Question

我正在尝试读取文本文件，并按字符串将其输入到矢量字符串中。我需要它停在每个句子的结尾，然后在句子中挑出关键词。我理解如何找到关键词，但不知道如何让它最后停止输入字符串。我正在使用while循环来检查每一行，我正在考虑使用一系列if语句，例如

if(std::vector<string>::iterator i == ".") i == "\0"

到目前为止我执行向量填充的代码是：

std::string c;
ifstream infile;
infile.open("example.txt");
while(infile >> c){
    a.push_back(c);
}

好的，所以我已经开始考虑将文本文件的每个单词加载到标记中的方法，考虑到“”作为分隔符，并且有一个特殊的单词列表：

    const int MAX_PER_LINE = 512;
    const int MAX_TOK = 20;
    const char* const DELIMETER = " -";
    const char* const SPECIAL ="!?.";
    const char* const ignore[]  = {"Mr.", "Ms.","Mrs.","sr.", "Ave.", "Rd."};

然后

             if(!file.good()){
         return 1;
     }
     //parsing algorithm paraphrased from cs.dvc.edu/HowTo_Parse.html
     while(!file.eof()){
     char line[MAX_PER_LINE];

     file.getline(line, MAX_PER_LINE);
     int n = 0;
     const char* token[MAX_TOK] = {};
     token[0] = strtok(line, DELIMETER);
     if(token[0]){
         for(n = 1; n < MAX_TOK; ++n){
             token[n] = strtok(0, DELIMETER);
             if(!token[n]) break;
         }
     }
     //for(int i = 0; i < n; ++i){
     for(int i = 0; i < n; ++i){
         cout << "Token[" << i << "] =" << token[i] << endl;
         cout << endl; 
     }
     }

现在我正在寻找一个if语句中的内容，以便它检查每个标记的特殊情况，或者如果它们遵循具有特殊情况的标记，则将它们加载到新的集合标记中。我知道psuedo代码大部分，但我不知道要用什么样的语法，如果（token [i]包含特殊情况或令牌[i]之前没有任何东西（对于第一个令牌）或大写，并使用特殊情况下的令牌将其加载到新令牌中。

任何帮助将不胜感激。

Answer 1

编写自己的句子分隔符对于没有国际化的小项目或项目是可以的。对于文本边界的基于高级文本的解决方案，我建议ICU BreakIterator。基于unicode.org标准化，它们提供字符，单词，换行符和句子边界。他们有C ++的开源库（我认为也是Java）。请参阅this page，它有链接到库的下载页面。

这样可以避免重新发明轮子并避免以后出现潜在问题。大多数领先的出版软件产品如QuarkXPress等都使用这个库。

编辑：我试图在句子边界上找到ICU的BreakIterator用法的快速教程，但我找到了单词边界的例子 - （句子边界计算将非常相似，可能需要在下面用createWordInstance替换createSentenceInstance ）

void listWordBoundaries(const UnicodeString& s) {
    UErrorCode status = U_ZERO_ERROR;
    BreakIterator* bi = BreakIterator::createWordInstance(Locale::getUS(), status);


    bi->setText(s);
    int32_t p = bi->first();
    while (p != BreakIterator::DONE) {
        printf("Boundary at position %d\n", p);
        p = bi->next();
    }
    delete bi;
}

Answer 2

查找以句点结尾的单词非常简单，只需检查word.back() == '.'。您还需要首先检查word.empty()，因为如果字符串为空，back()是未定义的行为。如果您的编译器不支持C ++ 11，那么您也可以使用word[word.size() - 1] == '.'来完成更长的工作。

这是一个基本的例子，它使用任何以“。”结尾的单词天真地分割句子：

#include <iostream>
#include <string>
#include <vector>

int main(int argc, char** argv) {
    if (argc == 1) {
        std::cerr << "Usage: " << argv[0] << " [text to split]\n"
            << "Splits the input text into one sentence per line." << std::endl;
        return 1;
    }

    std::vector<std::string> sentences;
    std::string current_sentence;
    for (int i = 1; i < argc; ++i) {
        std::string word(argv[i]);
        current_sentence.append(word);
        current_sentence.push_back(' ');
        /* use word.back() == '.' for C++11 */
        if (!word.empty() && word[word.size() - 1] == '.') {
            sentences.push_back(current_sentence);
            current_sentence.clear();
        }
    }
    if (!current_sentence.empty()) {
        sentences.push_back(current_sentence);
    }

    for (size_t i = 0; i < sentences.size(); ++i) {
        std::cout << sentences[i] << std::endl;
    }
    return 0;
}

运行如：

$ g++ test.cpp
$ ./a.out This is a test. And a second sentence. So we meet again Mr. Bond.
This is a test. 
And a second sentence. 
So we meet again Mr. 
Bond.

