如果我有一个包含以逗号分隔的数字列表的std :: string,那么解析数字并将它们放在整数数组中的最简单方法是什么?
我不想将其概括为解析其他任何内容。只是一个逗号分隔整数的简单字符串,如“1,1,1,1,2,1,1,1,0”。
答案 0 :(得分:135)
#include <vector>
#include <string>
#include <sstream>
#include <iostream>
int main()
{
std::string str = "1,2,3,4,5,6";
std::vector<int> vect;
std::stringstream ss(str);
int i;
while (ss >> i)
{
vect.push_back(i);
if (ss.peek() == ',')
ss.ignore();
}
for (i=0; i< vect.size(); i++)
std::cout << vect.at(i)<<std::endl;
}
答案 1 :(得分:88)
不那么冗长,std并用逗号分隔任何东西。
stringstream ss( "1,1,1,1, or something else ,1,1,1,0" );
vector<string> result;
while( ss.good() )
{
string substr;
getline( ss, substr, ',' );
result.push_back( substr );
}
答案 2 :(得分:59)
另一种相当不同的方法:使用将逗号视为空格的特殊区域设置:
#include <locale>
#include <vector>
struct csv_reader: std::ctype<char> {
csv_reader(): std::ctype<char>(get_table()) {}
static std::ctype_base::mask const* get_table() {
static std::vector<std::ctype_base::mask> rc(table_size, std::ctype_base::mask());
rc[','] = std::ctype_base::space;
rc['\n'] = std::ctype_base::space;
rc[' '] = std::ctype_base::space;
return &rc[0];
}
};
要使用此功能,您imbue()
使用包含此构面的区域设置的流。完成后,您可以读取数字,就好像逗号根本不存在一样。例如,我们将从输入中读取逗号分隔的数字,然后在标准输出上逐行写出:
#include <algorithm>
#include <iterator>
#include <iostream>
int main() {
std::cin.imbue(std::locale(std::locale(), new csv_reader()));
std::copy(std::istream_iterator<int>(std::cin),
std::istream_iterator<int>(),
std::ostream_iterator<int>(std::cout, "\n"));
return 0;
}
答案 3 :(得分:43)
C++ String Toolkit Library (Strtk)针对您的问题提供了以下解决方案:
#include <string>
#include <deque>
#include <vector>
#include "strtk.hpp"
int main()
{
std::string int_string = "1,2,3,4,5,6,7,8,9,10,11,12,13,14,15";
std::vector<int> int_list;
strtk::parse(int_string,",",int_list);
std::string double_string = "123.456|789.012|345.678|901.234|567.890";
std::deque<double> double_list;
strtk::parse(double_string,"|",double_list);
return 0;
}
可以找到更多示例Here
答案 4 :(得分:17)
使用通用算法和Boost.Tokenizer的替代解决方案:
struct ToInt
{
int operator()(string const &str) { return atoi(str.c_str()); }
};
string values = "1,2,3,4,5,9,8,7,6";
vector<int> ints;
tokenizer<> tok(values);
transform(tok.begin(), tok.end(), back_inserter(ints), ToInt());
答案 5 :(得分:6)
您也可以使用以下功能。
void tokenize(const string& str, vector<string>& tokens, const string& delimiters = ",")
{
// Skip delimiters at beginning.
string::size_type lastPos = str.find_first_not_of(delimiters, 0);
// Find first non-delimiter.
string::size_type pos = str.find_first_of(delimiters, lastPos);
while (string::npos != pos || string::npos != lastPos) {
// Found a token, add it to the vector.
tokens.push_back(str.substr(lastPos, pos - lastPos));
// Skip delimiters.
lastPos = str.find_first_not_of(delimiters, pos);
// Find next non-delimiter.
