标签：set 打开 har 知识 void googl stdio.h pos zip
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。
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：
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：
// 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 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 <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);
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;
}
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：
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 <stdio.h>
#include <string.h>
#include <math.h>
#include <malloc.h>
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 <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);
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;
}
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：
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标签：set 打开 har 知识 void googl stdio.h pos zip
原文地址：http://www.cnblogs.com/jhcelue/p/7353425.html
