Word2Vec代码注解-distance

      暂且把文章类型设为“原创”了，如果Mikolov看到的话，还望勿怪。毕竟代码都是别人写的，自己只是往上面加了一些注释，所以，如有什么问题，还请及时通知我改为更合适的选项！

      好了，话入正题！

      其实，distance的主要功能是计算两个词(组)之间的距离(相似度)。程序将该功能表现为用户输入一条字符串，里面可以包含以空格分割的多个词，然后程序从训练出的模型(demo-word.sh训练出的模型文件名为vectors.bin，二进制文件)中与之最为相似的40个词并输出。以下是添加了注释的源代码，为了与原有代码保持对齐，所有注释均放在代码行后，未另起新行：

//  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 <stdio.h>#include <string.h>#include <math.h>#include <malloc.h>const long long max_size = 2000;         // max length of stringsconst long long N = 40;                  // number of closest words that will be shownconst long long max_w = 50;              // max length of vocabulary entriesint 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];//dist: distance; len: norm of vector; bestd: best distance  long long words, size, a, b, c, d, cn, bi[100];  char ch;  float *M;  char *vocab;  if (argc < 2) {    printf("Usage: ./distance <FILE>\nwhere FILE contains word projections in the BINARY FORMAT\n");    return 0;  }  strcpy(file_name, argv[1]);//copy filename to the variable  f = fopen(file_name, "rb");//open the file in BINARY mode  if (f == NULL) {    printf("Input file not found\n");    return -1;  }  fscanf(f, "%lld", &words);//word count  fscanf(f, "%lld", &size);//dimension count  vocab = (char *)malloc((long long)words * max_w * sizeof(char));//memory alloc, store words, max_w chars per word  for (a = 0; a < N; a++) bestw[a] = (char *)malloc(max_size * sizeof(char));//memory alloc, store closest vocabs  M = (float *)malloc((long long)words * (long long)size * sizeof(float));//memory alloc, store vectors  if (M == NULL) {    printf("Cannot allocate memory: %lld MB    %lld  %lld\n", (long long)words * size * sizeof(float) / 1048576, words, size);//1048576 = 1024 * 1024    return -1;  }  for (b = 0; b < words; b++) {    fscanf(f, "%s%c", &vocab[b * max_w], &ch);    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);//denominator, [ZH_CN]FenMu    for (a = 0; a < size; a++) M[a + b * size] /= len;//unit vector, [ZH_CN]DanWei XiangLiang  }  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) {//multiple words separated by BLANK symbol      st[cn][b] = st1[c];      b++;      c++;      st[cn][b] = 0;//assign '\0' after the words, if new charactor found, this value can be replaced.      if (st1[c] == 0) break;      if (st1[c] == ' ') {        cn++;        b = 0;        c++;      }    }    cn++;    for (a = 0; a < cn; a++) {//check word position in vocab array.      for (b = 0; b < words; b++) if (!strcmp(&vocab[b * max_w], st[a])) break;//found it, b store the position      if (b == words) b = -1;//not found, b = -1      bi[a] = b;//position array, store all words input this time      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;//if not found, continue; otherwise, search for closest words    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;//if not found, pass it      for (a = 0; a < size; a++) vec[a] += M[a + bi[b] * size];//add all vectors of words found this time    }    len = 0;    for (a = 0; a < size; a++) len += vec[a] * vec[a];    len = sqrt(len);//is it needful? Yes, different vector makes different len value    for (a = 0; a < size; a++) vec[a] /= len;//unit vector    for (a = 0; a < N; a++) bestd[a] = -1;//best distances, init with -1    for (a = 0; a < N; a++) bestw[a][0] = 0;//best words, init with '\0'    for (c = 0; c < words; c++) {//traverse all words      a = 0;      for (b = 0; b < cn; b++) if (bi[b] == c) a = 1;      if (a == 1) continue;//pass words in vocab      dist = 0;      for (a = 0; a < size; a++) dist += vec[a] * M[a + c * size];      for (a = 0; a < N; a++) {//sort dist, Insertion Sort, DESC        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]);//display results  }  return 0;}

      从代码中可以看出，Mikolov使用的方法原理上还是很简单的。将输入的所有(在模型中存在的)词的向量进行加和(行107)，然后进行向量的单位化(行112)，最后与vocab中所有在输入序列中出现的word对应的(单位)向量取cosine值。此过程中需要注意的是，Mikolov默认跳过了所有在输入序列中出现的word(行118)。尽管在计算过程中曾经得到过所有词与输入串的dist，但是由于N = 40(行21)，因此只有前40条记录会被保存并进行最后的输出，有需要的话可以自行设定该值。

      另外，由于max_w = 50(行22)，因此对于训练出的结果中存在长词(≥50字节)的情况，还需修改该常量的值。

      附上测试截图：