pos = str.find_first_of(delimiters, lastPos);
}
}
答案 6 :(得分:5)
std::string input="1,1,1,1,2,1,1,1,0";
std::vector<long> output;
for(std::string::size_type p0=0,p1=input.find(',');
p1!=std::string::npos || p0!=std::string::npos;
(p0=(p1==std::string::npos)?p1:++p1),p1=input.find(',',p0) )
output.push_back( strtol(input.c_str()+p0,NULL,0) );
当然,检查strtol()
中的转换错误是个好主意。也许代码也可能从其他一些错误检查中受益。
答案 7 :(得分:4)
这里有很多可怕的答案,所以我会添加我的(包括测试程序):
#include <string>
#include <iostream>
#include <cstddef>
template<typename StringFunction>
void splitString(const std::string &str, char delimiter, StringFunction f) {
std::size_t from = 0;
for (std::size_t i = 0; i < str.size(); ++i) {
if (str[i] == delimiter) {
f(str, from, i);
from = i + 1;
}
}
if (from <= str.size())
f(str, from, str.size());
}
int main(int argc, char* argv[]) {
if (argc != 2)
return 1;
splitString(argv[1], ',', [](const std::string &s, std::size_t from, std::size_t to) {
std::cout << "`" << s.substr(from, to - from) << "`\n";
});
return 0;
}
不错的属性:
std::stringview
然后它不会进行任何分配,而且应该非常快。您可能希望更改一些设计选择:
示例输入和输出:
"" -> {""}
"," -> {"", ""}
"1," -> {"1", ""}
"1" -> {"1"}
" " -> {" "}
"1, 2," -> {"1", " 2", ""}
" ,, " -> {" ", "", " "}
答案 8 :(得分:2)
#include <sstream>
#include <vector>
const char *input = "1,1,1,1,2,1,1,1,0";
int main() {
std::stringstream ss(input);
std::vector<int> output;
int i;
while (ss >> i) {
output.push_back(i);
ss.ignore(1);
}
}
错误的输入(例如连续的分隔符)会搞砸了,但你说的很简单。
答案 9 :(得分:2)
我很惊讶没有人使用std::regex
提出解决方案:
#include <string>
#include <algorithm>
#include <vector>
#include <regex>
void parse_csint( const std::string& str, std::vector<int>& result ) {
typedef std::regex_iterator<std::string::const_iterator> re_iterator;
typedef re_iterator::value_type re_iterated;
std::regex re("(\\d+)");
re_iterator rit( str.begin(), str.end(), re );
re_iterator rend;
std::transform( rit, rend, std::back_inserter(result),
[]( const re_iterated& it ){ return std::stoi(it[1]); } );
}
此函数在输入向量的后面插入所有整数。您可以调整正则表达式以包括负整数或浮点数等。
答案 10 :(得分:1)
string exp = "token1 token2 token3";
char delimiter = ' ';
vector<string> str;
string acc = "";
for(int i = 0; i < exp.size(); i++)
{
if(exp[i] == delimiter)
{
str.push_back(acc);
acc = "";
}
else
acc += exp[i];
}
答案 11 :(得分:1)
我还不能发表评论(在网站上开始),但在他的帖子中添加了更为通用的Jerry Coffin奇妙的ctype派生类。
感谢Jerry的超级想法。
(因为必须经过同行评审,暂时在此处添加)
struct SeparatorReader: std::ctype<char>
{
template<typename T>
SeparatorReader(const T &seps): std::ctype<char>(get_table(seps), true) {}
template<typename T>
std::ctype_base::mask const *get_table(const T &seps) {
auto &&rc = new std::ctype_base::mask[std::ctype<char>::table_size]();
for(auto &&sep: seps)
rc[static_cast<unsigned char>(sep)] = std::ctype_base::space;
return &rc[0];
}
};
答案 12 :(得分:0)
bool GetList (const std::string& src, std::vector<int>& res)
{
using boost::lexical_cast;
using boost::bad_lexical_cast;
bool success = true;
typedef boost::tokenizer<boost::char_separator<char> > tokenizer;
boost::char_separator<char> sepa(",");
tokenizer tokens(src, sepa);
for (tokenizer::iterator tok_iter = tokens.begin();
tok_iter != tokens.end(); ++tok_iter) {
try {
res.push_back(lexical_cast<int>(*tok_iter));
}
catch (bad_lexical_cast &) {
success = false;
}
}
return success;
}
答案 13 :(得分:0)
std::string stringIn = "my,csv,,is 10233478,separated,by commas";
std::vector<std::string> commaSeparated(1);
int commaCounter = 0;
for (int i=0; i<stringIn.size(); i++) {
if (stringIn[i] == ",") {
commaSeparated.push_back("");
commaCounter++;
} else {
commaSeparated.at(commaCounter) += stringIn[i];
}
}
最后你将得到一个字符串向量,句子中的每个元素用空格分隔。空字符串保存为单独的项目。
答案 14 :(得分:0)
简单复制/粘贴功能,基于boost tokenizer。
void strToIntArray(std::string string, int* array, int array_len) {
boost::tokenizer<> tok(string);
int i = 0;
for(boost::tokenizer<>::iterator beg=tok.begin(); beg!=tok.end();++beg){
if(i < array_len)
array[i] = atoi(beg->c_str());
i++;
}
答案 15 :(得分:0)
这是最简单的方法,我经常使用。它适用于任何一个字符的定界符。
#include<bits/stdc++.h>
using namespace std;
int main() {
string str;
cin >> str;
int temp;
vector<int> result;
char ch;
stringstream ss(str);
do
{
ss>>temp;
result.push_back(temp);
}while(ss>>ch);
for(int i=0 ; i < result.size() ; i++)
cout<<result[i]<<endl;
return 0;
}
答案 16 :(得分:-1)
void ExplodeString( const std::string& string, const char separator, std::list<int>& result ) {
if( string.size() ) {
std::string::const_iterator last = string.begin();
for( std::string::const_iterator i=string.begin(); i!=string.end(); ++i ) {
if( *i == separator ) {
const std::string str(last,i);
int id = atoi(str.c_str());
result.push_back(id);
last = i;
++ last;
}
}
if( last != string.end() ) result.push_back( atoi(&*last) );
}
}
答案 17 :(得分:-4)
#include <sstream>
#include <vector>
#include <algorithm>
#include <iterator>
const char *input = ",,29870,1,abc,2,1,1,1,0";
int main()
{
std::stringstream ss(input);
std::vector<int> output;
int i;
while ( !ss.eof() )
{
int c = ss.peek() ;
if ( c < '0' || c > '9' )
{
ss.ignore(1);
continue;
}
if (ss >> i)
{
output.push_back(i);
}
}
std::copy(output.begin(), output.end(), std::ostream_iterator<int> (std::cout, " ") );
return 0;
}