I am attempting to create an efficient algorithm to pull all of the similar elements between two lists. The problem is two fold. First, I can not seem to find any similar algorithms online. Second, there should be a more efficient way.
By 'similar', I mean a predesigned way to find all similar elements between two lists in a timely fashion.
Currently, I am taking a greedy approach by:
- Sorting the lists that are being compared,
- Comparing each element in the shorter list to each element in the larger list,
- Since the
largeList
andsmallList
are sorted we can save the last index that was visited, - Continue from the previous index (
largeIndex
).
Currently, the run-time seems to be average of O(nlog(n))
. This can be seen by running the test cases listed after this block of code.
This is being run on Python 3.3.
Right now, my code looks as such:
def compare(small,large,largeStart,largeEnd):
for i in range(largeStart, largeEnd):
if small==large[i]:
return [1,i]
if small<large[i]:
if i!=0:
return [0,i-1]
else:
return [0, i]
return [0,largeStart]
def determineLongerList(aList, bList):
if len(aList)>len(bList):
return (aList, bList)
elif len(aList)<len(bList):
return (bList, aList)
else:
return (aList, bList)
def compareElementsInLists(aList, bList):
import time
startTime = time.time()
holder = determineLongerList(aList, bList)
sameItems = []
iterations = 0
##########################################
smallList = sorted(holder[1])
smallLength = len(smallList)
smallIndex = 0
largeList = sorted(holder[0])
largeLength = len(largeList)
largeIndex = 0
while (smallIndex<smallLength):
boolean = compare(smallList[smallIndex],largeList,largeIndex,largeLength)
if boolean[0]==1:
#`compare` returns 1 as True
sameItems.append(smallList[smallIndex])
oldIndex = largeIndex
largeIndex = boolean[1]
else:
#else no match and possible new index
oldIndex = largeIndex
largeIndex = boolean[1]
smallIndex+=1
iterations =largeIndex-oldIndex+iterations+1
print('RAN {it} OUT OF {mathz} POSSIBLE'.format(it=iterations, mathz=smallLength*largeLength))
print('RATIO:\t\t'+str(iterations/(smallLength*largeLength))+'\n')
return sameItems
, and here are some test cases:
def testLargest():
import time
from random import randint
print('\n\n******************************************\n')
start_time = time.time()
lis = []
for i in range(0,1000000):
ran = randint(0,1000000)
lis.append(ran)
lis2 = []
for i in range(0,1000000):
ran = randint(0,1000000)
lis2.append(ran)
timeTaken = time.time()-start_time
print('CREATING LISTS TOOK:\t\t'+str(timeTaken))
print('\n******************************************')
start_time = time.time()
c = compareElementsInLists(lis, lis2)
timeTaken = time.time()-start_time
print('COMPARING LISTS TOOK:\t\t'+str(timeTaken))
print('NUMBER OF SAME ITEMS:\t\t'+str(len(c)))
print('\n******************************************')
#testLargest()
'''
One rendition of testLargest:
******************************************
CREATING LISTS TOOK: 21.009342908859253
******************************************
RAN 999998 OUT OF 1000000000000 POSSIBLE
RATIO: 9.99998e-07
COMPARING LISTS TOOK: 13.99990701675415
NUMBER OF SAME ITEMS: 632328
******************************************
'''
def testLarge():
import time
from random import randint
print('\n\n******************************************\n')
start_time = time.time()
lis = []
for i in range(0,1000000):
ran = randint(0,100)
lis.append(ran)
lis2 = []
for i in range(0,1000000):
ran = randint(0,100)
lis2.append(ran)
timeTaken = time.time()-start_time
print('CREATING LISTS TOOK:\t\t'+str(timeTaken))
print('\n******************************************')
start_time = time.time()
c = compareElementsInLists(lis, lis2)
timeTaken = time.time()-start_time
print('COMPARING LISTS TOOK:\t\t'+str(timeTaken))
print('NUMBER OF SAME ITEMS:\t\t'+str(len(c)))
print('\n******************************************')
testLarge()
set
once the data becomes large enough. Also,set
does not return he entire collection, but just similar elements. I am looking intocollections.Counter
now... – T.Woody Apr 6 at 1:07