word2vec中文类似词计算和聚类的使用说明及c语言源代码

word2vec相关基础知识、下载安装參考前文：word2vec词向量中文文本相似度计算
文件夹：

• word2vec使用说明及源代码介绍
  
  • 1.下载地址
  • 2.中文语料
  • 3.參数介绍
  • 4.计算相似词语
  • 5.三个词预測语义语法关系
  • 6.关键词聚类

1、下载地址

官网C语言下载地址：http://word2vec.googlecode.com/svn/trunk/

执行 make 编译word2vec工具：
Makefile的编译代码在makefile.txt文件里，先改名makefile.txt 为Makefile，然后在当前文件夹下运行make进行编译，生成可运行文件(编译过程中报出非常出Warning，gcc不支持pthread多线程命令。凝视就可以)。
再执行演示样例脚本：./demo-word.sh 和 ./demo-phrases.sh：
a). 从http://mattmahoney.net/dc/text8.zip 在线下载了一个文件text8 ( 一个解压后不到100M的txt文件，可自己下载并解压放到同级文件夹下)。可替换为自己的语料
b). 运行word2vec生成词向量到 vectors.bin文件里
c). 假设运行 sh demo-word.sh 训练生成vectors.bin文件后，下次能够直接调用已经训练好的词向量，如命令 ./distance vectors.bin

2、中文语料

语料是我使用Selenium爬取的三大百科（百度、互动、维基）文本信息。当中每一个百科有100个国家。总共300个国家（0001.txt~0300.txt），然后使用Jieba工具进行中文分词处理。

最后输出Result_Country.txt文件。它把全部文本合并。共300行，每行相应一个国家的分词文本信息。

3、參数介绍

下图參数源自文章：Windows下使用Word2vec继续词向量训练 - 一仅仅鸟的天空
Java推荐參考文章：word2vec使用指导

demo-word.sh文件，參考：http://jacoxu.com/?p=1084

make  
#if [ ! -e text8 ]; then  
#  wget http://mattmahoney.net/dc/text8.zip -O text8.gz  
#  gzip -d text8.gz -f  
#fi  
time ./word2vec -train Result_Country.txt -output vectors.bin -cbow 1 -size 200 -window 8 -negative 25 -hs 0 -sample 1e-4 -threads 20 -binary 1 -iter 15  
./distance vectors.bin  

详细命令解释例如以下：
-train Result_Country.txt 表示的是输入文件是Result_Country.txt
-output vectors.bin 输出文件是vectors.bin
-cbow 0 表示不使用cbow模型，默觉得Skip-Gram模型
-size 200 每一个单词的向量维度是200
-window 8 训练的窗体大小为8，就是考虑一个词前八个和后八个词语（实际代码中另一个随机选窗体的过程，窗体大小小于等于5）
-negative 0 表示是否使用NEG方，0表示不使用
-hs 1 是否使用HS方法，0表示不使用，1表示使用HS方法
-sample 指的是採样的阈值，假设一个词语在训练样本中出现的频率越大。那么就越会被採样
-binary 1 为1指的是结果二进制存储，为0是普通存储（普通存储的时候是能够打开看到词语和相应的向量的）
除了以上命令中的參数，word2vec还有几个參数对我们比較实用比方：
-alpha 设置学习速率。默认的为0.025
–min-count 设置最低频率，默认是5。假设一个词语在文档中出现的次数小于5。那么就会丢弃
-classes 设置聚类个数，看了一下源代码用的是k-means聚类的方法
要注意-threads 20 线程数也会对结果产生影响。

4、计算相似词语

命令：sh demo-word.sh
demo-word.sh 中指令：

make
#if [ ! -e text8 ]; then
#  wget http://mattmahoney.net/dc/text8.zip -O text8.gz
#  gzip -d text8.gz -f
#fi
time ./word2vec -train Result_Country.txt -output vectors.bin -cbow 1 -size 200 -window 8 -negative 25 -hs 0 -sample 1e-4 -threads 20 -binary 1 -iter 15
./distance vectors.bin

执行结果例如以下图所看到的：

假设想要不训练调用上次训练的vectors.bin文件。则输入 ./distance vectors.bin
输入"阿富汗"输出相似次及相似距离，如"喀布尔"阿富汗首都，"坎大哈"阿富汗城市，类似中东国家"伊拉克"等。

输入"国歌"输出相似词例如以下图所看到的：

不只名词能够获取相似词，动词也能够。如输入"位于"，输出例如以下：

distance.c 源代码：

//  Copyright 2013 Google Inc. All Rights Reserved.
//
//  Licensed under the Apache License, Version 2.0 (the "License");
//  you may not use this file except in compliance with the License.
//  You may obtain a copy of the License at
//
//      http://www.apache.org/licenses/LICENSE-2.0
//
//  Unless required by applicable law or agreed to in writing, software
//  distributed under the License is distributed on an "AS IS" BASIS,
//  WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
//  See the License for the specific language governing permissions and
//  limitations under the License.

#include 
#include 
#include 
#include 

const long long max_size = 2000;         // max length of strings
const long long N = 40;                  // number of closest words that will be shown
const long long max_w = 50;              // max length of vocabulary entries

int main(int argc, char **argv) {
  FILE *f;
  char st1[max_size];
  char *bestw[N];
  char file_name[max_size], st[100][max_size];
  float dist, len, bestd[N], vec[max_size];
  long long words, size, a, b, c, d, cn, bi[100];
  char ch;
  float *M;
  char *vocab;
  if (argc < 2) {
    printf("Usage: ./distance \nwhere FILE contains word projections in the BINARY FORMAT\n");
    return 0;
  }
  strcpy(file_name, argv[1]);
  f = fopen(file_name, "rb");
  if (f == NULL) {
    printf("Input file not found\n");
    return -1;
  }
  fscanf(f, "%lld", &words);
  fscanf(f, "%lld", &size);
  vocab = (char *)malloc((long long)words * max_w * sizeof(char));
  for (a = 0; a < N; a++) bestw[a] = (char *)malloc(max_size * sizeof(char));
  M = (float *)malloc((long long)words * (long long)size * sizeof(float));
  if (M == NULL) {
    printf("Cannot allocate memory: %lld MB    %lld  %lld\n", (long long)words * size * sizeof(float) / 1048576, words, size);
    return -1;
  }
  for (b = 0; b < words; b++) {
    a = 0;
    while (1) {
      vocab[b * max_w + a] = fgetc(f);
      if (feof(f) || (vocab[b * max_w + a] == ' ')) break;
      if ((a < max_w) && (vocab[b * max_w + a] != '\n')) a++;
    }
    vocab[b * max_w + a] = 0;
    for (a = 0; a < size; a++) fread(&M[a + b * size], sizeof(float), 1, f);
    len = 0;
    for (a = 0; a < size; a++) len += M[a + b * size] * M[a + b * size];
    len = sqrt(len);
    for (a = 0; a < size; a++) M[a + b * size] /= len;
  }
  fclose(f);
  while (1) {
    for (a = 0; a < N; a++) bestd[a] = 0;
    for (a = 0; a < N; a++) bestw[a][0] = 0;
    printf("Enter word or sentence (EXIT to break): ");
    a = 0;
    while (1) {
      st1[a] = fgetc(stdin);
      if ((st1[a] == '\n') || (a >= max_size - 1)) {
        st1[a] = 0;
        break;
      }
      a++;
    }
    if (!strcmp(st1, "EXIT")) break;
    cn = 0;
    b = 0;
    c = 0;
    while (1) {
      st[cn][b] = st1[c];
      b++;
      c++;
      st[cn][b] = 0;
      if (st1[c] == 0) break;
      if (st1[c] == ' ') {
        cn++;
        b = 0;
        c++;
      }
    }
    cn++;
    for (a = 0; a < cn; a++) {
      for (b = 0; b < words; b++) if (!strcmp(&vocab[b * max_w], st[a])) break;
      if (b == words) b = -1;
      bi[a] = b;
      printf("\nWord: %s  Position in vocabulary: %lld\n", st[a], bi[a]);
      if (b == -1) {
        printf("Out of dictionary word!\n");
        break;
      }
    }
    if (b == -1) continue;
    printf("\n                                              Word       Cosine distance\n------------------------------------------------------------------------\n");
    for (a = 0; a < size; a++) vec[a] = 0;
    for (b = 0; b < cn; b++) {
      if (bi[b] == -1) continue;
      for (a = 0; a < size; a++) vec[a] += M[a + bi[b] * size];
    }
    len = 0;
    for (a = 0; a < size; a++) len += vec[a] * vec[a];
    len = sqrt(len);
    for (a = 0; a < size; a++) vec[a] /= len;
    for (a = 0; a < N; a++) bestd[a] = -1;
    for (a = 0; a < N; a++) bestw[a][0] = 0;
    for (c = 0; c < words; c++) {
      a = 0;
      for (b = 0; b < cn; b++) if (bi[b] == c) a = 1;
      if (a == 1) continue;
      dist = 0;
      for (a = 0; a < size; a++) dist += vec[a] * M[a + c * size];
      for (a = 0; a < N; a++) {
        if (dist > bestd[a]) {
          for (d = N - 1; d > a; d--) {
            bestd[d] = bestd[d - 1];
            strcpy(bestw[d], bestw[d - 1]);
          }
          bestd[a] = dist;
          strcpy(bestw[a], &vocab[c * max_w]);
          break;
        }
      }
    }
    for (a = 0; a < N; a++) printf("%50s\t\t%f\n", bestw[a], bestd[a]);
  }
  return 0;
}

