如何迭代由空格分隔的单词组成的字符串中的单词?

注意,我对C字符串函数或那种字符操作/访问不感兴趣。比起效率,我更喜欢优雅。我当前的解决方案:

#include <iostream>
#include <sstream>
#include <string>

using namespace std;

int main() {
    string s = "Somewhere down the road";
    istringstream iss(s);

    do {
        string subs;
        iss >> subs;
        cout << "Substring: " << subs << endl;
    } while (iss);
}

当前回答

我喜欢将boost/regex方法用于此任务,因为它们为指定拆分条件提供了最大的灵活性。

#include <iostream>
#include <string>
#include <boost/regex.hpp>

int main() {
    std::string line("A:::line::to:split");
    const boost::regex re(":+"); // one or more colons

    // -1 means find inverse matches aka split
    boost::sregex_token_iterator tokens(line.begin(),line.end(),re,-1);
    boost::sregex_token_iterator end;

    for (; tokens != end; ++tokens)
        std::cout << *tokens << std::endl;
}

其他回答

#include <iostream>
#include <regex>

using namespace std;

int main() {
   string s = "foo bar  baz";
   regex e("\\s+");
   regex_token_iterator<string::iterator> i(s.begin(), s.end(), e, -1);
   regex_token_iterator<string::iterator> end;
   while (i != end)
      cout << " [" << *i++ << "]";
}

IMO,这是最接近python的re.split()。有关regex_token_iterator的更多信息,请参阅cplusplus.com。-1(regex_token_iterator tor中的第四个参数)是序列中不匹配的部分,使用匹配作为分隔符。

#include <iostream>
#include <string>
#include <deque>

std::deque<std::string> split(
    const std::string& line, 
    std::string::value_type delimiter,
    bool skipEmpty = false
) {
    std::deque<std::string> parts{};

    if (!skipEmpty && !line.empty() && delimiter == line.at(0)) {
        parts.push_back({});
    }

    for (const std::string::value_type& c : line) {
        if (
            (
                c == delimiter 
                &&
                (skipEmpty ? (!parts.empty() && !parts.back().empty()) : true)
            )
            ||
            (c != delimiter && parts.empty())
        ) {
            parts.push_back({});
        }

        if (c != delimiter) {
            parts.back().push_back(c);
        }
    }

    if (skipEmpty && !parts.empty() && parts.back().empty()) {
        parts.pop_back();
    }

    return parts;
}

void test(const std::string& line) {
    std::cout << line << std::endl;

    std::cout << "skipEmpty=0 |";
    for (const std::string& part : split(line, ':')) {
        std::cout << part << '|';
    }
    std::cout << std::endl;

    std::cout << "skipEmpty=1 |";
    for (const std::string& part : split(line, ':', true)) {
        std::cout << part << '|';
    }
    std::cout << std::endl;

    std::cout << std::endl;
}

int main() {
    test("foo:bar:::baz");
    test("");
    test("foo");
    test(":");
    test("::");
    test(":foo");
    test("::foo");
    test(":foo:");
    test(":foo::");

    return 0;
}

输出:

foo:bar:::baz
skipEmpty=0 |foo|bar|||baz|
skipEmpty=1 |foo|bar|baz|


skipEmpty=0 |
skipEmpty=1 |

foo
skipEmpty=0 |foo|
skipEmpty=1 |foo|

:
skipEmpty=0 |||
skipEmpty=1 |

::
skipEmpty=0 ||||
skipEmpty=1 |

:foo
skipEmpty=0 ||foo|
skipEmpty=1 |foo|

::foo
skipEmpty=0 |||foo|
skipEmpty=1 |foo|

:foo:
skipEmpty=0 ||foo||
skipEmpty=1 |foo|

:foo::
skipEmpty=0 ||foo|||
skipEmpty=1 |foo|

我用这个分隔符分隔字符串。第一个将结果放入预先构建的向量中,第二个返回新向量。

#include <string>
#include <sstream>
#include <vector>
#include <iterator>

template <typename Out>
void split(const std::string &s, char delim, Out result) {
    std::istringstream iss(s);
    std::string item;
    while (std::getline(iss, item, delim)) {
        *result++ = item;
    }
}

std::vector<std::string> split(const std::string &s, char delim) {
    std::vector<std::string> elems;
    split(s, delim, std::back_inserter(elems));
    return elems;
}

请注意,此解决方案不会跳过空令牌,因此下面将找到4项,其中一项为空:

std::vector<std::string> x = split("one:two::three", ':');

这类似于堆栈溢出问题:如何在C++中标记字符串?。需要Boost外部库

#include <iostream>
#include <string>
#include <boost/tokenizer.hpp>

using namespace std;
using namespace boost;

int main(int argc, char** argv)
{
    string text = "token  test\tstring";

    char_separator<char> sep(" \t");
    tokenizer<char_separator<char>> tokens(text, sep);
    for (const string& t : tokens)
    {
        cout << t << "." << endl;
    }
}

LazyString拆分器:

#include <string>
#include <algorithm>
#include <unordered_set>

using namespace std;

class LazyStringSplitter
{
    string::const_iterator start, finish;
    unordered_set<char> chop;

public:

    // Empty Constructor
    explicit LazyStringSplitter()
    {}

    explicit LazyStringSplitter (const string cstr, const string delims)
        : start(cstr.begin())
        , finish(cstr.end())
        , chop(delims.begin(), delims.end())
    {}

    void operator () (const string cstr, const string delims)
    {
        chop.insert(delims.begin(), delims.end());
        start = cstr.begin();
        finish = cstr.end();
    }

    bool empty() const { return (start >= finish); }

    string next()
    {
        // return empty string
        // if ran out of characters
        if (empty())
            return string("");

        auto runner = find_if(start, finish, [&](char c) {
            return chop.count(c) == 1;
        });

        // construct next string
        string ret(start, runner);
        start = runner + 1;

        // Never return empty string
        // + tail recursion makes this method efficient
        return !ret.empty() ? ret : next();
    }
};

我将此方法称为LazyStringSplitter是因为一个原因——它不会一次性拆分字符串。本质上,它的行为类似于python生成器它公开了一个名为next的方法,该方法返回从原始字符串拆分的下一个字符串我使用了c++11STL中的无序集,因此查找分隔符的速度要快得多下面是它的工作原理

测试程序

#include <iostream>
using namespace std;

int main()
{
    LazyStringSplitter splitter;

    // split at the characters ' ', '!', '.', ','
    splitter("This, is a string. And here is another string! Let's test and see how well this does.", " !.,");

    while (!splitter.empty())
        cout << splitter.next() << endl;
    return 0;
}

输出,输出

This
is
a
string
And
here
is
another
string
Let's
test
and
see
how
well
this
does

改进这一点的下一个计划是实施开始和结束方法,以便可以执行以下操作:

vector<string> split_string(splitter.begin(), splitter.end());