std::map - erase last element

问题

My map is defined as such: map<string, LocationStruct> myLocations; where the key is a time string

I am only keeping 40 items in this map, and would like to drop off the last item in the map when i reach 40 items. I know that i can't do myLocations.erase(myLocations.end()), so how do i go about this?

I do intend for the last item in the map to be the oldest, and therefore FIFO. The data will be coming in rather quick (about 20Hz), so i'm hoping that the map can keep up with it. I do need to look up the data based on time, so i really do need it to be the key, but i am open to alternate methods of accomplishing this.

The format of the string is a very verbose "Thursday June 21 18:44:21:281", though i can pare that down to be the seconds since epoch for simplicity. It was my first go at it, and didn't think too much about the format yet.

回答1:

I assume when you say "erase last element", you mean "erase oldest element".

I wouldn't use a string for times, use a date/time type instead (like unix timestamp). Then they'll be sorted by time, instead of lexicographically, and you can myLocations.erase(myLocations.begin()), since the oldest would always be at the beginning.

Even better, use a boost::circular_buffer<std::pair<timetype, LocationStruct>>, and use std::lower_bound to find elements by time. This will automatically remove the oldest for you, and has the same logorithmic complexity on finding an element by time. It's also faster when adding data. It's pretty much win all around for your situation. If you really want to avoid boost, then a std::deque fits your needs best, and gives great performance, but if you already have a working map, then staying with a std::map is probably best.

Here's how to do the find in a deque:

typedef ???? timetype;
typedef std::pair<Timetype, LocationStruct> TimeLocPair
typedef std::deque<TimeLocPair> LocationContainer;
typedef LocationContainer::const_iterator LocationIterator;

bool compareTimeLocPair(const TimeLocPair& lhs, const TimeLocPair& rhs)
{return lhs.first < rhs.first;}

LocationIterator find(const LocationContainer& cont, timetype time) {
    TimeLocPair finder(time, LocationStruct());
    LocationIterator it = std::lower_bound(cont.begin(), cont.end(), finder, compareTimeLocPair);
    if (it == cont.end() || it->first != time)
        return cont.end();
    return it;
}

回答2:

The most idiomatic way would be:

myLocations.erase( std::prev( myLocations.end() ) );

If you don't ha ve C++11, use the corresponding function from your toolbox.

回答3:

Try this, it works:

map<string, LocationStruct>::iterator it = myLocations.end();
it--;
myLocations.erase(it);

回答4:

Well, a quick check on g++ 4.4 suggests that this works just fine:

myLocations.erase(myLocations.rbegin()->first);

though I must confess I don't know why it doesn't like accepting only the iterator itself.

回答5:

Since you are Storing the time as key String . The last element (earliest by time in a day considering time from 00:00 to 24:00) will be a lower bound element and hence You can fetch the iterator like this

     `map<string, LocationStruct>::iterator it;`
      it=myLocations.lower_bound ('00:00');
      myLocations.erase ( it, it+1);

But if it belongs to different dates then you need to even consider the day and manupilate your code accordingly . As you mentioned data is coming quick enough you dont need to take the date into consideration . But The safe way here would be take the entire date in terms of second and remove the lowest one as mentioned above . That would take care even if frequency of new data arriving is pretty slow.

回答6:

map::erase for last element without TS bindings, simply use following:

myLocations.erase ((--myLocations.end()));

来源：https://stackoverflow.com/questions/11143214/stdmap-erase-last-element

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c++

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erase