std::string human_format::comma_format(std::string name) {
// Split on commas. If there are no commas, return.
std::deque < std::string > split_string = split_parts(name, ",");
if(split_string.size() < 2) {
return name;
}
std::string output;
std::string holding;
// Comma formatting
while(split_string.size() > 0) {
unsigned int split_size = (split_string.size() - 1);
if(match_component(split_string[split_size], suffixes)) {
if(output.size() == 0) {
output.append(split_string[split_size]);
} else {
output.append(" " + split_string[split_size]);
}
} else {
if(output.size() == 0) {
}
holding.append(split_string[split_size]);
}
split_string.pop_back();
}
if(holding.size() > 0) {
output = holding + output;
}
return output;
}
static inline rational<T> apply(std::string const& source)
{
T before, after;
bool negate;
std::string::size_type len;
// Note: decimal comma is not (yet) supported, it does (and should) not
// occur in a WKT, where points are comma separated.
std::string::size_type p = source.find(".");
if (p == std::string::npos)
{
p = source.find("/");
if (p == std::string::npos)
{
return rational<T>(atol(source.c_str()));
}
split_parts(source, p, before, after, negate, len);
return negate
? -rational<T>(before, after)
: rational<T>(before, after)
;
}
split_parts(source, p, before, after, negate, len);
T den = 1;
for (std::string::size_type i = 0; i < len; i++)
{
den *= 10;
}
return negate
? -rational<T>(before) - rational<T>(after, den)
: rational<T>(before) + rational<T>(after, den)
;
}
std::vector < std::string > human_parse::parse_single(std::string name){
// If it's literally just an empty string, throw things.
if(name.size() == 0){
throw std::range_error("You have provided an empty string");
}
// Split and create output object.
std::deque < std::string > split_name = split_parts(name, " ");
std::vector < std::string > output(5);
// If there's only one element we assume it is a first name and return it.
if(split_name.size() == 1){
output[1] = split_name[0];
return output;
}
// If there are > 1 element and we find a salutation in the first, pop those two elements.
if(split_name.size() > 1 && match_component(split_name[0], salutations)){
output[0] = split_name[0];
split_name.pop_front();
output[1] = split_name[0];
split_name.pop_front();
} else {
output[1] = split_name[0];
split_name.pop_front();
}
// If there's still > 1 elemenent, what about suffixes?
if(split_name.size() > 1){
while(split_name.size() > 1 && match_component(split_name[split_name.size() - 1], suffixes)){
if(output[4] == ""){
output[4] = split_name[split_name.size() - 1];
} else {
output[4] = split_name[split_name.size() - 1] + " " + output[4];
}
split_name.pop_back();
}
}
// If there's only >1 element, surnames
if(split_name.size() > 0){
output[3] = split_name[split_name.size() - 1];
split_name.pop_back();
} else {
return output;
}
// If there is still 1 or more elements we test for compounds
while(split_name.size() > 0 && match_component(split_name[split_name.size() - 1], compounds)){
output[3] = split_name[split_name.size() - 1] + " " + output[3];
split_name.pop_back();
}
// If we still have elements, those are middle names.
if(split_name.size() > 0){
output[2].append(split_name[0]);
for(unsigned int i = 1; i < split_name.size(); i++){
output[2].append(" " + split_name[i]);
}
}
return output;
}
