Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
711 views
in Technique[技术] by (71.8m points)

dictionary - C++ Help finding the max value in a map

I've been doing a basic program to find the max, min, median, variance, mode etc. of a vector. Everything went fine until I got to the mode.

The way I see it, I should be able to loop through the vector, and for each number that occurs I increment a key on the map. Finding the key with the highest value would then be the one that occured the most. Comparing to other keys would tell me if it's a single multiple or no mode answer.

Here's the chunk of code that's been causing me so much trouble.

map<int,unsigned> frequencyCount;
// This is my attempt to increment the values
// of the map everytime one of the same numebers 
for(size_t i = 0; i < v.size(); ++i)
    frequencyCount[v[i]]++;

unsigned currentMax = 0;
unsigned checked = 0;
unsigned maax = 0;
for(auto it = frequencyCount.cbegin(); it != frequencyCount.cend(); ++it )
    //checked = it->second;
    if (it ->second > currentMax)
    {
        maax = it->first;
    }
    //if(it ->second > currentMax){
    //v = it->first

cout << " The highest value within the map is: " << maax << endl;

The entire program can be seen here. http://pastebin.com/MzPENmHp

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Answer

0 votes
by (71.8m points)

You can use std::max_element to find the highest map value (the following code requires C++11):

std::map<int, size_t> frequencyCount;
using pair_type = decltype(frequencyCount)::value_type;

for (auto i : v)
    frequencyCount[i]++;

auto pr = std::max_element
(
    std::begin(frequencyCount), std::end(frequencyCount),
    [] (const pair_type & p1, const pair_type & p2) {
        return p1.second < p2.second;
    }
);
std::cout << "A mode of the vector: " << pr->first << '
';

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

2.1m questions

2.1m answers

60 comments

57.0k users

...