5、三个词预測语义语法关系

命令：sh demo-analogy.sh
demo-analogy.sh 中指令：

make
#if [ ! -e text8 ]; then
#  wget http://mattmahoney.net/dc/text8.zip -O text8.gz
#  gzip -d text8.gz -f
#fi
echo -------------------------------------------------------------------------------------
echo Note that for the word analogy to perform well, the model should be trained on much larger data set
echo Example input: paris france berlin
echo -------------------------------------------------------------------------------------
time ./word2vec -train Result_Country.txt -output vectors.bin -cbow 1 -size 200 -window 8 -negative 25 -hs 0 -sample 1e-4 -threads 20 -binary 1 -iter 15
./word-analogy vectors.bin

执行结果例如以下图所看到的：

输入"韩国、首尔、日本"能够预測其首都"东京"：
韩国的首都是首尔  <==>  日本的首都是东京

输入"中国 亚洲 德国"能够预測语义语法关系"欧洲"：
中国位于亚洲 <==> 德国位于欧洲

假设输入只2个词体会提示错误。同一时候输入"EXIT"可推出继续输入。

word-analogy.c 源代码：

//  Copyright 2013 Google Inc. All Rights Reserved.
//
//  Licensed under the Apache License, Version 2.0 (the "License");
//  you may not use this file except in compliance with the License.
//  You may obtain a copy of the License at
//
//      http://www.apache.org/licenses/LICENSE-2.0
//
//  Unless required by applicable law or agreed to in writing, software
//  distributed under the License is distributed on an "AS IS" BASIS,
//  WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
//  See the License for the specific language governing permissions and
//  limitations under the License.

#include 
#include 
#include 
#include 

const long long max_size = 2000;         // max length of strings
const long long N = 40;                  // number of closest words that will be shown
const long long max_w = 50;              // max length of vocabulary entries

