It seems that numpy.resize is not supported in numba.
What is the best way to use dynamically growing arrays with numba.jit in nopython mode?
So far the best I could do is define and resize the arrays outside the jitted function, is there a better (and neater) option?
numpy.resize is a pure python function:
import numpy as np
def resize(a, new_shape):
"""I did some minor changes so it all works with just `import numpy as np`."""
if isinstance(new_shape, (int, np.core.numerictypes.integer)):
new_shape = (new_shape,)
a = np.ravel(a)
Na = len(a)
if not Na:
return np.zeros(new_shape, a.dtype)
total_size = np.multiply.reduce(new_shape)
n_copies = int(total_size / Na)
extra = total_size % Na
if total_size == 0:
return a[:0]
if extra != 0:
n_copies = n_copies+1
extra = Na-extra
a = np.concatenate((a,)*n_copies)
if extra > 0:
a = a[:-extra]
return np.reshape(a, new_shape)
For 1D arrays this would be straight-forward to implement yourself. Unfortunatly it's a lot more complicated for ND arrays because some operations aren't supported in nopython numba functions: isinstance, reshape and the tuple-multiplication. Here is the 1D equivalent:
import numpy as np
import numba as nb
#nb.njit
def resize(a, new_size):
new = np.zeros(new_size, a.dtype)
idx = 0
while True:
newidx = idx + a.size
if newidx > new_size:
new[idx:] = a[:new_size-newidx]
break
new[idx:newidx] = a
idx = newidx
return new
when you don't want that "repeat input" behaviour and only use it for increasing the size it's even easier:
#nb.njit
def resize(a, new_size):
new = np.zeros(new_size, a.dtype)
new[:a.size] = a
return new
These functions are decorated with numba.njit and can therefore be called in any numba functions in nopython mode.
A word of caution though: Generally you don't want to resize - or if you do then make sure you choose an approach that has amoritzed O(1) cost (Wikipedia link). If you can estimate the maximum length then it's better to pre-allocate a correctly sized (or slightly overallocated) array immediatly.
Typically the strategy I employ is to just allocate more than enough array storage to accommodate the calculation and then keep track of the final index/indices used, and then slice the array down to the actual size before returning. This assumes that you know beforehand what the maximum size you could possibly grow the array to is. The thought is that in most of my own applications, memory is cheap but resizing and switching between python and jitted functions a lot is expensive.
To dynamically increase the size of an existing array (and therefore do it in-place), numpy.ndarray.resize must be used instead of numpy.resize. This method is NOT implemented in Python, and is not available in Numba, so it just cannot be done.
Related
I can use a for loop to loop over two byte sequences and return the index at the first difference of course:
bytes1 = b'12345'
bytes2 = b'1F345'
for index, pair in enumerate(zip(bytes1, bytes2)):
if pair[0] != pair[1]:
print(index)
break
But I don't think that's a smart and fast way to do it. I would hope a native method exists that I can call to get this done. Is there something that can help me here? I can also use numpy if it helps.
I also want to clarify that this will run many times, with medium sequences. Approximately 300MB is expected, chunked by 100kB. I might be able to change that for larger if it helps significantly.
a solution with numpy is to convert them to an array of uint8 then xor them and use argmax to get the first non-zero.
import numpy as np
bytes1 = b'12345'
bytes2 = b'1F345'
bytes3 = np.frombuffer(bytes1,dtype=np.uint8)
bytes4 = np.frombuffer(bytes2,dtype=np.uint8)
max_loc = np.flatnonzero(bytes3 ^ bytes4)[0]
print(max_loc)
1
problem is that this still iterates to the end of the string on all functions, it's done in C so it is not too slow, slicing long array into multiple smaller ones can reduce the overhead for very long arrays.
Edit: modified argmax to the correct flatnonzero as pointed by #jasonharper, which throws an indexError if both inputs are equal.
If using numba is ok:
import numba
#numba.jit()
def method2(bytes1, bytes2):
idx = 0
while idx < len(bytes1):
if bytes1[idx] != bytes2[idx]:
return idx
idx += 1
return idx
Note that first run of this function will be significantly slower (due to compilation performed by numba). Takes like 2 seconds.
