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# Parse wiktionary.xml with pure python, such that it can be run with pypy (python just in time compiler)
# optimization would be possible through cython and assembler loops etc
# on a linux system, get the first n lines of a document with:
# head -n1000000 dewiktionary-20181201-pages-articles.xml > wiktionaryFirstMio.xml
import sys
import os
import re
class Parser(object):
def __init__(self, InputDokument, OutputDokument):
self.Indok = InputDokument
self.Outdok = OutputDokument
def GetSeparators(self):
with open(self.Indok) as xmldok:
with open(self.Outdok , 'w') as getsepdok:
seperators = []
counter = 0
for line in xmldok:
counter += 1
#print(counter)
if (counter % 10) == 0:
print(counter)
seperator =[]
val = 0
#if counter == 10000:
#seperatorsSet = []
#getsepdok.write('[' + '\n')
#for element in seperators:
#seperatorsSet.append(''.join(element))
#for element in set(seperatorsSet):
#getsepdok.write(str(''.join(element)) + '\n')
#getsepdok.write(']')
for letter in line:
#print(letter)
if letter == '>':
val = 0
seperators.append(seperator)
seperator = []
if val == 1:
seperator.append(letter)
else:
pass
if letter == '<':
val = 1
seperatorsSet = []
getsepdok.write('[' + '\n')
for element in seperators:
seperatorsSet.append(''.join(element))
seperatorsSet = set(seperatorsSet)
for element in set(seperatorsSet):
getsepdok.write(str(''.join(element)) + '\n')
getsepdok.write(']')
return seperatorsSet
def GetPayloadBetweenTwoSymbols(self, SymbolA, SymbolB , LogLineNumber=False, Doc = True):
with open(self.Indok) as xmldok:
with open(self.Outdok , 'w') as payloaddok:
seperators = []
counter = 0
valA = 0
valB = 0
seperator =[]
for line in xmldok:
#print(line)
counter += 1
if LogLineNumber == True:
if (counter % 10000) == 0:
print(counter)
wait1letterA = False
wait1letterB = False
#for letter in line.decode('utf-8'):
for letter in line:
#print(letter)
#print(set(range(1, len(SymbolA))))
if valA % len(SymbolA) in set(range(1, len(SymbolA) )):
#print('jo')
if wait1letterA == True:
#print('joo')
#print(letter)
valA -= valA % len(SymbolA)
wait1letterA = False
wait1letterA = True
if valB in set(range(1, len(SymbolB) )):
if wait1letterB == True:
valB = 0
wait1letterB = False
wait1letterB = True
for n in range(len(SymbolB)):
if valA >= len(SymbolA) and valB == n and letter == SymbolB[n]:
valB = n + 1
wait1letterB = False
else:
pass
if valB == len(SymbolB) and valA >= len(SymbolA):
valB = 0
#print(letter)
#print(valA)
valA -= len(SymbolA)
#print(valA)
#print(seperators)
if valA >= len(SymbolA):
seperator.append(letter)
else:
pass
#print(valA)
#print(SymbolA[6])
#print(len(SymbolA))
#print(range(len(SymbolA)))
if valA == 0:
if len(seperator[:-(len(SymbolB)-1)]) >= 1:
seperators.append(seperator[:-(len(SymbolB)-1)])
seperator = []
for n in range(len(SymbolA)):
#print(n)
if valA % len(SymbolA) == n and letter == SymbolA[n]:
valA += 1
#print(valA)
wait1letterA = False
break
else:
pass
seperatorsSet = []
#getsepdok.write('[' + '\n')
for element in seperators:
seperatorsSet.append(''.join(element))
seperatorsSet = set(seperatorsSet)
output = []
ID = 0
## Set has a probabilistic factor in it!!!! thats why the nmbers change
for element in seperatorsSet:
output.append([element, ID])
ID += 1
return output
def GetPayloadBetweenTwoSymbolsInPayload(self, Payload, SymbolA, SymbolB, LogElementNumber):
seperators = []
counter = 0
for element in Payload:
counter += 1
if LogElementNumber == True:
if (counter % 1000) == 0:
print(counter)
seperator =[]
wait1letterA = False
wait1letterB = False
valA = 0
valB = 0
for letter in element[0]:
#print(letter)