int main(int argc, char **argv) {
  FILE *f;
  char st1[max_size];
  char bestw[N][max_size];
  char file_name[max_size], st[100][max_size];
  float dist, len, bestd[N], vec[max_size];
  long long words, size, a, b, c, d, cn, bi[100];
  char ch;
  float *M;
  char *vocab;
  if (argc < 2) {
    printf("Usage: ./word-analogy \nwhere FILE contains word projections in the BINARY FORMAT\n");
    return 0;
  }
  strcpy(file_name, argv[1]);
  f = fopen(file_name, "rb");
  if (f == NULL) {
    printf("Input file not found\n");
    return -1;
  }
  fscanf(f, "%lld", &words);
  fscanf(f, "%lld", &size);
  vocab = (char *)malloc((long long)words * max_w * sizeof(char));
  M = (float *)malloc((long long)words * (long long)size * sizeof(float));
  if (M == NULL) {
    printf("Cannot allocate memory: %lld MB    %lld  %lld\n", (long long)words * size * sizeof(float) / 1048576, words, size);
    return -1;
  }
  for (b = 0; b < words; b++) {
    a = 0;
    while (1) {
      vocab[b * max_w + a] = fgetc(f);
      if (feof(f) || (vocab[b * max_w + a] == ' ')) break;
      if ((a < max_w) && (vocab[b * max_w + a] != '\n')) a++;
    }
    vocab[b * max_w + a] = 0;
    for (a = 0; a < size; a++) fread(&M[a + b * size], sizeof(float), 1, f);
    len = 0;
    for (a = 0; a < size; a++) len += M[a + b * size] * M[a + b * size];
    len = sqrt(len);
    for (a = 0; a < size; a++) M[a + b * size] /= len;
  }
  fclose(f);
  while (1) {
    for (a = 0; a < N; a++) bestd[a] = 0;
    for (a = 0; a < N; a++) bestw[a][0] = 0;
    printf("Enter three words (EXIT to break): ");
    a = 0;
    while (1) {
      st1[a] = fgetc(stdin);
      if ((st1[a] == '\n') || (a >= max_size - 1)) {
        st1[a] = 0;
        break;
      }
      a++;
    }
    if (!strcmp(st1, "EXIT")) break;
    cn = 0;
    b = 0;
    c = 0;
    while (1) {
      st[cn][b] = st1[c];
      b++;
      c++;
      st[cn][b] = 0;
      if (st1[c] == 0) break;
      if (st1[c] == ' ') {
        cn++;
        b = 0;
        c++;
      }
    }
    cn++;
    if (cn < 3) {
      printf("Only %lld words were entered.. three words are needed at the input to perform the calculation\n", cn);
      continue;
    }
    for (a = 0; a < cn; a++) {
      for (b = 0; b < words; b++) if (!strcmp(&vocab[b * max_w], st[a])) break;
      if (b == words) b = 0;
      bi[a] = b;
      printf("\nWord: %s  Position in vocabulary: %lld\n", st[a], bi[a]);
      if (b == 0) {
        printf("Out of dictionary word!\n");
        break;
      }
    }
    if (b == 0) continue;
    printf("\n                                              Word              Distance\n------------------------------------------------------------------------\n");
    for (a = 0; a < size; a++) vec[a] = M[a + bi[1] * size] - M[a + bi[0] * size] + M[a + bi[2] * size];
    len = 0;
    for (a = 0; a < size; a++) len += vec[a] * vec[a];
    len = sqrt(len);
    for (a = 0; a < size; a++) vec[a] /= len;
    for (a = 0; a < N; a++) bestd[a] = 0;
    for (a = 0; a < N; a++) bestw[a][0] = 0;
    for (c = 0; c < words; c++) {
      if (c == bi[0]) continue;
      if (c == bi[1]) continue;
      if (c == bi[2]) continue;
      a = 0;
      for (b = 0; b < cn; b++) if (bi[b] == c) a = 1;
      if (a == 1) continue;
      dist = 0;
      for (a = 0; a < size; a++) dist += vec[a] * M[a + c * size];
      for (a = 0; a < N; a++) {
        if (dist > bestd[a]) {
          for (d = N - 1; d > a; d--) {
            bestd[d] = bestd[d - 1];
            strcpy(bestw[d], bestw[d - 1]);
          }
          bestd[a] = dist;
          strcpy(bestw[a], &vocab[c * max_w]);
          break;
        }
      }
    }
    for (a = 0; a < N; a++) printf("%50s\t\t%f\n", bestw[a], bestd[a]);
  }
  return 0;
}

6、关键词聚类

命令：sh demo-classes.sh
demo-classes.sh 中指令：

make
#if [ ! -e text8 ]; then
#  wget http://mattmahoney.net/dc/text8.zip -O text8.gz
#  gzip -d text8.gz -f
#fi
time ./word2vec -train Result_Country.txt -output classes.txt -cbow 1 -size 200 -window 8 -negative 25 -hs 0 -sample 1e-4 -threads 20 -iter 15 -classes 100
sort classes.txt -k 2 -n > classes.sorted.txt
echo The word classes were saved to file classes.sorted.txt

执行结果例如以下图所看到的：

当中生词文件classes.txt和排序后的文件classes.sorted.txt：
聚类算法是Kmeans，类簇设置为100类。相应0~99，每类的关键词例如以下。可是怎样计算300行数据每行相应的类标。还不太清楚~

当中聚类代码见 word2vec.c 文件 void TrainModel() 函数：

demo-phrases.sh（word2phrase.c） 是就是将词语拼成短语。
希望文章对你有所帮助，尤其是正在学习word2vec基础文章的。

推荐文章：文本深度表示模型Word2Vec - 小唯THU大神
                 利用中文数据跑Google开源项目word2vec - hebin大神
                 Word2vec在事件挖掘中的调研 - 热点事件推荐 (思路不错)

(By:Eastmount 2016-02-20 深夜2点     http://blog.csdn.net/eastmount/   )