[英]Sorting string in order of high frequency terms from Inverted Index Elasticsearch
我是Elasticsearch的新手,我想知道这样做是否可行:
我有一堆地址字符串,我想对字符串中最重复的术语进行排序。
例如:
1. Shop no 1 ABC Lane City1 - Zipcode1
2. Shop no 2 EFG Lane City1 - Zipcode2
3. Shop no 1 XYZ Lane City2 - Zipcode3
4. Shop no 3 ABC Lane City1 - Zipcode1
我真正需要的是将它们放在字符串中最常见的术语上。
那么前面例子的排序输出应该是:
1. Shop no 1 ABC Lane City1 - Zipcode1
4. Shop no 3 ABC Lane City1 - Zipcode1 # Because 1 and 2 have the most common words in them.
2. Shop no 2 EFG Lane City1 - Zipcode2 # Second most common words with 1 and 4.
3. Shop no 1 XYZ Lane City2 - Zipcode3 # Not all that many common terms amongst them.
我不知道如何去做。 我知道我可以触发每个字符串作为查询,以获得最接近被触发的查询的结果。 但我有十万行本身,它似乎根本不是一个有效的选择。
如果我可以使用matchall()
并使用每个字符串中包含大量重复术语的term
过滤器进行sort
,那将非常有用。
可以对包含倒排索引中大多数相似单词的文档进行排序吗?
这是我的数据外观的示例pastebin: 示例地址
解
我使用https://stackoverflow.com/a/15174569/61903来计算两个字符串的余弦相似度 (@vpekar的信用)作为相似性的基本算法。 通常我将所有字符串放入列表中。 然后我将索引参数i设置为0并循环i,只要它在列表长度的范围内。 在该循环中,我将位置p从i + 1迭代到长度(列表)。 然后我找到list [i]和list [p]之间的最大余弦值。 两个文本字符串都将被列入一个列表中,因此在以后的相似度计算中不会将它们考虑在内。 两个文本字符串将与余弦值一起放入结果列表中,数据结构为VectorResult。
之后,列表按余弦值排序。 我们现在有唯一的字符串对,下降余弦,即相似值。 HTH。
import re
import math
import timeit
from collections import Counter
WORD = re.compile(r'\w+')
def get_cosine(vec1, vec2):
intersection = set(vec1.keys()) & set(vec2.keys())
numerator = sum([vec1[x] * vec2[x] for x in intersection])
sum1 = sum([vec1[x] ** 2 for x in vec1.keys()])
sum2 = sum([vec2[x] ** 2 for x in vec2.keys()])
denominator = math.sqrt(sum1) * math.sqrt(sum2)
if not denominator:
return 0.0
else:
return float(numerator) / denominator
def text_to_vector(text):
words = WORD.findall(text)
return Counter(words)
class VectorResult(object):
def __init__(self, cosine, text_1, text_2):
self.cosine = cosine
self.text_1 = text_1
self.text_2 = text_2
def __eq__(self, other):
if self.cosine == other.cosine:
return True
return False
def __le__(self, other):
if self.cosine <= other.cosine:
return True
return False
def __ge__(self, other):
if self.cosine >= other.cosine:
return True
return False
def __lt__(self, other):
if self.cosine < other.cosine:
return True
return False
def __gt__(self, other):
if self.cosine > other.cosine:
return True
return False
def main():
start = timeit.default_timer()
texts = []
with open('data.txt', 'r') as f:
texts = f.readlines()
cosmap = []
i = 0
out = []
while i < len(texts):
max_cosine = 0.0
current = None
for p in range(i + 1, len(texts)):
if texts[i] in out or texts[p] in out:
continue
vector1 = text_to_vector(texts[i])
vector2 = text_to_vector(texts[p])
cosine = get_cosine(vector1, vector2)
if cosine > max_cosine:
current = VectorResult(cosine, texts[i], texts[p])
max_cosine = cosine
if current:
out.extend([current.text_1, current.text_2])
cosmap.append(current)
i += 1
cosmap = sorted(cosmap)
for item in reversed(cosmap):
print(item.cosine, item.text_1, item.text_2)
end = timeit.default_timer()
print("Similarity Sorting of {} strings lasted {} s.".format(len(texts), end - start))
if __name__ == '__main__':
main()
结果
我在http://pastebin.com/hySkZ4Pn上使用了你的光盘地址作为测试数据:
1.0000000000000002 NO 15& 16 1ST FLOOR,2ND MAIN ROAD,KHB COLONY,GANDINAGAR YELAHANKA
NO 15& 16 1ST FLOOR,2ND MAIN ROAD,KHB COLONY,GANDINAGAR YELAHANKA
1.0 # 51/3 AGRAHARA YELAHANKA
#51/3 AGRAHARA YELAHANKA
0.9999999999999999 # C M C ROAD,YALAHANKA
# C M C ROAD,YALAHANKA
0.8728715609439696 # 1002/B B B ROAD,YELAHANKA
0,B B ROAD,YELAHANKA
0.8432740427115678 # LAKSHMI COMPLEX C M C ROAD,YALAHANKA
# SRI LAKSHMAN COMPLEX C M C ROAD,YALAHANKA
0.8333333333333335 # 85/1 B B M P OFFICE ROAD,KOGILU YELAHANKA
#85/1 B B M P OFFICE NEAR KOGILU YALAHANKA
0.8249579113843053 # 689 3RD A CROSS SHESHADRIPURAM CALLEGE OPP YELAHANKA
# 715 3RD CROSS A SECTUR SHESHADRIPURAM CALLEGE OPP YELAHANKA
0.8249579113843053 # 10 RAMAIAIA COMPLEX B B ROAD,YALAHANKA
# JAMATI COMPLEX B B ROAD,YALAHANKA
[ SNIPPED ]
Similarity Sorting of 702 strings lasted 8.955146235887025 s.
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