Given two sorted arrays like the following:
a = array([1,2,4,5,6,8,9])
b = array([3,4,7,10])
I would like the output to be:
c = array([1,2,3,4,5,6,7,8,9,10])
or:
c = array([1,2,3,4,4,5,6,7,8,9,10])
I'm aware that I can do the following:
c = unique(concatenate((a,b))
I'm just wondering if there is a faster way to do it as the arrays I'm dealing with have millions of elements.
Any idea is welcomed. Thanks
Since you use numpy, I doubt that bisec helps you at all... So instead I would suggest two smaller things:
np.sort, use c.sort() method instead which sorts the array in place and avoids the copy.np.unique must use np.sort which is not in place. So instead of using np.unique do the logic by hand. IE. first sort (in-place) then do the np.unique method by hand (check also its python code), with flag = np.concatenate(([True], ar[1:] != ar[:-1])) with which unique = ar[flag] (with ar being sorted). To be a bit better, you should probably make the flag operation in place itself, ie. flag = np.ones(len(ar), dtype=bool) and then np.not_equal(ar[1:], ar[:-1], out=flag[1:]) which avoids basically one full copy of flag..sort has 3 different algorithms, since your arrays maybe are almost sorted already, changing the sorting method might make a speed difference.This would make the full thing close to what you got (without doing a unique beforehand):
def insort(a, b, kind='mergesort'):
# took mergesort as it seemed a tiny bit faster for my sorted large array try.
c = np.concatenate((a, b)) # we still need to do this unfortunatly.
c.sort(kind=kind)
flag = np.ones(len(c), dtype=bool)
np.not_equal(c[1:], c[:-1], out=flag[1:])
return c[flag]
Inserting elements into the middle of an array is a very inefficient operation as they're flat in memory, so you'll need to shift everything along whenever you insert another element. As a result, you probably don't want to use bisect. The complexity of doing so would be around O(N^2).
Your current approach is O(n*log(n)), so that's a lot better, but it's not perfect.
Inserting all the elements into a hash table (such as a set) is something. That's going to take O(N) time for uniquify, but then you need to sort which will take O(n*log(n)). Still not great.
The real O(N) solution involves allocated an array and then populating it one element at a time by taking the smallest head of your input lists, ie. a merge. Unfortunately neither numpy nor Python seem to have such a thing. The solution may be to write one in Cython.
It would look vaguely like the following:
def foo(numpy.ndarray[int, ndim=1] out,
numpy.ndarray[int, ndim=1] in1,
numpy.ndarray[int, ndim=1] in2):
cdef int i = 0
cdef int j = 0
cdef int k = 0
while (i!=len(in1)) or (j!=len(in2)):
# set out[k] to smaller of in[i] or in[j]
# increment k
# increment one of i or j
When curious about timings, it's always best to just timeit. Below, i've listed a subset of the various methods and their timings:
import numpy as np
import timeit
import heapq
def insort(a, x, lo=0, hi=None):
if hi is None: hi = len(a)
while lo < hi:
mid = (lo+hi)//2
if x < a[mid]: hi = mid
else: lo = mid+1
return lo, np.insert(a, lo, [x])
size=10000
a = np.array(range(size))
b = np.array(range(size))
def op(a,b):
return np.unique(np.concatenate((a,b)))
def martijn(a,b):
c = np.copy(a)
lo = 0
for i in b:
lo, c = insort(c, i, lo)
return c
def martijn2(a,b):
c = np.zeros(len(a) + len(b), a.dtype)
for i, v in enumerate(heapq.merge(a, b)):
c[i] = v
def larsmans(a,b):
return np.array(sorted(set(a) | set(b)))
def larsmans_mod(a,b):
return np.array(set.union(set(a),b))
def sebastian(a, b, kind='mergesort'):
# took mergesort as it seemed a tiny bit faster for my sorted large array try.
c = np.concatenate((a, b)) # we still need to do this unfortunatly.
c.sort(kind=kind)
flag = np.ones(len(c), dtype=bool)
np.not_equal(c[1:], c[:-1], out=flag[1:])
return c[flag]
Results:
martijn2 25.1079499722
OP 1.44831800461
larsmans 9.91507601738
larsmans_mod 5.87612199783
sebastian 3.50475311279e-05
My specific contribution here is larsmans_mod which avoids creating 2 sets -- it only creates 1 and in doing so cuts execution time nearly in half.
EDIT removed martijn as it was too slow to compete. Also tested for slightly bigger arrays (sorted) input. I also have not tested for correctness in output ...