注意它是怎么想的“先生”是句子的结尾。

我不确定一种聪明的处理方式，但是一个（脆弱的）选项是制作一个不是句子结尾的单词列表，然后检查单词是否在列表中，如这样：

#include <algorithm>
#include <iostream>
#include <set>
#include <string>
#include <vector>

const std::string tmp[] = {
    "dr.",
    "mr.",
    "mrs.",
    "ms.",
    "rd.",
    "st."
};
const std::set<std::string> ABBREVIATIONS(tmp, tmp + sizeof(tmp) / sizeof(tmp[0]));

bool has_period(const std::string& word) {
    return !word.empty() && word[word.size() - 1] == '.';
}

bool is_abbreviation(std::string word) {
    /* Convert to lowercase, so we don't need to check every possible
     * variation of each word. Remove this (and update the set initialization)
     * if you don't care about handling poor grammar. */
    std::transform(word.begin(), word.end(), word.begin(), ::tolower);

    /* Check if the word is an abbreviation. */
    return ABBREVIATIONS.find(word) != ABBREVIATIONS.end();
}

int main(int argc, char** argv) {
    if (argc == 1) {
        std::cerr << "Usage: " << argv[0] << " [text to split]\n"
            << "Splits the input text into one sentence per line." << std::endl;
        return 1;
    }

    std::vector<std::string> sentences;
    std::string current_sentence;
    for (int i = 1; i < argc; ++i) {
        std::string word(argv[i]);
        current_sentence.append(word);
        current_sentence.push_back(' ');
        if (has_period(word) && !is_abbreviation(word)) {
            sentences.push_back(current_sentence);
            current_sentence.clear();
        }
    }
    if (!current_sentence.empty()) {
        sentences.push_back(current_sentence);
    }

    for (size_t i = 0; i < sentences.size(); ++i) {
        std::cout << sentences[i] << std::endl;
    }
    return 0;
}

在C ++ 11中，您可以使用unordered_set提高效率，使用std::string::back更简单，更简单的初始化（std::set<std::string> PERIOD_WORDS = { "dr.", "mr.", "mrs." /*etc.*/ }）。

运行此版本：

$ g++ test.cpp
$ ./a.out This is a test. And a second sentence. So we meet again Mr. Bond.
This is a test. 
And a second sentence. 
So we meet again Mr. Bond.

但是，当然，它仍然没有发现我们没有明确编程的任何情况：

$ ./a.out Example Ave. is just north of here.
Example Ave. 
is just north of here.

即使我们添加了这些内容，也很难检测到“我居住在例如Ave.”这样的句子，其中句子以缩写结尾。我希望这是一个有用的开始。

编辑：我刚刚阅读了评论中链接的sentence breaking Wikipedia article，并且合并规则相对容易：

（c）如果下一个标记大写，则结束一个句子。

类似的东西：

#include <algorithm>
#include <iostream>
#include <set>
#include <string>
#include <vector>

const std::string tmp[] = {
    "ave.",
    "dr.",
    "mr.",
    "mrs.",
    "ms.",
    "rd.",
    "st."
};
const std::set<std::string> PERIOD_WORDS(tmp, tmp + sizeof(tmp) / sizeof(tmp[0]));

bool has_period(const std::string& word) {
    return !word.empty() && word[word.size() - 1] == '.';
}

bool is_abbreviation(std::string word) {
    /* Convert to lowercase, so we don't need to check every possible
     * variation of each word. Remove this (and update the set initialization)
     * if you don't care about handling poor grammar. */
    std::transform(word.begin(), word.end(), word.begin(), ::tolower);

    /* Check if the word is a word that ends with a period. */
    return PERIOD_WORDS.find(word) != PERIOD_WORDS.end();
}

bool is_capitalized(const std::string& word) {
    return !word.empty() && std::isupper(word[0]);
}

int main(int argc, char** argv) {
    if (argc == 1) {
        std::cerr << "Usage: " << argv[0] << " [text to split]\n"
            << "Splits the input text into one sentence per line." << std::endl;
        return 1;
    }

    std::vector<std::string> sentences;
    std::string current_sentence;
    for (int i = 1; i < argc; ++i) {
        std::string word(argv[i]);
        std::string next_word(i + 1 < argc ? argv[i + 1] : "");
        current_sentence.append(word);
        current_sentence.push_back(' ');
        if (next_word.empty()
            || has_period(word)
            && (!is_abbreviation(word) || is_capitalized(next_word))) {
            sentences.push_back(current_sentence);
            current_sentence.clear();
        }
    }

    for (size_t i = 0; i < sentences.size(); ++i) {
        std::cout << sentences[i] << std::endl;
    }
    return 0;
}

然后甚至像这样的工作：

$ ./a.out Example Ave. is just north of here. I live on Example Ave. Test test test.
Example Ave. is just north of here. 
I live on Example Ave. 
Test test test.

但它仍然无法处理某些情况：

$ ./a.out Mr. Adams lives on Example Ave. Example Ave. is just north of here. I live on Example Ave. Test test test.
Mr. 
Adams lives on Example Ave. 
Example Ave. is just north of here. 
I live on Example Ave. 
Test test test.

认识到句子的结束

2 个答案: