I am attempting to find the most performant method to find unique values from a NumPy array. NumPy's unique
function is very slow and sorts the values first before finding the unique. Pandas hashes the values using the klib C library which is much faster. I am looking for a Cython solution.
The simplest solution seems to just iterate through the array and use a Python set to add each element like this:
from numpy cimport ndarray
from cpython cimport set
@cython.wraparound(False)
@cython.boundscheck(False)
def unique_cython_int(ndarray[np.int64_t] a):
cdef int i
cdef int n = len(a)
cdef set s = set()
for i in range(n):
s.add(a[i])
return s
I also tried an unordered_set from c++
from libcpp.unordered_set cimport unordered_set
@cython.wraparound(False)
@cython.boundscheck(False)
def unique_cpp_int(ndarray[np.int64_t] a):
cdef int i
cdef int n = len(a)
cdef unordered_set[int] s
for i in range(n):
s.insert(a[i])
return s
Performance
# create array of 1,000,000
a = np.random.randint(0, 50, 1000000)
# Pure Python
%timeit set(a)
86.4 ms ± 2.58 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
# Convert to list first
a_list = a.tolist()
%timeit set(a_list)
10.2 ms ± 74.8 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
# NumPy
%timeit np.unique(a)
32 ms ± 1.17 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
# Pandas
%timeit pd.unique(a)
5.3 ms ± 257 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
# Cython
%timeit unique_cython_int(a)
13.4 ms ± 1.02 ms per loop (mean ± std. dev. of 7 runs, 100 loops each)
# Cython - c++ unordered_set
%timeit unique_cpp_int(a)
17.8 ms ± 158 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
So pandas is about 2.5x faster than a cythonized set. Its lead increases when there are more distinct elements. Surprisingly, a pure python set (on a list) beats out a cythonized set.
My question here - is there a faster way to do this in Cython than just use the add
method repeatedly? And could the c++ unordered_set be improved?
The story changes when we use unicode strings. I believe I have to convert the numpy array to an object
data type to properly add its type for Cython.
@cython.wraparound(False)
@cython.boundscheck(False)
def unique_cython_str(ndarray[object] a):
cdef int i
cdef int n = len(a)
cdef set s = set()
for i in range(n):
s.add(a[i])
return s
And again I tried an unordered_set
from c++
@cython.wraparound(False)
@cython.boundscheck(False)
def unique_cpp_str(ndarray[object] a):
cdef int i
cdef int n = len(a)
cdef unordered_set[string] s
for i in range(n):
s.insert(a[i])
return s
Performance
Create an array of 1 million strings with 1,000 distinct values
s_1000 = []
for i in range(1000):
s = np.random.choice(list('abcdef'), np.random.randint(5, 50))
s_1000.append(''.join(s))
s_all = np.random.choice(s_1000, 1000000)
# s_all has numpy unicode as its data type. Must convert to object
s_unicode_obj = s_all.astype('O')
# c++ does not easily handle unicode. Convert to bytes and then to object
s_bytes_obj = s_all.astype('S').astype('O')
# Pure Python
%timeit set(s_all)
451 ms ± 5.94 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
%timeit set(s_unicode_obj)
71.9 ms ± 5.91 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
# using set on a list
s_list = s_all.tolist()
%timeit set(s_list)
63.1 ms ± 7.38 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
# NumPy
%timeit np.unique(s_unicode_obj)
1.69 s ± 97.5 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
%timeit np.unique(s_all)
633 ms ± 3.99 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
# Pandas
%timeit pd.unique(s_unicode_obj)
97.6 ms ± 6.61 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
# Cython
%timeit unique_cython_str(s_unicode_obj)
60 ms ± 5.81 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
# Cython - c++ unordered_set
%timeit unique_cpp_str2(s_bytes_obj)
247 ms ± 8.45 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
So, it appears that Python's set outperforms pandas for unicode strings but not on integers. And again, iterating through the array in Cython doesn't really help us at all.
It's possible to circumvent sets if you know the range of your integers isn't too crazy. You can simply create a second array of all zeros/False
and turn their position True
when you encounter each one and append that number to a list. This is extremely fast since no hashing is done.
The following works for positive integer arrays. If you had negative integers, you would have to add a constant to shift the numbers up to 0.
@cython.wraparound(False)
@cython.boundscheck(False)
def unique_bounded(ndarray[np.int64_t] a):
cdef int i, n = len(a)
cdef ndarray[np.uint8_t, cast=True] unique = np.zeros(n, dtype=bool)
cdef list result = []
for i in range(n):
if not unique[a[i]]:
unique[a[i]] = True
result.append(a[i])
return result
%timeit unique_bounded(a)
1.18 ms ± 21.3 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
The downside is of course memory usage since your largest integer could force an extremely large array. But this method could work for floats too if you knew precisely how many significant digits each number had.
Integers 50 unique of 1,000,000 total
Strings 1,000 unique of 1,000,000 total
Appreciate all the help making these faster.
Using an object or Map and initialising only the values that appear in the array might be faster and won't fail if the assumptions (like age being an integer between 0 and 200) are not met.