In addition to the other answer on using bisect.insort, if you are not content with performance, you may try using blist module with bisect. It should improve the performance.
Traditional list insertion complexity is O(n), while blist's complexity on insertion is O(log(n)).
Also, you arrays seem to be sorted. If so, you can use merge function from heapq mudule to utilize the fact that both arrays are presorted. This approach will take an overhead because of crating a new array in memory. It may be an option to consider as this solution's time complexity is O(n+m), while the solutions with insort are O(n*m) complexity (n elements * m insertions)
import heapq
a = [1,2,4,5,6,8,9]
b = [3,4,7,10]
it = heapq.merge(a,b) #iterator consisting of merged elements of a and b
L = list(it) #list made of it
print(L)
Output:
[1, 2, 3, 4, 4, 5, 6, 7, 8, 9, 10]
If you want to delete repeating values, you can use groupby:
import heapq
import itertools
a = [1,2,4,5,6,8,9]
b = [3,4,7,10]
it = heapq.merge(a,b) #iterator consisting of merged elements of a and b
it = (k for k,v in itertools.groupby(it))
L = list(it) #list made of it
print(L)
Output:
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
You could use the bisect module for such merges, merging the second python list into the first.
The bisect* functions work for numpy arrays but the insort* functions don't. It's easy enough to use the module source code to adapt the algorithm, it's quite basic:
from numpy import array, copy, insert
def insort(a, x, lo=0, hi=None):
if hi is None: hi = len(a)
while lo < hi:
mid = (lo+hi)//2
if x < a[mid]: hi = mid
else: lo = mid+1
return lo, insert(a, lo, [x])
a = array([1,2,4,5,6,8,9])
b = array([3,4,7,10])
c = copy(a)
lo = 0
for i in b:
lo, c = insort(c, i, lo)
Not that the custom insort is really adding anything here, the default bisect.bisect works just fine too:
import bisect
c = copy(a)
lo = 0
for i in b:
lo = bisect.bisect(c, i)
c = insert(c, i, lo)
Using this adapted insort is much more efficient than a combine and sort. Because b is sorted as well, we can track the lo insertion point and search for the next point starting there instead of considering the whole array each loop.
If you don't need to preserve a, just operate directly on that array and save yourself the copy.
More efficient still: because both lists are sorted, we can use heapq.merge:
from numpy import zeros
import heapq
c = zeros(len(a) + len(b), a.dtype)
for i, v in enumerate(heapq.merge(a, b)):
c[i] = v
Use the bisect module for this:
import bisect
a = array([1,2,4,5,6,8,9])
b = array([3,4,7,10])
for i in b:
pos = bisect.bisect(a, i)
insert(a,[pos],i)
I can't test this right now, but it should work
The sortednp package implements an efficient merge of sorted numpy-arrays, just sorting the values, not making them unique:
import numpy as np
import sortednp
a = np.array([1,2,4,5,6,8,9])
b = np.array([3,4,7,10])
c = sortednp.merge(a, b)
I measured the times and compared them in this answer to a similar post where it outperforms numpy's mergesort (v1.17.4).
Seems like no one mentioned union1d (union1d). Currently, it is a shortcut for unique(concatenate((ar1, ar2))), but its a short name to remember and it has a potential to be optimized by numpy developers since its a library function. It performs very similar to insort from seberg's accepted answer for large arrays. Here is my benchmark:
import numpy as np
def insort(a, b, kind='mergesort'):
# took mergesort as it seemed a tiny bit faster for my sorted large array try.
c = np.concatenate((a, b)) # we still need to do this unfortunatly.
c.sort(kind=kind)
flag = np.ones(len(c), dtype=bool)
np.not_equal(c[1:], c[:-1], out=flag[1:])
return c[flag]
size = int(1e7)
a = np.random.randint(np.iinfo(np.int).min, np.iinfo(np.int).max, size)
b = np.random.randint(np.iinfo(np.int).min, np.iinfo(np.int).max, size)
np.testing.assert_array_equal(insort(a, b), np.union1d(a, b))
import timeit
repetitions = 20
print("insort: %.5fs" % (timeit.timeit("insort(a, b)", "from __main__ import a, b, insort", number=repetitions)/repetitions,))
print("union1d: %.5fs" % (timeit.timeit("np.union1d(a, b)", "from __main__ import a, b; import numpy as np", number=repetitions)/repetitions,))
Output on my machine:
insort: 1.69962s
union1d: 1.66338s