Then, for each next run of the function:
easy case you posted, index = 1 -> numba is 2x faster,
for index = 100 -> numba is 33x faster
I need a much faster code to remove values of an 1D array (array length ~ 10-15) that are common with another 1D array (array length ~ 1e5-5e5 --> rarely up to 7e5), which are index arrays contain integers. There is no duplicate in the arrays, and they are not sorted and the order of the values must be kept in the main array after modification. I know that can be achieved using such np.setdiff1d or np.in1d (which both are not supported for numba jitted in no-python mode), and other similar posts (e.g. this) have not much more efficient way to do so, but performance is important here because all the values in the main index array will be gradually be removed in loops.
import numpy as np
import numba as nb
n = 500000
r = 10
arr1 = np.random.permutation(n)
arr2 = np.random.randint(0, n, r)
# #nb.jit
def setdif1d_np(a, b):
return np.setdiff1d(a, b, assume_unique=True)
# #nb.jit
def setdif1d_in1d_np(a, b):
return a[~np.in1d(a, b)]
There is another related post that proposed by norok2 for 2D arrays, that is ~15 times faster solution (hashing-like way using numba) than usual methods described there. This solution may be the best if it could be prepared for 1D arrays:
#nb.njit
def mul_xor_hash(arr, init=65537, k=37):
result = init
for x in arr.view(np.uint64):
result = (result * k) ^ x
return result
#nb.njit
def setdiff2d_nb(arr1, arr2):
# : build `delta` set using hashes
delta = {mul_xor_hash(arr2[0])}
for i in range(1, arr2.shape[0]):
delta.add(mul_xor_hash(arr2[i]))
# : compute the size of the result
n = 0
for i in range(arr1.shape[0]):
if mul_xor_hash(arr1[i]) not in delta:
n += 1
# : build the result
result = np.empty((n, arr1.shape[-1]), dtype=arr1.dtype)
j = 0
for i in range(arr1.shape[0]):
if mul_xor_hash(arr1[i]) not in delta:
result[j] = arr1[i]
j += 1
return result
I tried to prepare that for 1D arrays, but I have some problems/question with that.
At first, IDU what does mul_xor_hash exactly do, and if init and k are arbitrary selected or not
Why mul_xor_hash will not work without nb.njit:
File "C:/Users/Ali/Desktop/test - Copy - Copy.py", line 21, in mul_xor_hash
result = (result * k) ^ x
TypeError: ufunc 'bitwise_xor' not supported for the input types, and the inputs could not be safely coerced to any supported types according to the casting rule ''safe''
IDK how to implement mul_xor_hash on 1D arrays (if it could), which I guess may make it faster more than for 2Ds, so I broadcast the input arrays to 2D by [None, :], which get the following error just for arr2:
print(mul_xor_hash(arr2[0]))
ValueError: new type not compatible with array
and what does delta do
I am searching the most efficient way in this regard. In the absence of better method than norok2 solution, how to prepare this solution for 1D arrays?
Understanding the hash-based solution
At first, IDU what does mul_xor_hash exactly do, and if init and k are arbitrary selected or not
mul_xor_hash is a custom hash function. Functions mixing xor and multiply (possibly with shifts) are known to be relatively fast to compute the hash of a raw data buffer. The multiplication tends to shuffle bits and the xor is used to somehow combine/accumulate the result in a fixed size small value (ie. the final hash). There are many different hashing functions. Some are faster than others, some cause more collisions than other in a given context. A fast hashing function causing too many collisions can be useless in practice as it would result in a pathological situation where all conflicting values needs to be compared. This is why fast hash functions are hard to implement.
init and k are parameter certainly causing the hash to be pretty balance. This is pretty common in such a hash function. k needs to be sufficiently big for the multiplication to shuffle bits and it should typically also be a prime number (values like power of two tends to increase collisions due to modular arithmetic behaviours). init plays a significant role only for very small arrays (eg. with 1 item): it helps to reduce collisions by xoring the final hash by a non-trivial constant. Indeed, if arr.size = 1, then result = (init * k) ^ arr[0] where init * k is a constant. Having an identity hash function equal to arr[0] is known to be bad since it tends to result in many collisions (this is a complex topic, but put it shortly, arr[0] can be divided by the number of buckets in the hash table for example). Thus, init should be a relatively big number and init * k should also be a big non-trivial value (a prime number is a good target value).