#print(set(range(1, len(SymbolA))))
if valA % len(SymbolA) in set(range(1, len(SymbolA) )):
#print(valA)
#print('jo')
if wait1letterA == True:
#print('joo')
valA -= valA % len(SymbolA)
wait1letterA = False
wait1letterA = True
if valB in set(range(1, len(SymbolB) )) and valA >= len(SymbolA):
if wait1letterB == True:
valB = 0
wait1letterB = False
wait1letterB = True
#for n in range(len(SymbolB)):
#if valB == n and letter == SymbolB[n]:
#valB = n + 1
#wait1letterB = False
#else:
#pass
if letter == SymbolB[valB % len(SymbolB)] and valA >= len(SymbolA):
valB += 1
wait1letterB = False
else:
pass
if valB == len(SymbolB) and valA >= len(SymbolA):
valB = 0
#print(valA)
valA -= len(SymbolA)
#print(valA)
#print(seperators)
if valA >= len(SymbolA):
##print(letter)
seperator.append(letter)
#print(seperator)
else:
pass
#print(valA)
#print(SymbolA[6])
#print(len(SymbolA))
#print(range(len(SymbolA)))
if valA == 0:
#print('seps')
if len(seperator[:-(len(SymbolB)-1)]) >= 1:
seperators.append([''.join(seperator[:-(len(SymbolB)-1)]), element[1]])
seperator = []
# Optimierungsmoeglichkeit: Hier kann die for schleife durch viele ifs ersetzt werden, sowas wie start for after zwei ifs.
# wuerde einiges an computation wegnehmen, auch da beide symbole
#for n in range(len(SymbolA)):
##print(n)
#if valA % len(SymbolA) == n and letter == SymbolA[n]:
##print(SymbolA[n])
#valA += 1
#wait1letterA = False
#else:
#pass
for n in range(len(SymbolA)):
#print(n)
if valA % len(SymbolA) == n and letter == SymbolA[n]:
valA += 1
#print(valA)
wait1letterA = False
break
else:
pass
return seperators
def GetPayloadBetweenTwoOneSymbolsInPayload(self, Payload, SymbolA, SymbolB, LogElementNumber, Payloadrow, IDrow):
counter = 0
seperator =[]
seperators = []
for payload in Payload:
val = 0
for letter in payload[Payloadrow]:
counter += 1
#print(counter)
if LogElementNumber == True:
if (counter % 10) == 0:
print(counter)
#print(letter)
if letter == SymbolB:
val -= 1
if val >= 1:
seperator.append(letter)
else:
pass
if val == 0 and len(seperator) >= 1:
seperators.append([''.join(seperator), payload[IDrow]])
seperator = []
if letter == SymbolA:
#print(val)
val += 1
return seperators
def CutTextAtSymbol(self, text, symbol):
itisthesymbol = 0
outtext = []
output = []
symbolisthere = 0
for letter in text:
outtext.append(letter)
#print(letter)
if letter != symbol[itisthesymbol]:
itisthesymbol = 0
if letter == symbol[itisthesymbol]:
itisthesymbol += 1
if itisthesymbol == len(symbol):
#print(outtext)
output.append(''.join(outtext))
itisthesymbol = 0
symbolisthere = 1
if symbolisthere == 0:
output.append(''.join(outtext))
return output[0]
def GetPayloadBetweenTwoSymbolsInText(self, text, SymbolA, SymbolB):
seperators = []
seperator =[]
wait1letterA = False
wait1letterB = False
valA = 0
valB = 0
for letter in text:
#print(letter)
#print(SymbolA)
if valA % len(SymbolA) in set(range(1, len(SymbolA) )):
if wait1letterA == True:
valA -= valA % len(SymbolA)
wait1letterA = False
wait1letterA = True
#print('B',valB)
#print(valA)
if valB in set(range(1, len(SymbolB) )):
if wait1letterB == True:
valB = 0
wait1letterB = False
wait1letterB = True
#print('B',valB)
#print(valA)
if letter == SymbolB[valB % len(SymbolB)]:
valB += 1
wait1letterB = False
else:
pass
if valB == len(SymbolB):
valB = 0
valA -= len(SymbolA)
#print('B',valB)
#print(valA)
if valA >= len(SymbolA):
#print('append')
seperator.append(letter)
else:
pass
if valA == 0:
if len(seperator[:-(len(SymbolB)-1)]) >= 1:
seperators.append([''.join(seperator[:-(len(SymbolB)-1)])])
seperator = []
# Optimierungsmoeglichkeit: Hier kann die for schleife durch viele ifs ersetzt werden, sowas wie start for after zwei ifs.
