首页 > 解决方案 > 用 levenahtein 将大文件聚类到 3 个组

问题描述

嗨,我有一个小文件和一个大文件,这里的代码甚至不适用于大文件,只适用于小文件,那么我如何读取大文件并对其执行操作?当我阅读并尝试在一个循环中进行聚类时,它不起作用,因为每次迭代都只在线。这是小文件的问题:文件行,我需要将它们分成3组。我已经尝试过亲和力传播,但它没有获得组大小参数,它给了我 4 个组,而第 4 组只有一个词与另一个组非常接近:

0
 - *Bras5emax Estates, L.T.D.
:* Bras5emax Estates, L.T.D.

1
 - *BOZEMAN Enterprises
:* BBAZEMAX ESTATES, LTD
, BOZEMAN Ent.
, BOZEMAN Enterprises
, BOZERMAN ENTERPRISES
, BRAZEMAX ESTATYS, LTD
, Bozeman Enterprises

2
 - *PC Adelman
:* John Smith
, Michele LTD
, Nadelman, Jr
, PC Adelman

3
 - *Gramkai, Inc.
:* Gramkai Books
, Gramkai, Inc.
, Gramkat Estates, Inc., Gramkat, Inc.

然后我尝试了 K-MEANS 但结果:

0
 - *Gramkai Books
, Gramkai, Inc.
, Gramkat Estates, Inc., Gramkat, Inc.
:*
1
 - *BBAZEMAX ESTATES, LTD
, BOZEMAN Enterprises
, BOZERMAN ENTERPRISES
, BRAZEMAX ESTATYS, LTD
, Bozeman Enterprises
, Bras5emax Estates, L.T.D.
:*
2
 - *BOZEMAN Ent.
, John Smith
, Michele LTD
, Nadelman, Jr
, PC Adelman
:*

如您所见,BOZEMAN Ent。属于第 2 组而不是第 1 组。

我的问题是:有没有办法做一个更好的集群?K-MEANS 中是否有 cluster_center ?

编码:

import numpy as np
import sklearn.cluster
import distance

f = open("names.txt", "r")
words = f.readlines()
words = np.asarray(words) #So that indexing with a list will work
lev_similarity = -1*np.array([[distance.levenshtein(w1,w2) for w1 in words] for w2 in words])
affprop = sklearn.cluster.KMeans(n_clusters=3)
affprop.fit(lev_similarity)
for cluster_id in np.unique(affprop.labels_):
    print(cluster_id)
    cluster = np.unique(words[np.nonzero(affprop.labels_==cluster_id)])
    cluster_str = ", ".join(cluster)
    print(" - *%s:*" % ( cluster_str))

标签: pythonfilemachine-learningcluster-computingk-means

解决方案


可以通过多种方式改进给定文本名称(企业)的聚类。

  1. 介绍一些文本清理和领域知识,例如去除点、常见的企业停用词和降低字符:
words = [re.sub(r"(,|\.|ltd|l\.t\.d|inc|estates|enterprises|ent|estatys)","", w.lower()).strip() for w in words]
  1. 使用“标准化”版本的distance.levenshtein距离可以进行有意义的比较,例如:
distance.nlevenshtein("abc", "acd", method=1)  # shortest alignment
distance.nlevenshtein("abc", "acd", method=2)  # longest alignment
  1. 尝试另一种距离度量:sorensen或者jaccard已经标准化。

下面是完整的代码示例:

words = \
["Gramkai Books",
"Gramkai, Inc.",
"Gramkat Estates, Inc.",
"Gramkat, Inc.",
"BBAZEMAX ESTATES, LTD",
"BOZEMAN Enterprises",
"BOZERMAN ENTERPRISES",
"BRAZEMAX ESTATYS, LTD",
"Bozeman Enterprises",
"Bras5emax Estates, L.T.D.",
"BOZEMAN Ent.",
"John Smith",
"Michele LTD",
"Nadelman, Jr",
"PC Adelman"]

import re
import sklearn
from sklearn import cluster
words = [re.sub(r"(,|\.|ltd|l\.t\.d|inc|estates|enterprises|ent|estatys)","", w.lower()).strip() for w in words]
words = np.asarray(words) #So that indexing with a list will work
lev_similarity = -1*np.array([[distance.nlevenshtein(w1,w2,method = 1) for w1 in words] for w2 in words])
affprop = sklearn.cluster.KMeans(n_clusters=3)
affprop.fit(lev_similarity)
for cluster_id in np.unique(affprop.labels_):
    print(cluster_id)
    cluster = np.unique(words[np.nonzero(affprop.labels_==cluster_id)])
    cluster_str = ", ".join(cluster)
    print(" - *%s:*" % ( cluster_str))

结果:

0
 - *john smith, michele, nadelman jr, pc adelman:*
1
 - *bbazemax, bozeman, bozerman, bras5emax, brazemax:*
2
 - *gramkai, gramkai books, gramkat:*

最后,您可能需要将更改后的名称与原始名称连接起来。


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