Why mul_xor_hash will not work without nb.njit
It depends of the input. The input needs to be a 1D array and have a raw size in byte divisible by 8 (eg. 64-bit items, 2n x 32-bit ones, 4n x 16-bit one or 8n 8-bit ones). Here is some examples:
mul_xor_hash(np.random.rand(10))
mul_xor_hash(np.arange(10)) # Do not work with 9
and what does delta do
It is a set containing the hash of the arr2 row so to find matching lines faster than comparing them without hashes.
how to prepare this solution for 1D arrays?
AFAIK, hashes are only use to avoid comparisons of rows but this is because the input is the 2D array. In 1D, there is no such a problem.
There is big catch with this method: it only works if there is no hash collisions. Otherwise, the implementation wrongly assumes that values are equal even if they are not! #norok explicitly mentioned it in the comments though:
Note that the collision handling for the hashings should also be implemented
Faster implementation
Using the 2D solution of #norok2 for 1D is not a good idea since hashes will not make it faster the way they are used. In fact, a set already use a hash function internally anyway. Not to mention collisions needs to be properly implemented (which is done by a set).
Using a set is a relatively good idea since it causes the complexity to be O(n + m) where n = len(arr1) and m = len(arr2). That being said, if arr1 is converted to a set, then it will be too big to fit in L1 cache (due to the size of arr1 in your case) resulting in slow cache misses. Additionally, the growing size of the set will cause values to be re-hashed which is not efficient. If arr2 is converted to a set, then the many hash table fetches will not be very efficient since arr2 is very small in your case. This is why this solution is sub-optimal.
One solution is to split arr1 in chunks and then build a set based on the target chunk. You can then check if a value is in the set or not efficiently. Building the set is still not very efficient due to the growing size. This problem is due to Python itself which do not provide a way to reserve some space for the data structure like other languages do (eg. C++). One solution to avoid this issue is simply to reimplement an hash-table which is not trivial and cumbersome. Actually, Bloom filters can be used to speed up this process since they can quickly find if there is no collision between the two sets arr1 and arr2 in average (though they are not trivial to implement).
Another optimization is to use multiple threads to compute the chunks in parallel since they are independent. That being said, the appending to the final array is not easy to do efficiently in parallel, especially since you do not want the order to be modified. One solution is to move away the copy from the parallel loop and do it serially but this is slow and AFAIK there is no simple way to do that in Numba currently (since the parallelism layer is very limited). Consider using native languages like C/C++ for an efficient parallel implementation.
In the end, hashing can be pretty complex and the speed up can be quite small compared to a naive implementation with two nested loops since arr2 only have few items and modern processors can compare values quickly using SIMD instructions (while hash-based method can hardly benefit from them on mainstream processors). Unrolling can help to write a pretty simple and fast implementation. Again, unfortunately, Numba use LLVM-Jit internally which appear to fail to vectorize such a simple code (certainly due to missing optimizations in either LLVM-Jit or even LLVM itself). As a result, the non vectorized code is finally a bit slower (rather than 4~10 times faster on a modern mainstream processor). One solution is to use a C/C++ code instead to do that (or possibly Cython).
Here is a serial implementation using basic Bloom filters:
#nb.njit('uint32(int32)')
def hash_32bit_4k(value):
return (np.uint32(value) * np.uint32(27_644_437)) & np.uint32(0x0FFF)
#nb.njit(['int32[:](int32[:], int32[:])', 'int32[:](int32[::1], int32[::1])'])
def setdiff1d_nb_faster(arr1, arr2):
out = np.empty_like(arr1)
bloomFilter = np.zeros(4096, dtype=np.uint8)
for j in range(arr2.size):
bloomFilter[hash_32bit_4k(arr2[j])] = True
cur = 0
for i in range(arr1.size):
