竞赛圈   > 分享公交准点预测的代码(0.048+)

qmll

数据挖掘工程师   |   C/C++
  • 关注者 2
  • 关注了

qmll

数据挖掘工程师   |   C/C++

/*
基本思路为:从历史数据中求得“公交车在某个时间段之内从第i站到第j站的平均时间”,然后用于待预测线路的预测

注:本方法用到的关键信息有:时间戳、上下行、下一站点, 没有考虑工作日与周末的差异,也没有考虑车门、瞬时速度等信息;也没有根据当天情况校正首个预测站点的到站时间

1. 从待预测线路的文件中,取出每一条线路的  busid-predTime-UP
2. 根据busid-predTime-UP,每一天单独处理,从训练集中取出数据,满足3个条件:对应busid,在predTime附近,UP字段吻合
2.1 将一天之中,举例说明,若待预测线路中有这个数据,111-8-1,即公交车111号在早上8点附近且UP字段为1,那么就需要从这一天的数据中,找到车辆111在8点左右的所有符合条件的数据
对于左右的定义:以8点为例,7点40~9点20之间的,都算是左右,只要满足上述3个条件,都会取出
2.2 仍然是针对一天的数据,对每一条busid-predTime-UP记录,按时间戳进行排序;然后判断公交车从第i站到第j站的耗时,记作 busid-predTime-UP-Stop的耗时
2.3 当所有训练集中的数据都统计完毕,可以求得 busid-predTime-UP-Stop对应的平均时间(中位数时间的结果要好一些,看来会有一些异常值啊)
3. 最后,生成提交文件

*/

#include <iostream>
#include <vector>
#include <map>
#include <functional>
#include <fstream>
#include <sstream>
#include <string>
#include <set>
#include <sstream>
using namespace std;


// 分割字符串的函数……
vector<string> SplitLine(string stringplit, string strs)
{
	vector<string> res;
	//string::size_type pos;
	int pos(-1);
	strs += stringplit;
	int size = int(strs.size());

	for (int i = 0; i < size; i++){
		pos = strs.find(stringplit, i);
		if (pos<size){
			string tempstr = strs.substr(i, pos - i);
			res.push_back(tempstr);
			i = pos + int(stringplit.size()) - 1;
		}
	}
	return res;
}



// 读测试集,然后把待预测线路搞出来,目的是为了从训练集中找数据
// set<string>的string内容为  busid-predTime-up。
set<string> loadTargetBusTimeUP(string datapath);

// 然后,根据待预测的线路,把所有的这些训练集中对应的数据都找出来
map<string, double> loadTrainBusTimeUPStop(string basicpath, set<string> targetBusTimeUp);

// 再次读测试集!给出预测的时间!
map<string, vector<double>> predPathTimes(string basicpath, map<string, double> busTimeUpStopCost, string respath);

int main()
{
	string predPath = "C:/Users/Administrator/Desktop/toBePredicted_update_0523.csv";
	set<string> targetBusTimeUp = loadTargetBusTimeUP(predPath);

	string basicpath = "C:/Users/Administrator/Desktop/train1-24/train2017";
	map<string, double> busTimeUpStopCost = loadTrainBusTimeUPStop(basicpath, targetBusTimeUp);

	string respath = "C:/Users/Administrator/Desktop/submitted.csv";
	map<string, vector<double>> results = predPathTimes(predPath, busTimeUpStopCost, respath);

	//string finalpath = "C:/Users/Administrator/Desktop/submitted_revised.csv";
	//adjustPathTimes(basicpath, results, finalpath);


	system("pause");
	return 1;
}





set<string> loadTargetBusTimeUP(string datapath)
{
	set<string> res;
	ifstream fin(datapath.c_str());
	string str, substri;
	getline(fin, str);
	while (getline(fin, str)){
		vector<string> thisline = SplitLine(",", str);
		// 提取待预测的小时
		vector<string> hourMinSec = SplitLine(":", thisline[3]);
		string infokey = thisline[2] + "-" + hourMinSec[0] + "-" + thisline[6];
		res.insert(infokey);
	}
	fin.close();

	cout << "目标线路已读入完毕!" << endl;

	string respath = "C:/Users/Administrator/Desktop/targetPath.csv";
	ofstream fou(respath.c_str());
	for (set<string>::iterator it = res.begin(); it != res.end(); ++it)
		fou << *it << endl;
	fou.close();

	return res;
}



map<string, double> loadTrainBusTimeUPStop(string basicpath, set<string> targetBusTimeUp)
{

	map<string, vector<int>> results;

	cout << "开始处理训练集!" << endl;

	for (int i = 1008; i < 1025; i++){
		// 存放这一天的时间、站点的信息
		map<string, map<int, int>> todayRecords;

		stringstream isis; isis << i; string stri; isis >> stri;
		string datapath = basicpath + stri + ".csv";
		
		cout << "	...读入数据 "<<datapath << endl;

		ifstream fin(datapath.c_str());
		if (!fin){ cout << "Can not open file" << datapath << endl; system("pause"); }
		string str, substri;
		// 跳过表头
		getline(fin, str);

		while (getline(fin, str)){
			vector<string> thisline = SplitLine(",", str);

			// 若时间段不对,直接跳过
			// 删除掉	晚上23点~早上6点以前的数据;
			vector<string> vecHourMinSec = SplitLine(":", thisline[2]);
			int thishour = atoi(vecHourMinSec[0].c_str());
			if (thishour >= 23) continue;
			if (thishour < 6) continue;

			int thismin = atoi(vecHourMinSec[1].c_str());
			int thissec = atoi(vecHourMinSec[2].c_str());
			
			// 生成当天的时间戳。
			int thisTime = thishour * 3600 + thismin * 60 + thissec;

			// 字段顺序
			// O_LINENO-0,O_TERMINALNO-1,O_Time-2,O_LONGITUDE-3,O_LATITUDE-4,O_SPEED-5,O_MIDDOOR-6,O_REARDOOR-7,O_FRONTDOOR-8,O_UP-9,O_RUN-10,O_NEXTSTATIONNO-11

			// 判断此记录,是否在“待预测线路”的附近时间之内
			{
				// 先放到本小时里面
				string strhour; stringstream isis; isis << thishour; isis >> strhour;
				string newkey = thisline[1] + "-" + strhour + "-" + thisline[9];
				
				if (targetBusTimeUp.count(newkey)){
					todayRecords[newkey].insert(make_pair(thisTime, atoi(thisline[11].c_str())));
				}

			}
			if (thismin < 20){
				//可以算作是上一个小时的记录
				string strhour; stringstream isis; isis << thishour-1; isis >> strhour;
				string newkey = thisline[1] + "-" + strhour + "-" + thisline[9];

				if (targetBusTimeUp.count(newkey)){
					todayRecords[newkey].insert(make_pair(thisTime, atoi(thisline[11].c_str())));
				}
			}
			if (thismin > 40){
				//可以算作是下一个小时的记录
				string strhour; stringstream isis; isis << thishour+1; isis >> strhour;
				string newkey = thisline[1] + "-" + strhour + "-" + thisline[9];

				if (targetBusTimeUp.count(newkey)){
					todayRecords[newkey].insert(make_pair(thisTime, atoi(thisline[11].c_str())));
				}
			}
		}
		fin.close();


		// 这一天的内容搞定了~~ 那么,接下来应该是根据这些信息,去判断到站时间

		cout << "	...估计到站时间 " << endl;
		// map<string, map<int, string>> todayRecords;
		for (map<string, map<int, int>>::iterator itbustime = todayRecords.begin(); itbustime != todayRecords.end(); ++itbustime){

			// 处理这一bus-time-up的数据
			// 记录上一个站点
			int lastStop = itbustime->second.begin()->second;

			map<int, int> correctTimeStop;	//用于存放“站点切换”的时刻

			for (map<int, int>::iterator ittime = itbustime->second.begin(); ittime != itbustime->second.end(); ++ittime){
				// 仅考察“nextstop递增的情况”,且需要判断站点是否连续
				if (ittime->second > lastStop){
					if (ittime->second = lastStop + 1){
						// 表示stop是连续的,但还需要判断,时间间隔是否大于30秒;
						// 此处记录的时间是“结束lastStop时刻的时间”,切记,之后会用到
						if (correctTimeStop.empty()){
							correctTimeStop[ittime->first] = lastStop;
						}
						else{
							map<int, int>::iterator ittime2 = correctTimeStop.end();
							ittime2--;
							if (ittime->first - ittime2->first > 30)
								correctTimeStop[ittime->first] = lastStop;
						}
					}
					// 如果不连续,那么就粗暴地舍弃掉了!因为不知道是否是正确的!
				}
				lastStop = ittime->second;
			}


			//// 写一下这个correctTimeStop,用于观察
			//string path = basicpath + "tempCorrectTimePath.csv";
			//ofstream fou(path.c_str(), ios::app);
			//for (map<int, int>::iterator it = correctTimeStop.begin(); it != correctTimeStop.end(); ++it){
			//	fou << itbustime->first << "," << it->first << "," << it->second << endl;
			//}
			//fou.close();
				


			// 然后,需要计算每个站点的到站时间。在correctTimeStop里面,站点为x,表示从第x-1站到达第x站所需要的时间。
			// 在访问correctTimeStop的时候,需要注意,若遇到站点不连续的情况,比如 223,3;  434, 5;   556,6
			//      那么,无法计算从第2到第3站的时间消耗;也无法计算从第4站到第5站的时间消耗; 但是,可以计算从第5站到第6站的时间消耗
			//      据此规则,判断所有有效的“时间消耗”;然后,写入文件到results中
			map<int, int>::iterator ittime1 = correctTimeStop.begin(); 
			map<int, int>::iterator ittime2 = ittime1; ittime2++;

			// 如果第一个站是2,且是在整点时刻左右,就说明到达第二站的时间,应该算作是这个时刻减去整点
			if (ittime1->second == 2){
				if (ittime1->first % 3600 <= 120){
					//刚刚超过整点,那么就设置,站点2的到达时间为这个余数。
					string strstop; stringstream isis; isis << ittime1->second; isis >> strstop;
					string busTimeUpStop = itbustime->first + "-" + strstop;
					if (results.count(busTimeUpStop))
						results[busTimeUpStop].push_back(ittime1->first % 3600);
					else{
						vector<int> temp; temp.push_back(ittime1->first % 3600);
						results[busTimeUpStop] = temp;
					}
				}
			}

			// 根据上述连续站点的规则,判断有效的“时间消耗”
			while (ittime2 != correctTimeStop.end()){
				if (ittime2->second == ittime1->second + 1){
					// 只有当站点连续的时候,才保留此结果
					string strstop; stringstream isis; isis << ittime2->second; isis >> strstop;
					string busTimeUpStop = itbustime->first + "-" + strstop;
					if (results.count(busTimeUpStop))
						results[busTimeUpStop].push_back(ittime2->first - ittime1->first);
					else{
						vector<int> temp; temp.push_back(ittime2->first - ittime1->first);
						results[busTimeUpStop] = temp;
					}
						
				}

				
				ittime1++;
				ittime2++;
			}
		

		}// 当天的记录处理完毕! 

		
	}// 所有的日期都记录完毕!


	cout << "训练集处理完毕!--- " << endl;

	//// 最后,取中位数吧!
	//map<string, double> res;
	//for (map<string, vector<int>>::iterator itres = results.begin(); itres != results.end(); ++itres){
	//	multiset<int> temp;
	//	for (vector<int>::iterator ittime = itres->second.begin(); ittime != itres->second.end(); ++ittime){
	//		temp.insert(*ittime);
	//	}
	//	multiset<int>::iterator itit = temp.begin();
	//	for (size_t i = 0; i < temp.size() / 2; i++)
	//		itit++;
	//	res[itres->first] = *itit;
	//}

	// 最后,取平均数吧!
	map<string, double> res;
	for (map<string, vector<int>>::iterator itres = results.begin(); itres != results.end(); ++itres){
		int tempsum = 0;
		for (vector<int>::iterator ittime = itres->second.begin(); ittime != itres->second.end(); ++ittime){
			tempsum += *ittime;
		}
	
		res[itres->first] = double(tempsum)/double(itres->second.size());
	}


	// 最好把这个文件写出来!
	string respath = "C:/Users/Administrator/Desktop/busTimeUpStopCost.csv";
	ofstream fou(respath.c_str());
	for (map<string, double>::iterator it = res.begin(); it != res.end(); ++it)
		fou << it->first << "," << it->second << endl;
	fou.close();

	return res;
}



map<string, vector<double>> predPathTimes(string datapath, map<string, double> busTimeUpStopCost, string respath)
{
	cout << "开始预测最终结果 " << endl;

	map<string, vector<double>> results;
	
	ifstream fin(datapath.c_str());
	string str, substri;
	getline(fin, str);
	while (getline(fin, str)){
		vector<string> thisline = SplitLine(",", str);
		// 生成key
		vector<string> hourMinSec = SplitLine(":", thisline[3]);
		string infokey = thisline[2] + "-" + hourMinSec[0] + "-" + thisline[6];
		// 此时的infokey,是为了配合stop使用的
		int stStop = atoi(thisline[4].c_str()), enStop = atoi(thisline[5].c_str());
		
		vector<double> timeCosts;
		for (int stopi = stStop; stopi <= enStop; ++stopi){
			string strStopi; stringstream isis; isis << stopi; isis >> strStopi;
			string busTimeUpStop = infokey + "-" + strStopi;
			if (busTimeUpStopCost.count(busTimeUpStop)){
				// 找到记录了,甚好!
				timeCosts.push_back(busTimeUpStopCost[busTimeUpStop]);
			}
			else{
				// 没找到,怎么办呢?直接填充一个数字吧~~ yin叔等选手讲,平均值大约在109或者128什么的,我就填110吧(本来也想填上下游时间的,可是懒得搞了)
				timeCosts.push_back(110);
			}
		}
		// 最后,再累加即可
		for (size_t i = 1; i < timeCosts.size(); i++){
			timeCosts[i] += timeCosts[i - 1];
		}

		string thiskey;
		for (int i = 0; i < 6; i++)
			thiskey = thiskey + thisline[i] + ",";

		results[thiskey] = timeCosts;
	}
	fin.close();

	// 写出结果!
	ofstream fou(respath.c_str());
	fou << "O_DATA,O_LINENO,O_TERMINALNO,predHour,pred_start_stop_ID,pred_end_stop_ID,pred_timeStamps"<<endl;

	for (map<string, vector<double>>::iterator itres = results.begin(); itres != results.end(); ++itres)
	{
		fou << itres->first;
		vector<double>::iterator itval = itres->second.begin();
		fou << *itval;
		++itval;
		while (itval != itres->second.end()){
			fou << ";"<< *itval;
			++itval;
		} 
		fou << endl;
	}
	fou.close();
	

	return results;
}


//void adjustPathTimes(string basicpath, map<string, vector<double>> results, string respath)
//{
//	cout << "根据当天的情况,纠正首次到站时间 " << endl;
//
//	
//	for (int i = 1025; i < 1032; i++){
//		// 存放这一天的时间、站点的信息
//		map<string, map<int, int>> todayRecords;
//		
//	}
//
//	return;
//}


3条评论

分享

3条评论
意见反馈
关注微信公众号 关注微信公众号
  • © 2013-2018 DataCastle 蜀ICP备17028166号-1

扫一扫分享给周围朋友