# Put (word_index, matches_whole) to our dictĭ]. Superstring License Key 0 Comments It is a single unified theory that would be able to describe all of the forces at work in the universe.While twenty-six dimensions are far too many, they can be twisted or even folded into tiny tight compact regions in spacetime, thus possibly reducing the dimensions to four. # Building a big dict first, this may be a little slow and cosuming a lot memory However, a key property of these models, shown by A. So I changed the 'checkForSubstring' method:''' Schwarz that anomalies in type I string theory cancel sparked the first superstring revolution. Text = '''Sort all the strings by length, for each word, compare it only to the longer words. Taking the benefit of that index dict, the time-complexity of every following search will drop down to O(1), that may speed things up A LOT.īelow is my example code: # -*- coding: utf-8 -*. If the LARGE string list is not that big, you may build a HUGE dict with every possible contiguous substrings in it. This solves the earlier problem.īut it's not appreciably faster, could someone suggest a faster way to do this? At the moment, it's a bottleneck. Instead of sorting by length, this version splits the list of strings into many lists by length and then checks each of these lists against every list of larger words. SameLengthWordsLists = for num in set(len(i) for i in words)] Also discover the danceability, energy, liveness, instrumentalness, happiness and more musical analysis points on Musicstax. So I changed the checkForSubstring method: def checkForSubstring(words): Find the key and tempo for Superstring - RANK 1s Radio Edit By Cygnus X, Rank 1. Sort all the strings by length, for each word, compare it only to the longer words.īut this method has a flaw in that words are still being compared to words of the same length which are arbitrarily placed after it during the list sort. Print('%s is a substring of some other string' % currentWord) If superWord.find(subWord) != -1 and len(subWord) != len(superWord): Originally I had this method: def isSubstring(subWord, words): I'm working on a project that requires checking a string against a very large list of strings, searching for occasions where the string is a substring of one of the elements of the list.

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