# wuerde einiges an computation wegnehmen, auch da beide symbole
#for n in range(len(SymbolA)):
#print(SymbolA[valA % len(SymbolA)])
if letter == SymbolA[valA % len(SymbolA)]:
#print('oi')
valA += 1
wait1letterA = False
else:
pass
return seperators
def GetPayloadBetweenTwoSameSymbolsInText(self, text, Symbol):
seperators = []
seperator =[]
wait1letter = False
nowendit = False
val = 0
for letter in text:
#print(letter)
#print(SymbolA)
if nowendit == False and letter == Symbol[val % len(Symbol)]:
val += 1
if nowendit == True and letter == Symbol[val % len(Symbol)]:
val -= 1
if val == len(Symbol):
seperator.append(letter)
nowendit = True
#print('append')
if val == 0 and len(seperator) >= 1:
seperators.append(' '.join(seperator))
seperator = []
nowendit = False
return seperators
def ParseWordswithSymbolFromSymbolongoing(self, text, Symbol):
seperators = []
#print(text.split())
for word in text.split():
val = 0
waitoneletter = False
seperator = []
for letter in word:
#print(letter)
#print(val)
if val < len(Symbol):
if letter == Symbol[val]:
val += 1
#print(letter)
#print(len(Symbol))
#print(val)
if val >= len(Symbol):
val = len(Symbol)
if val < len(Symbol):
if letter != Symbol[val]:
val = 0
if val == len(Symbol):
seperator.append(letter)
#print('itsappending')
if len(seperator) >= 1:
seperators.append(''.join(seperator))
seperator = []
return seperators
def ParseWithHighestLetterAccordance(self, inputtext, Letters):
# first check if there is a word that has all letters
short = False
lettervect = []
Lettervector = []
wordscores = []
text = inputtext.lower()
if '.' in set(Letters):
short = True
if short == True:
for letter in re.sub("[^a-zA-Züäö.]", " ", Letters):
letter = letter.lower()
#print(re.sub("[^a-züäö.]", " ", Letters))
if letter != '.' and letter != ' ':
lettervect.append(letter)
if letter == '.':
Lettervector.append(lettervect)
lettervect = []
if len(lettervect) >= 1:
Lettervector.append(lettervect)
else:
for letter in re.sub("[^a-zA-Züäö.]", " ", Letters):
letter = letter.lower()
Lettervector.append([letter])
#print(text)
#print(Lettervector)
from copy import deepcopy
for word in text.split():
lettervector = deepcopy(Lettervector)
#print(word)
#print(Lettervector)
wordscore = []
for n in range(len(lettervector)):
wordscore.append([word, 0])
#wordscore = len(lettervector) * [[word, 0 ]]
#print(wordscore)
firstletter = 0
usedletters = []
for letter in word:
firstletter += 1
#print(set(Letters))
#print(wordscore)
#print(lettervector[n])
if firstletter == 1:
if letter == lettervector[0][0]:
#print('oi')
#print(lettervector)
#print(len(lettervector[2]))
wordscore[0][1] += 1
lettervector[0].remove(letter)
#print(usedletters)
else:
lettervector[0].remove(lettervector[0][0])
for n in range(len(lettervector)):
#print('1' ,letter)
#print(lettervector[n][0])
if letter in set(lettervector[n]):
#print('ooioi',usedletters)
if letter not in set(usedletters):
#print('something was added', letter)
wordscore[n][1] += 1
lettervector[n].remove(letter)
#print('angesprungen')
wordscores.append(wordscore)
#print(wordscores)
#checkbest_firstlettervector = []
#for n in range(len(wordscores)):
