np.matrix([1 if x == id else 0 for x in results])
100000 loops, best of 3: 7.25 µs per loop
Edit:
But in my opinion, if you want to compute the neural networks error. You should just use np.argmax and compute whether it was successful or not. That error calculation may give you more noise than it is useful. You can make a confusion matrix if you feel your network is prone to similarities.
A few other methods that also seem to be slower than #umutto's above:
%timeit a = np.zeros((n,)); a[id] = 1 #umutto's method
The slowest run took 45.34 times longer than the fastest. This could mean that an intermediate result is being cached.
1000000 loops, best of 3: 1.53 µs per loop
Boolean construction:
%timeit a = np.arange(n) == id
The slowest run took 13.98 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 3.76 µs per loop
Boolean construction to integer:
%timeit a = (np.arange(n) == id).astype(int)
The slowest run took 15.31 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 5.47 µs per loop
List construction:
%timeit a = [0]*n; a[id] = 1; a=np.asarray(a)
10000 loops, best of 3: 77.3 µs per loop
Using scipy.sparse
%timeit a = sparse.coo_matrix(([1], ([id],[0])), shape=(n,1))
10000 loops, best of 3: 51.1 µs per loop
Now what's actually faster may depend on what's being cached, but it seems like constructing the zero array is probably fastest, especially if you can use np.zeros_like(result) instead of np.zeros(len(result))
One liner:
x = np.identity(n)[id]
How would you vectorize the evaluation of arrays of lambda functions?
Here's an example to understand what I'm talking about. (And even though I'm using numpy arrays, I'm not limiting myself to only using numpy.)
Let's say I have the following numpy arrays.
array1 = np.array(["hello", 9])
array2 = np.array([lambda s: s == "hello", lambda num: num < 10])
(You could store these kinds of objects in numpy without throwing an error, believe it or not.) What I want is something akin to the following.
array2 * array1
# Return np.array([True, True]). PS: An explanation of how to `AND` all of
# booleans together quickly would be nice too.
Of course, this seems impractical for arrays of size 2, but for arrays of arbitrary sizes, I'll assume this would yield a performance boost because of all of the low level optimizations.
So, anyone know how to write this weird kind of python code?
The simple answer, of course, is that you can't easily do this with numpy (or with standard Python, for that matter). Numpy doesn't actually vectorize most operations itself, to my knowledge: it uses libraries like BLAS/ATLAS/etc that do for certain situations. Even if it did, it would do so in C for specific situations: it certainly can't vectorize Python function execution.
If you want to involve multiprocessing in this, it is possible, but it depends on your situation. Are your individual function applications time-consuming, making them feasible to send out one-by-one, or do you need a very large number of fast function executions, in which case you'd probably want to send batches of them to each process?
In general, because of what could be argued as poor fundamental design (eg, the Global Interpreter Lock), it's very difficult with standard Python to have lightweight parallelization as you're hoping for here. There are significantly heavier methods, like the multiprocessing module or Ipython.parallel, but these require some work to use.
Alright guys, I have an answer: numpy's vectorize.
Please read the edited section though. You'll discover that python actually optimizes code for you, which actually defeats the purpose of using numpy arrays in this case. (But using numpy arrays does not decrease the performance.)
The last test really shows is that python lists are as efficient as they could be, and so this vectorization procedure is unnecessary. This is why I didn't mark this question as the "best answer".
Setup code:
def factory(i): return lambda num: num==i
array1 = list()
for i in range(10000): array1.append(factory(i))
array1 = np.array(array1)
array2 = np.array(xrange(10000))
The "unvectorized" version:
def evaluate(array1, array2):
return [func(val) for func, val in zip(array1, array2)]
%timeit evaluate(array1, array2)
# 100 loops, best of 3: 10 ms per loop
The vectorized version
def evaluate2(func, b): return func(b)
vec_evaluate = np.vectorize(evaluate2)
vec_evaluate(array1, array2)
# 100 loops, best of 3: 2.65 ms per loop
EDIT
Okay, I just wanted to paste more benchmarks that I received using the above tests, except with different test cases.
I made a third edit, showing what happens if you simply use python lists. The long story short, you actually won't regret much. This test case is on the very bottom.
Test cases only involving integers
In summary, if n is small, then the unvectorized version is better. Otherwise, vectorized is the way to go.
With n = 30
%timeit evaluate(array1, array2)
# 10000 loops, best of 3: 35.7 µs per loop
%timeit vec_evaluate(array1, array2)
# 10000 loops, best of 3: 27.6 µs per loop
With n = 7
%timeit evaluate(array1, array2)
100000 loops, best of 3: 9.93 µs per loop
%timeit vec_evaluate(array1, array2)
10000 loops, best of 3: 21.6 µs per loop
Test cases involving strings
Vectorization wins.