The cimport statement is used in a definition or implementation file to gain access to names declared in another definition file. Its syntax exactly parallels that of the normal Python import statement. When pure python syntax is used, the same effect can be done by importing from special cython.
There are a few ways to get a list of unique values in Python. This article will show you how. Using a set one way to go about it. A set is useful because it contains unique elements. You can use a set to get the unique elements. Then, turn the set into a list.
The idea is to traverse the given array from left to right and keep track of visited elements in a hash table. Finally, print the element with count 1. The hashing-based solution requires O (n) extra space. We can use bitwise AND to find the unique element in O (n) time and constant extra space.
Write Python Program to Find and print the Unique Items in an Array. The unique function in the Numpy module returns the unique array items. This Python example uses the unique function and returns the unique array items.
The hashing-based solution requires O (n) extra space. We can use bitwise AND to find the unique element in O (n) time and constant extra space. Create an array count [] of size equal to number of bits in binary representations of numbers. Fill count array such that count [i] stores count of array elements with i-th bit set.
I think the answer to you question "what is the fastest way to find unique elements" is "it depends". It depends on your data set and on your hardware.
For your scenarios (I mostly looked at integer case) pandas (and used khash
) does a pretty decent job. I was not able to match this performance using std::unordered_map
.
However, google::dense_hash_set
was slightly faster in my experiments than the pandas-solution.
Please read on for a more detailed explanation.
I would like to start out by explaining the results you are observing and use these insights later on.
I start with your int-example: there are only 50
unique elements but 1,000,000
in the array:
import numpy as np
import pandas as pd
a=np.random.randint(0,50, 10**6, dtype=np.int64)
As baseline the timings of np.unique()
and pd.unique()
for my machine:
%timeit np.unique(a)
>>>82.3 ms ± 539 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
%timeit pd.unique(a)
>>>9.4 ms ± 110 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
pandas approach with the set (O(n)
) is about 10 times faster than numpy's approach with sorting (O(nlogn)
). log n = 20
for n=10**6
, so the factor 10 is about the expected difference.
Another difference is, that np.unique
returns a sorted array, so one could use binary search to look up the elements. pd.unique
returns an unsorted array so we need either to sort it (which might be O(n log n)
if there are not many duplicates in the original data) or to transform it to a set-like structure.
Let's take a look at the simple Python-Set:
%timeit set(a)
>>> 257 ms ± 21.9 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
First thing we must be aware here: we are comparing apples and oranges. The previous unique
-functions return numpy arrays, which consists out of lowly c-integers. This one returns a set of full-fledged Python-integers. Quite a different thing!
That means for every element in the numpy-array we must first create a python-object - quite an overhead and only then can we add it to the set.
The conversion to Python-integers can be done in a preprocessing step - your version with list
:
A=list(a)
%timeit set(A)
>>> 104 ms ± 952 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
%timeit set(list(a))
>>> 270 ms ± 23.3 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
More than 100 ms are needed for the creation of the Python-integers. However, the python-integers are more complex than the lowly C-ints and thus handling them costs more. Using pd.unique
on C-int and than promoting to Python-set is much faster.
And now your Cython version:
%timeit unique_cython_int(a)
31.3 ms ± 630 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
That I don't really understand. I would expect it to perform similar to set(a)
-cython would cut out the interpreter, but that would not explain the factor 10. However, we have only 50 different integers (which are even in the integers-pool because they are smaller than 256
), so there is probably some optimization, which plays a role/difference.
Let's try another data-set (there are now 10**5
different numbers):
b=np.random.randint(0, 10**5,10**6, dtype=np.int64)
%timeit unique_cython_int(b)
>>> 236 ms ± 31.1 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
%timeit set(b)
>>> 388 ms ± 15.3 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
A speed-up less than 2 is something what I would expect.
Let's take a look at cpp-version:
%timeit unique_cpp_int(a)
>>> 25.4 ms ± 534 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
%timeit unique_cpp_int(b)
>>> 100 ms ± 4.8 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
There is some overhead in copying the data from the cpp-set to the Python set (as DavidW have pointed out), but otherwise the behavior as I would expect given my experience with it: std::unordered_map
is somewhat faster than Python, but not the greatest implementation around - panda seems to beat it:
%timeit set(pd.unique(b))
>>> 45.8 ms ± 3.48 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
So it looks like, that in the situation, where there are many duplicated and the hash-function is cheap, the pandas-solution is hard to beat.
One maybe could try out the google data structures.
However, when the data has only very few duplicates, the numpy's sorting solution may become the faster one. The main reason is, that numpy's unique
needs only twice the memory - the original data and the output, while pandas hash-set-solution needs much more memory: the original data, the set and the output. For huge datasets it might become the difference between having enough RAM and not having enough RAM.
It depends on the set-implementation how much memory-overhead is needed and it is always about the trade-off between memory and speed. For example std::unordered_set
needs at least 32
byte to save a 8
-byte integer. Some google's data structures can do better.