#checkbest_firstlettervector.append([ n , wordscores[n][0][1]])
#print('wordscores', wordscores)
#best_n_lettervectors = sorted(checkbest_firstlettervector[::-1], key=lambda tup: tup[1], reverse=True)
#print(best_n_lettervectors)
#for wordscore in wordscores:
ntupelscores = []
ntupelscoresm = []
for o in range(len(wordscores)):
#print('newlettervectorindex')
lastletterexistentindex = 1
lastlettercame = False
if wordscores[o][0][1] >= 1:
for m in range(1, len(lettervector) + 1):
#print(m)
if o <= len(text.split()) - (m):
triplescore = []
for q in range(len(wordscores[o])):
triplescore.append(0)
#print(len(lettervector))
for n in range(m):
#print(wordscores[lettervectorindex[0] + n][n][1])
#wordscores[lettervectorindex[0] + 1][1][1] + wordscores[lettervectorindex[0] + 2][2][1]
for p in range(len(wordscores[o])):
#print(wordscore[o + n][p][1])
#print(len(Lettervector[p]))
if wordscores[o + n][p][1] == len(Lettervector[p]):
triplescore[p] += wordscores[o + n][p][1]
letterlength = 0
for r in range(len(lettervector)):
letterlength += len(Lettervector[r])
#print(wordscore)
#print(sum(triplescore))
if p == len(wordscores[o]) - 1 and wordscores[o + n][p][1] == len(Lettervector[p]) and lastlettercame == False and sum(triplescore) == letterlength:
#print('oioioioioioioooioioioiiiiiiiiiiiiiiiiiiiiiiiiiii')
lastletterexistentindex = n
lastlettercame = True
#triplescore += wordscores[o + n][p][1]
ntupelscores.append([[o , m, lastletterexistentindex], sum(triplescore)])
#ntupelscoresm.append([m , triplescore])
#print(text.split())
#print('bliblablub', ntupelscores)
for tupel in ntupelscores:
if text.split()[tupel[0][0]][0] == Lettervector[0][0]:
tupel[1] += 3
#print('b',text.split()[tupel[0][0] + tupel[0][1] - 1][0])
#print('a',Lettervector[-1][0])
if text.split()[tupel[0][0] + tupel[0][1] - 1][0] == Lettervector[-1][0]:
tupel[1] += 3
# Bestrafe laengere Tupel, sprich wenn durch weitere worte kein score dazukommt
tupel[1] -= tupel[0][1] * 0.1
bestntupelscoresorted = sorted(ntupelscores[::-1], key=lambda tup: tup[1], reverse=True)
#bestntupelscoresortedm = sorted(ntupelscoresm[::-1], key=lambda tup: tup[1], reverse=True)
#print('oioioioioioioioioioi',bestntupelscoresorted)
outputntupel = []
#print(bestntupelscoresorted)
for s in range(bestntupelscoresorted[0][0][1] ):
outputntupel.append(text.split()[bestntupelscoresorted[0][0][0] + s])
#print(outputntupel)
return outputntupel
#def parseWordsContainingCertainSymbols(self, text, symbols):
#print()
#fooSeparator = 'title'
#cwd = os.getcwd()
#with open('dewiktionary-20181201-pages-articles.xml') as xmldok:
#with open(cwd + '/' + 'classes.txt', 'w') as Outdok:
#n = 0
#done = False
#while done == False:
#for line in xmldok:
#n += 1
##print(line)
##print(dok_to_token(line))
##print(n)
#for word in line:
#print(word)
#try:
#if dok_to_token(line)[:(len(fooSeparator) + 2)] == '<' + fooSeparator + '>':
#Outdok.write(dok_to_token(line)[len(fooSeperator):-len(fooSeperator)] + '\n')
#except:
#pass
#if n >= 100000:
#quit()

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parse1.pyc View File


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