# If the bloom-filter value is true, we know arr1[i] is not in arr2.
# Otherwise, there is maybe a false positive (conflict) and we need to check to be sure.
if bloomFilter[hash_32bit_4k(arr1[i])] and arr1[i] in arr2:
continue
out[cur] = arr1[i]
cur += 1
return out[:cur]
Here is an untested variant that should work for 64-bit integers (floating point numbers need memory views and possibly a prime constant too):
#nb.njit('uint64(int64)')
def hash_32bit_4k(value):
return (np.uint64(value) * np.uint64(67_280_421_310_721)) & np.uint64(0x0FFF)
Note that if all the values in the small array are contained in the main array in each loop, then we can speed up the arr1[i] in arr2 part by removing values from arr2 when we find them. That being said, collisions and findings should be very rare so I do not expect this to be significantly faster (not to mention it adds some overhead and complexity). If items are computed in chunks, then the last chunks can be directly copied without any check but the benefit should still be relatively small. Note that this strategy can be effective for the naive (C/C++) SIMD implementation previously mentioned though (it can be about 2x faster).
Generalization and parallel implementation
This section focus on the algorithm to use regarding the input size. It particularly details an SIMD-based implementation and discuss about the use of multiple threads.
First of all, regarding the value r, the best algorithm to use can be different. More specifically:
when r is 0, the best thing to do is to return the input array arr1 unmodified (possibly a copy to avoid issue with in-place algorithms);
when r is 1, we can use one basic loop iterating over the array, but the best implementation is likely to use np.where of Numpy which is highly optimized for that
when r is small like <10, then using a SIMD-based implementation should be particularly efficient, especially if the iteration range of the arr2-based loop is known at compile-time and is unrolled
for bigger r values that are still relatively small (eg. r < 1000 and r << n), the provided hash-based solution should be one of the best;
for larger r values with r << n, the hash-based solution can be optimized by packing boolean values as bits in bloomFilter and by using multiple hash-functions instead of one so to better handle collisions while being more cache-friendly (in fact, this is what actual bloom filters does); note that multi-threading can be used so speed up the lookups when r is huge and r << n;
when r is big and not much smaller than n, then the problem is pretty hard to solve efficiently and the best solution is certainly to sort both arrays (typically with a radix sort) and use a merge-based method to remove the duplicates, possibly with multiple threads when both r and n are huge (hard to implement).
Let's start with the SIMD-based solution. Here is an implementation:
#nb.njit('int32[:](int32[::1], int32[::1])')
def setdiff1d_nb_simd(arr1, arr2):
out = np.empty_like(arr1)
limit = arr1.size // 4 * 4
limit2 = arr2.size // 2 * 2
cur = 0
z32 = np.int32(0)
# Tile (x4) based computation
for i in range(0, limit, 4):
f0, f1, f2, f3 = z32, z32, z32, z32
v0, v1, v2, v3 = arr1[i], arr1[i+1], arr1[i+2], arr1[i+3]
# Unrolled (x2) loop searching for a match in `arr2`
for j in range(0, limit2, 2):
val1 = arr2[j]
val2 = arr2[j+1]
f0 += (v0 == val1) + (v0 == val2)
f1 += (v1 == val1) + (v1 == val2)
f2 += (v2 == val1) + (v2 == val2)
f3 += (v3 == val1) + (v3 == val2)
# Remainder of the previous loop
if limit2 != arr2.size:
val = arr2[arr2.size-1]
f0 += v0 == val
f1 += v1 == val
f2 += v2 == val
f3 += v3 == val
if f0 == 0: out[cur] = arr1[i+0]; cur += 1
if f1 == 0: out[cur] = arr1[i+1]; cur += 1
if f2 == 0: out[cur] = arr1[i+2]; cur += 1
if f3 == 0: out[cur] = arr1[i+3]; cur += 1
# Remainder
for i in range(limit, arr1.size):
if arr1[i] not in arr2:
out[cur] = arr1[i]
cur += 1
return out[:cur]
It turns out this implementation is always slower than the hash-based one on my machine since Numba clearly generate an inefficient for the inner arr2-based loop and this appears to come from broken optimizations related to the ==: Numba simply fail use SIMD instructions for this operation (for no apparent reasons). This prevent many alternative SIMD-related codes to be fast as long as they are using Numba.