Running /usr/bin/time -fpeak_used_memory:%M python check_mem.py
with pandas/numpy unique:
#check_mem.py
import numpy as np
import pandas as pd
c=np.random.randint(0, 2**63,5*10**7, dtype=np.int64)
#pd.unique(c)
np.unique(c)
shows 1.2 GB for numpy
and 2.0GB for pandas
.
Actually, on my Windows machine np.unique
is faster than pd.unique
if there are (next to) only unique elements in the array, even for "only" 10^6
elements (probably because of the needed rehashes as the used set grows). This is however not the case for my Linux machine.
Another scenario in which pandas
doesn't shine is when the calculation of the hash function is not cheap: Consider long strings (let's say of 1000
characters) as objects.
To calculate the hash-value one needs to consider all 1000
characters (which means a lot of data-> a lot of hash misses), the comparison of two strings is mostly done after one or two characters - the probability is then already very high, that we know that the strings are different. So the log n
factor of the numpy's unique
doesn't look that bad anymore.
It could be better to use a tree-set instead of a hash-set in this case.
Improving on cpp-unordered set:
The method using cpp's unordered set could be improved due to its method reserve()
, which would eliminate the need for rehashing. But it is not imported to cython, so the usage is quite cumbersome from Cython.
The reserving however would not have any impact on the runtimes for data with only 50 unique elements and at most factor 2 (amortized costs due to the used resize-strategy) for the data with almost all elements unique.
The hash-function for ints
is identity (at least for gcc), so not much to gain here (I don't think using a more fancy hash-function would help here).
I see no way how cpp's unordered-set could be tweaked to beat the khash-implementation used by pandas, which seems to be quite good for this type of tasks.
Here are for example these pretty old benchmarks, which show that khash
is somewhat faster than std::unordered_map
with only google_dense being even faster.
Using google dense map:
In my experiments, google dense map (from here) was able to beat khash
- benchmark code can be found at the end of the answer.
It was faster if there were only 50 unique elements:
#50 unique elements:
%timeit google_unique(a,r)
1.85 ms ± 8.26 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%timeit pd.unique(a)
3.52 ms ± 33.9 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
but also faster if there were only unique elements:
%timeit google_unique(c,r)
54.4 ms ± 375 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
In [3]: %timeit pd.unique(c)
75.4 ms ± 499 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
My few experiments have also shown, that google_hash_set
uses maybe more memory (up to 20%) than khash, but more tests are needed to see whether this is really the case.
I'm not sure my answer helped you at all. My take-aways are:
set(pd.unique(...))
seems to be a good starting point. Listings for google-tests:
#google_hash.cpp
#include <cstdint>
#include <functional>
#include <sparsehash/dense_hash_set>
typedef int64_t lli;
void cpp_unique(lli *input, int n, lli *output){
google::dense_hash_set<lli, std::hash<lli> > set;
set.set_empty_key(-1);
for (int i=0;i<n;i++){
set.insert(input[i]);
}
int cnt=0;
for(auto x : set)
output[cnt++]=x;
}
the corresponding pyx-file:
#google.pyx
cimport numpy as np
cdef extern from "google_hash.cpp":
void cpp_unique(np.int64_t *inp, int n, np.int64_t *output)
#out should have enough memory:
def google_unique(np.ndarray[np.int64_t,ndim=1] inp, np.ndarray[np.int64_t,ndim=1] out):
cpp_unique(&inp[0], len(inp), &out[0])
the setup.py-file:
from distutils.core import setup, Extension
from Cython.Build import cythonize
import numpy as np
setup(ext_modules=cythonize(Extension(
name='google',
language='c++',
extra_compile_args=['-std=c++11'],
sources = ["google.pyx"],
include_dirs=[np.get_include()]
)))
Ipython-benchmark script, after calling python setup.py build_ext --inplace
:
import numpy as np
import pandas as pd
from google import google_unique
a=np.random.randint(0,50,10**6,dtype=np.int64)
b=np.random.randint(0, 10**5,10**6, dtype=np.int64)
c=np.random.randint(0, 2**63,10**6, dtype=np.int64)
r=np.zeros((10**6,), dtype=np.int64)
%timeit google_unique(a,r
%timeit pd.unique(a)
Other listings
Cython version after fixes:
%%cython
cimport cython
from numpy cimport ndarray
from cpython cimport set
cimport numpy as np
@cython.wraparound(False)
@cython.boundscheck(False)
def unique_cython_int(ndarray[np.int64_t] a):
cdef int i
cdef int n = len(a)
cdef set s = set()
for i in range(n):
s.add(a[i])
return s
C++ version after fixes:
%%cython -+ -c=-std=c++11
cimport cython
cimport numpy as np
from numpy cimport ndarray
from libcpp.unordered_set cimport unordered_set
@cython.wraparound(False)
@cython.boundscheck(False)
def unique_cpp_int(ndarray[np.int64_t] a):
cdef int i
cdef int n = len(a)
cdef unordered_set[int] s
for i in range(n):
s.insert(a[i])
return s
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