標籤:word2vec 機器學習
#include <stdio.h>#include <string.h>#include <math.h>//#include <malloc.h>#include <stdlib.h>const long long max_size = 2000; // max length of stringsconst long long N = 5; // 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];//指標數組,大小為N,其中的每一個元素指向char型的指標。 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 <FILE>\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);//vocab_size fscanf(f, "%lld", &size); //維數 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;//將座標歸一化 } 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++; } printf("st1:%s words:%lld \n",st1,words); if (!strcmp(st1, "EXIT")) break; cn = 0; b = 0; c = 0; while (1) {//把st中的每個單詞分別儲存到st1二維數組中,共cn個。 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; //只要有一個詞不在詞彙表都終止for迴圈 } } 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++) {//遍曆每個詞,如果輸入多個詞vec[a]是各個詞向量的累加和 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;//將vec歸一化,當只輸入一個詞時,不起作用。 for (a = 0; a < N; a++) bestd[a] = -1; for (a = 0; a < N; a++) bestw[a][0] = 0; //由於查詢詞和詞彙表的詞向量都做了歸一化,所以餘弦相似性等價於向量的內積,內積越大越相似 for (c = 0; c < words; c++) {//遍曆詞彙表 a = 0; for (b = 0; b < cn; b++) //a的作用:如果遍曆詞和查詢詞相同,則跳過此詞 if (bi[b] == c) a = 1; if (a == 1) continue; dist = 0; for (a = 0; a < size; a++) //求向量的內積 dist += vec[a] * M[a + c * size]; for (a = 0; a < N; a++) { //為dist尋找插入位置 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;}
word2vec中 distence.c 檔案源碼分析