Another issue with Numba is that np.where is slow since it use a naive implementation while the one of Numpy has been heavily optimized. The optimization done in Numpy can hardly be applied to the Numba implementation due to the previous issue. This prevent any speed up using np.where in a Numba code.
In practice, the hash-based implementation is pretty fast and the copy takes a significant time on my machine already. The computing part can be speed up using multiple thread. This is not easy since the parallelism model of Numba is very limited. The copy cannot be easily optimized with Numba (one can use non-temporal store but this is not yet supported by Numba) unless the computation is possibly done in-place.
To use multiple threads, one strategy is to first split the range in chunk and then:
build a boolean array determining, for each item of arr1, whether the item is found in arr2 or not (fully parallel)
count the number of item found by chunk (fully parallel)
compute the offset of the destination chunk (hard to parallelize, especially with Numba, but fast thanks to chunks)
copy the chunk to the target location without copying found items (fully parallel)
Here is an efficient parallel hash-based implementation:
#nb.njit('int32[:](int32[:], int32[:])', parallel=True)
def setdiff1d_nb_faster_par(arr1, arr2):
# Pre-computation of the bloom-filter
bloomFilter = np.zeros(4096, dtype=np.uint8)
for j in range(arr2.size):
bloomFilter[hash_32bit_4k(arr2[j])] = True
chunkSize = 1024 # To tune regarding the kind of input
chunkCount = (arr1.size + chunkSize - 1) // chunkSize
# Find for each item of `arr1` if the value is in `arr2` (parallel)
# and count the number of item found for each chunk on the fly.
# Note: thanks to page fault, big parts of `found` are not even written in memory if `arr2` is small
found = np.zeros(arr1.size, dtype=nb.bool_)
foundCountByChunk = np.empty(chunkCount, dtype=nb.uint16)
for i in nb.prange(chunkCount):
start, end = i * chunkSize, min((i + 1) * chunkSize, arr1.size)
foundCountInChunk = 0
for j in range(start, end):
val = arr1[j]
if bloomFilter[hash_32bit_4k(val)] and val in arr2:
found[j] = True
foundCountInChunk += 1
foundCountByChunk[i] = foundCountInChunk
# Compute the location of the destination chunks (sequential)
outChunkOffsets = np.empty(chunkCount, dtype=nb.uint32)
foundCount = 0
for i in range(chunkCount):
outChunkOffsets[i] = i * chunkSize - foundCount
foundCount += foundCountByChunk[i]
# Parallel chunk-based copy
out = np.empty(arr1.size-foundCount, dtype=arr1.dtype)
for i in nb.prange(chunkCount):
srcStart, srcEnd = i * chunkSize, min((i + 1) * chunkSize, arr1.size)
cur = outChunkOffsets[i]
# Optimization: we can copy the whole chunk if there is nothing found in it
if foundCountByChunk[i] == 0:
out[cur:cur+(srcEnd-srcStart)] = arr1[srcStart:srcEnd]
else:
for j in range(srcStart, srcEnd):
if not found[j]:
out[cur] = arr1[j]
cur += 1
return out
This implementation is the fastest for the target input on my machine. It is generally fast when n is quite big and the overhead to create threads is relatively small on the target platform (eg. on PCs but typically not computing servers with many cores). The overhead of the parallel implementation is significant so the number of core on the target machine needs to be at least 4 so the implementation can be significantly faster than the sequential implementation.
It may be useful to tune the chunkSize variable for the target inputs. If r << n, it is better to use a pretty big chunkSize. That being said, the number of chunk needs to be sufficiently big for multiple thread to operate on many chunks. Thus, chunkSize should be significantly smaller than n / numberOfThreads.
On my machine most of the time (65-70%) is spent in the final copy which is mostly memory-bound and can hardly be optimized further with Numba.