Setup code:
def factory(i): return lambda num: str(num)==str(i)
array1 = list()
for i in range(7):
array1.append(factory(i))
array1 = np.array(array1)
array2 = np.array(xrange(7))
With n = 10000
%timeit evaluate(array1, array2)
10 loops, best of 3: 36.7 ms per loop
%timeit vec_evaluate(array1, array2)
100 loops, best of 3: 6.57 ms per loop
With n = 7
%timeit evaluate(array1, array2)
10000 loops, best of 3: 28.3 µs per loop
%timeit vec_evaluate(array1, array2)
10000 loops, best of 3: 27.5 µs per loop
Random tests
Just to see how branch prediction played a role. From what I'm seeing, it didn't really change much. Vectorization still usually wins.
Setup code.
def factory(i):
if random() < 0.5:
return lambda num: str(num) == str(i)
return lambda num: num == i
When n = 10000
%timeit evaluate(array1, array2)
10 loops, best of 3: 25.7 ms per loop
%timeit vec_evaluate(array1, array2)
100 loops, best of 3: 4.67 ms per loop
When n = 7
%timeit evaluate(array1, array2)
10000 loops, best of 3: 23.1 µs per loop
%timeit vec_evaluate(array1, array2)
10000 loops, best of 3: 23.1 µs per loop
Using python lists instead of numpy arrays
I ran this test to see what happened when I chose not to use the "optimized" numpy arrays, and I received some very surprising results.
The setup code is almost the same, except I'm choosing not to use numpy arrays. I'm also doing this test for only the "random" case.
def factory(i):
if random() < 0.5:
return lambda num: str(num) == str(i)
return lambda num: num == i
array1 = list()
for i in range(10000): array1.append(factory(i))
array2 = range(10000)
And the "unvectorized" version:
%timeit evaluate(array1, array2)
100 loops, best of 3: 4.93 ms per loop
You could see this is actually pretty surprising, because this is almost the same benchmark I was receiving with my random test case involving the vectorized evaluate.
%timeit vec_evaluate(array1, array2)
10 loops, best of 3: 19.8 ms per loop
Likewise, if you change these into numpy arrays before using vec_evaluate, you get the same 4.5 ms benchmark.
I know that in Python, the in-place operators use the __iadd__ method for in-place operators. For immutable types, the __iadd__ is a workaround using the __add__, e.g., like tmp = a + b; a = tmp, but mutable types (like lists) are modified in-place, which causes a slight speed boost.
However, if I have a NumPy array where I modify its contained immutable types, e.g., integers or floats, there is also an even more significant speed boost. How does this work? I did some example benchmarks below:
import numpy as np
def inplace(a, b):
a += b
return a
def assignment(a, b):
a = a + b
return a
int1 = 1
int2 = 1
list1 = [1]
list2 = [1]
npary1 = np.ones((1000,1000))
npary2 = np.ones((1000,1000))
print('Python integers')
%timeit inplace(int1, 1)
%timeit assignment(int2, 1)
print('\nPython lists')
%timeit inplace(list1, [1])
%timeit assignment(list2, [1])
print('\nNumPy Arrays')
%timeit inplace(npary1, 1)
%timeit assignment(npary2, 1)
What I would expect is a similar difference as for the Python integers when I used the in-place operators on NumPy arrays, however the results are completely different:
Python integers
1000000 loops, best of 3: 265 ns per loop
1000000 loops, best of 3: 249 ns per loop
Python lists
1000000 loops, best of 3: 449 ns per loop
1000000 loops, best of 3: 638 ns per loop
NumPy Arrays
100 loops, best of 3: 3.76 ms per loop
100 loops, best of 3: 6.6 ms per loop
Each call to assignment(npary2, 1) requires creating a new one million element array. Consider how much time it takes just to allocate a (1000, 1000)-shaped array of ones:
In [21]: %timeit np.ones((1000, 1000))
100 loops, best of 3: 3.84 ms per loop
This allocation of a new temporary array requires on my machine about 3.84 ms, and is on the right order of magnitude to explain the entire difference between inplace(npary1, 1) and assignment(nparay2, 1):
In [12]: %timeit inplace(npary1, 1)
1000 loops, best of 3: 1.8 ms per loop
In [13]: %timeit assignment(npary2, 1)
100 loops, best of 3: 4.04 ms per loop
So, given that allocation is a relatively slow process, it makes sense that in-place addition is significantly faster than assignment to a new array.
