python - Fastest way of comparing two numpy arrays -
i have 2 arrays:
>>> import numpy np >>> a=np.array([2, 1, 3, 3, 3]) >>> b=np.array([1, 2, 3, 3, 3]) what fastest way of comparing these 2 arrays equality of elements, regardless of order?
edit measured execution times of following functions:
def compare1(): #works arrays without redundant elements a=np.array([1,2,3,5,4]) b=np.array([2,1,3,4,5]) temp=0 in a: temp+=len(np.where(b==i)[0]) if temp==5: val=true else: val=false return 0 def compare2(): a=np.array([1,2,3,3,3]) b=np.array([2,1,3,3,3]) val=np.all(np.sort(a)==np.sort(b)) return 0 def compare3(): #thx odiogosilva a=np.array([1,2,3,3,3]) b=np.array([2,1,3,3,3]) val=set(a)==set(b) return 0 import numpy.lib.arraysetops aso def compare4(): #thx tom10 a=np.array([1,2,3,3,3]) b=np.array([2,1,3,3,3]) val=len(aso.setdiff1d(a,b))==0 return 0 the results are:
>>> import timeit >>> timeit.timeit(compare1,number=1000) 0.0166780948638916 >>> timeit.timeit(compare2,number=1000) 0.016178131103515625 >>> timeit.timeit(compare3,number=1000) 0.008063077926635742 >>> timeit.timeit(compare4,number=1000) 0.03257489204406738 seems "set"-method odiogosilva fastest.
do know other methods can test well?
edit2 runtime above not right measure comparing arrays, explained in comment user2357112.
#test.py import numpy np import numpy.lib.arraysetops aso #without duplicates n=10000 a=np.arange(n,0,step=-2) b=np.arange(n,0,step=-2) def compare1(): temp=0 in a: temp+=len(np.where(b==i)[0]) if temp==len(a): val=true else: val=false return val def compare2(): val=np.all(np.sort(a)==np.sort(b)) return val def compare3(): val=set(a)==set(b) return val def compare4(): val=len(aso.setdiff1d(a,b))==0 return val the output is:
>>> test import * >>> import timeit >>> timeit.timeit(compare1,number=1000) 101.16708397865295 >>> timeit.timeit(compare2,number=1000) 0.09285593032836914 >>> timeit.timeit(compare3,number=1000) 1.425955057144165 >>> timeit.timeit(compare4,number=1000) 0.44780397415161133 now compare2 fastest. there still method outgun this?
numpy collection of set operations.
import numpy np import numpy.lib.arraysetops aso a=np.array([2, 1, 3, 3, 3]) b=np.array([1, 2, 3, 3, 3]) print aso.setdiff1d(a, b) 
Comments
Post a Comment