Split a string into words by multiple delimiters

Question

I have some text (meaningful text or arithmetical expression) and I want to split it into words.
If I had a single delimiter, I'd use:

std::stringstream stringStream(inputString);
std::string word;
while(std::getline(stringStream, word, delimiter)) 
{
    wordVector.push_back(word);
}

How can I break the string into tokens with several delimiters?

Answer 1

Assuming one of the delimiters is newline, the following reads the line and further splits it by the delimiters. For this example I've chosen the delimiters space, apostrophe, and semi-colon.

std::stringstream stringStream(inputString);
std::string line;
while(std::getline(stringStream, line)) 
{
    std::size_t prev = 0, pos;
    while ((pos = line.find_first_of(" ';", prev)) != std::string::npos)
    {
        if (pos > prev)
            wordVector.push_back(line.substr(prev, pos-prev));
        prev = pos+1;
    }
    if (prev < line.length())
        wordVector.push_back(line.substr(prev, std::string::npos));
}

Answer 2

If you have boost, you could use:

#include <boost/algorithm/string.hpp>
std::string inputString("One!Two,Three:Four");
std::string delimiters("|,:");
std::vector<std::string> parts;
boost::split(parts, inputString, boost::is_any_of(delimiters));

Answer 3

I don't know why nobody pointed out the manual way, but here it is:

const std::string delims(";,:. \n\t");
inline bool isDelim(char c) {
    for (int i = 0; i < delims.size(); ++i)
        if (delims[i] == c)
            return true;
    return false;
}

and in function:

std::stringstream stringStream(inputString);
std::string word; char c;

while (stringStream) {
    word.clear();

    // Read word
    while (!isDelim((c = stringStream.get()))) 
        word.push_back(c);
    if (c != EOF)
        stringStream.unget();

    wordVector.push_back(word);

    // Read delims
    while (isDelim((c = stringStream.get())));
    if (c != EOF)
        stringStream.unget();
}

This way you can do something useful with the delims if you want.

Answer 4

If you interesting in how to do it yourself and not using boost.

Assuming the delimiter string may be very long - let say M, checking for every char in your string if it is a delimiter, would cost O(M) each, so doing so in a loop for all chars in your original string, let say in length N, is O(M*N).

I would use a dictionary (like a map - "delimiter" to "booleans" - but here I would use a simple boolean array that has true in index = ascii value for each delimiter).

Now iterating on the string and check if the char is a delimiter is O(1), which eventually gives us O(N) overall.

Here is my sample code:

const int dictSize = 256;    

vector<string> tokenizeMyString(const string &s, const string &del)
{
    static bool dict[dictSize] = { false};

    vector<string> res;
    for (int i = 0; i < del.size(); ++i) {      
        dict[del[i]] = true;
    }

    string token("");
    for (auto &i : s) {
        if (dict[i]) {
            if (!token.empty()) {
                res.push_back(token);
                token.clear();
            }           
        }
        else {
            token += i;
        }
    }
    if (!token.empty()) {
        res.push_back(token);
    }
    return res;
}


int main()
{
    string delString = "MyDog:Odie, MyCat:Garfield  MyNumber:1001001";
//the delimiters are " " (space) and "," (comma) 
    vector<string> res = tokenizeMyString(delString, " ,");

    for (auto &i : res) {

        cout << "token: " << i << endl;
    }
return 0;
}

Note: tokenizeMyString returns vector by value and create it on the stack first, so we're using here the power of the compiler >>> RVO - return value optimization :)