NumPy operations on NumPy arrays may be fast, but creation of NumPy arrays is relatively slow. Consider, for example, how much more time it takes to create a NumPy array than a Python list:
In [14]: %timeit list()
10000000 loops, best of 3: 106 ns per loop
In [15]: %timeit np.array([])
1000000 loops, best of 3: 563 ns per loop
This is one reason why it is generally better to use one large NumPy array (allocated once) rather than thousands of small NumPy arrays.
I have a problem where I have to do the following calculation.
I wanted to avoid the loop version, so I vectorized it.
Why is the loop version actually fast than the vectorized version?
Does anybody have an explanation for this.
thx
import numpy as np
from numpy.core.umath_tests import inner1d
num_vertices = 40000
num_pca_dims = 1000
num_vert_coords = 3
a = np.arange(num_vert_coords * num_vertices * num_pca_dims).reshape((num_pca_dims, num_vertices*num_vert_coords)).T
#n-by-3
norms = np.arange(num_vertices * num_vert_coords).reshape(num_vertices,-1)
#Loop version
def slowversion(a,norms):
res_list = []
for c_idx in range(a.shape[1]):
curr_col = a[:,c_idx].reshape(-1,3)
res = inner1d(curr_col, norms)
res_list.append(res)
res_list_conc = np.column_stack(res_list)
return res_list_conc
#Fast version
def fastversion(a,norms):
a_3 = a.reshape(num_vertices, 3, num_pca_dims)
fast_res = np.sum(a_3 * norms[:,:,None], axis=1)
return fast_res
res_list_conc = slowversion(a,norms)
fast_res = fastversion(a,norms)
assert np.all(res_list_conc == fast_res)
Your "slow code" is likely doing better because inner1d is a single optimized C++ function that can* make use of your BLAS implementation. Lets look at comparable timings for this operation:
np.allclose(inner1d(a[:,0].reshape(-1,3), norms),
np.sum(a[:,0].reshape(-1,3)*norms,axis=1))
True
%timeit inner1d(a[:,0].reshape(-1,3), norms)
10000 loops, best of 3: 200 µs per loop
%timeit np.sum(a[:,0].reshape(-1,3)*norms,axis=1)
1000 loops, best of 3: 625 µs per loop
%timeit np.einsum('ij,ij->i',a[:,0].reshape(-1,3), norms)
1000 loops, best of 3: 325 µs per loop
Using inner is quite a bit faster then the pure numpy operations. Note that einsum is almost twice as fast as pure numpy expressions and for good reason. As your loop is not that large and most of the FLOPS are in the inner computations the saving for the inner operation outweigh the cost of looping.
%timeit slowversion(a,norms)
1 loops, best of 3: 991 ms per loop
%timeit fastversion(a,norms)
1 loops, best of 3: 1.28 s per loop
#Thanks to DSM for writing this out
%timeit np.einsum('ijk,ij->ik',a.reshape(num_vertices, num_vert_coords, num_pca_dims), norms)
1 loops, best of 3: 488 ms per loop
Putting this back together we can see the overall advantage of the "slow version" wins out; however, using an einsum implementation, which is fairly optimized for this sort of thing, gives us a further speed increase.
*I don't see it right off in the code, but it is clearly threaded.
I'm really confused with functions call speed in Python. First and second cases, nothing unexpected:
%timeit reduce(lambda res, x: res+x, range(1000))
10000 loops, best of 3: 150 µs per loop
def my_add(res, x):
return res + x
%timeit reduce(my_add, range(1000))
10000 loops, best of 3: 148 µs per loop
But third case looks strange for me:
from operator import add
%timeit reduce(add, range(1000))
10000 loops, best of 3: 80.1 µs per loop
At the same time:
%timeit add(10, 100)
%timeit 10 + 100
10000000 loops, best of 3: 94.3 ns per loop
100000000 loops, best of 3: 14.7 ns per loop
So, why the third case gives speed up about 50%?
add is implemented in C.
>>> from operator import add
>>> add
<built-in function add>
>>> def my_add(res, x):
... return res + x
...
>>> my_add
<function my_add at 0x18358c0>
The reason that a straight + is faster is that add still has to call the Python VM's BINARY_ADD instruction as well as perform some other work due to it being a function, while + is only a BINARY_ADD instruction.
The operator module exports a set of efficient functions corresponding to the intrinsic operators of Python. For example, operator.add(x, y) is equivalent to the expression x+y. The function names are those used for special class methods; variants without leading and trailing __ are also provided for convenience.
From Python docs (emphasis mine)
The operator module is an efficient (native I'd assume) implementation. IMHO, calling a native implementation should be quicker than calling a python function.
You could try calling the interpreter with -O or -OO to compile the python core and check the timing again.