Results
Here are results on my i5-9600KF-based machine (with 6 cores):
setdif1d_np: 2.65 ms
setdif1d_in1d_np: 2.61 ms
setdiff1d_nb: 2.33 ms
setdiff1d_nb_simd: 1.85 ms
setdiff1d_nb_faster: 0.73 ms
setdiff1d_nb_faster_par: 0.49 ms
The best provided implementation is about 4~5 time faster than the other ones.
What I found is that hashing does not help,. It is just trick for 2D case, to convert 1d arrays to single numbers and put them as such in a set.
Below is method of norok2 I converted to 1d arrays (and added annotations for faster compilation).
Note that this is only slightly (20-30%) faster than the methods you already have. And of course after second function call, on first due to compilation it is slightly slower.
#nb.njit('int32[:](int32[:], int32[:])')
def setdiff1d_nb(arr1, arr2):
delta = set(arr2)
# : build the result
result = np.empty(len(arr1), dtype=arr1.dtype)
j = 0
for i in range(arr1.shape[0]):
if arr1[i] not in delta:
result[j] = arr1[i]
j += 1
return result[:j]
I am writing code which utilizes Numba to JIT compile my python code.
The function takes in two arrays of same length as input, randomly selects a slicing point and returns a tuple with two Frankenstein array formed by parts of the two input strings.
Numba however does not yet support the numpy.concatenate function (don't know if it ever will). As I am unwilling to drop Numpy, does anyone know a performant solution for concatenating two Numpy arrays without the concatenate function?
def randomSlice(str1, str2):
lenstr = len(str1)
rnd = np.random.randint(1, lenstr)
return (np.concatenate((str1[:rnd], str2[rnd:])), np.concatenate((str2[:rnd], str1[rnd:])))
This might work for you:
import numpy as np
import numba as nb
#nb.jit(nopython=True)
def randomSlice_nb(str1, str2):
lenstr = len(str1)
rnd = np.random.randint(1, lenstr)
out1 = np.empty_like(str1)
out2 = np.empty_like(str1)
out1[:rnd] = str1[:rnd]
out1[rnd:] = str2[rnd:]
out2[:rnd] = str2[:rnd]
out2[rnd:] = str1[rnd:]
return (out1, out2)
On my machine, using Numba 0.27 and timing via the timeit module to make sure I'm not counting the jit time in the stats (or you could run it once, and then time subsequent calls), the numba version gives a small but non-negligible performance increase on various size input arrays of ints or floats. If the arrays have a dtype of something like |S1, then numba is significantly slower. The Numba team has spent very little time optimizing non-numeric usecases so this isn't terribly surprising. I'm a little unclear about the exact form of your input arrays str1 and str2, so I can't exactly guarantee that the code will work for your specific usecase.
I have come across the following issue when multiplying numpy arrays. In the example below (which is slightly simplified from the real version I am dealing with), I start with a nearly empty array A and a full array C. I then use a recursive algorithm to fill in A.
Below, I perform this algorithm in two different ways. The first method involves the operations
n_array = np.arange(0,c-1)
temp_vec= C[c-n_array] * A[n_array]
A[c] += temp_vec.sum(axis=0)
while the second method involves the for loop
for m in range(0, c - 1):
B[c] += C[c-m] * B[m]
Note that the arrays A and B are identical, but they are filled in using the two different methods.
In the example below I time how long it takes to perform the computation using each method. I find that, for example, with n_pix=2 and max_counts = 400, the first method is much faster than the second (that is, time_np is much smaller than time_for). However, when I then switch to, for example, n_pix=1000 and max_counts = 400, instead I find method 2 is much faster (time_for is much smaller than time_np). I would have thought that method 1 would always be faster since method 2 explicitly runs over a loop while method 1 uses np.multiply.
So, I have two questions:
Why does the timing behave this way as a function of n_pix for a fixed max_counts?
What is optimal method for writing this code so that it behaves quickly for all n_pix?
That is, can anyone suggest a method 3? In my project, it is very important for this piece of code to perform quickly over a range of large and small n_pix.
import numpy as np
import time
def return_timing(n_pix,max_counts):
A=np.zeros((max_counts+1,n_pix))
A[0]=np.random.random(n_pix)*1.8
A[1]=np.random.random(n_pix)*2.3
B=np.zeros((max_counts+1,n_pix))
B[0]=A[0]
B[1]=A[1]
C=np.outer(np.random.random(max_counts+1),np.random.random(n_pix))*3.24
time_np=0
time_for=0
for c in range(2, max_counts + 1):
t0 = time.time()
n_array = np.arange(0,c-1)
temp_vec= C[c-n_array] * A[n_array]
A[c] += temp_vec.sum(axis=0)
time_np += time.time()-t0
t0 = time.time()
for m in range(0, c - 1):
B[c] += C[c-m] * B[m]
time_for += time.time()-t0
return time_np, time_for
First of all, you can easily replace:
n_array = np.arange(0,c-1)
temp_vec= C[c-n_array] * A[n_array]
A[c] += temp_vec.sum(axis=0)
with:
A[c] += (C[c:1:-1] * A[:c-1]).sum(0)
This is much faster because indexing with an array is much slower than slicing. But the temp_vec is still hidden in there, created before summing is done. This leads to the idea of using einsum, which is the fastest because it doesn't make the temp array.
A[c] = np.einsum('ij,ij->j', C[c:1:-1], A[:c-1])
Timing. For small arrays:
>>> return_timing(10,10)
numpy OP 0.000525951385498
loop OP 0.000250101089478
numpy slice 0.000246047973633
einsum 0.000170946121216
For large:
>>> return_timing(1000,100)
numpy OP 0.185983896255
loop OP 0.0458009243011
numpy slice 0.038364648819
einsum 0.0167834758759
It is probably because your numpy-only version requires creation/allocation of new ndarrays (temp_vec and n_array), while your other method does not.
Creation of new ndarrays is very slow and if you can modify your code in such a way that it no longer have to continuously create them, I would expect that you could get better performance out of that method.
I am trying to convert part of a native python function to cython to improve the compute time. I would like to write a cython function just for the loop component that is taking up the time (as ipython lprun kindly told me). However this function takes in variably sized matrices .. and I can't see how to bring that across easily to statically typed cython.
for index1 in range(0,num_products):
for index2 in range(0,num_products):
cond_prob = (data[index1] * data[index2]).sum() / max(col_sums[index1], col_sums[index2])
prox[index1][index2] = cond_prob
This issue is that num_products changes year to year, so the matrix (data) size is variable.
What is the best strategy here?
Should I write two C functions. One to create a matrix of a certain dimension using memalloc, and then One to do the loops over the created matrix?
Is there some fancy cython/numpy wizardry to help in this scenario? Can I write a C function that takes in a variably sized Numpy Array in memory and pass the size?
Cython code is (strategically) statically typed, but that doesn't mean that arrays must have a fixed size. In straight C passing a multidimensional array to a function can be a little awkward maybe, but in Cython you should be able to do something like the following:
Note I took the function and variable names from your follow-up question.
import numpy as np
cimport numpy as np
cimport cython
#cython.boundscheck(False)
#cython.cdivision(True)
def cooccurance_probability_cy(double[:,:] X):
cdef int P, i, j, k
P = X.shape[0]
cdef double item
cdef double [:] CS = np.sum(X, axis=1)
cdef double [:,:] D = np.empty((P, P), dtype=np.float)
for i in range(P):
for j in range(P):
item = 0
for k in range(P):
item += X[i,k] * X[j,k]
D[i,j] = item / max(CS[i], CS[j])
return D
On the other hand, using just Numpy should also be quite fast for this problem, if you use the right functions and some broadcasting. In fact, as the calculation complexity is dominated by the matrix multiplication, I found the following is much faster than the Cython code above (np.inner uses a highly optimized BLAS routine):
def new(X):
CS = np.sum(X, axis=1, keepdims=True)
D = np.inner(X,X) / np.maximum(CS, CS.T)
return D
Have you tried getting rid of the for loops in numpy?
for the first part of your equation you could for example try:
(data[ np.newaxis,:] * data[:,np.newaxis]).sum(2)
if memory is an issue you can also use the np.einsum() function.
For the second part one could probably also cook up a numpy expression (bit more difficult) if you've not already tried that.