最近根据以前的计步算法基础,改进了一下,可以识别出是在走路还是跑步(骑行暂未实现)。
大体上,还是在计步算法上的改进,在步点的时候增加时间的判断。
之前的计步算法识别:http://blog.csdn.net/finnfu/article/details/45273183
直接提贴出核心代码:
package com.research.sensor.finnfu.gesturedetector.detector; import android.util.Log; /** * Created by finnfu on 2017/6/3. * 步行或者跑步 */ public class PaceAndRunDetector { //存放三轴数据 float[] oriValues = new float[3]; final int ValueNum = 4; //用于存放计算阈值的波峯波谷差值 float[] tempValue = new float[ValueNum]; int tempCount = 0; //是否上升的标志位 boolean isDirectionUp = false; //持续上升次数 int continueUpCount = 0; //上一点的持续上升的次数,为了记录波峯的上升次数 int continueUpFormerCount = 0; //上一点的状态,上升还是下降 boolean lastStatus = false; //波峯值 float peakOfWave = 0; //波谷值 float valleyOfWave = 0; //此次波峯的时间 long timeOfThisPeak = 0; //上次波峯的时间 long timeOfLastPeak = 0; //当前的时间 long timeOfNow = 0; //当前传感器的值 float gravityNew = 0; //上次传感器的值 float gravityOld = 0; //动态阈值需要动态的数据,这个值用于这些动态数据的阈值 final float InitialValue = (float) 1.3; //初始阈值 float ThreadValue = (float) 2.0; //波峯波谷时间差 int TimeInterval = 250; private int stepCount = 0; private long timeOfLastStep = 0; private long timeOfThisStep = 0; private long averageTimeOfEveryStep = 0; //数据的输入口 public void inputValue(float v){ detectorNewStep(v); } /* * 检测步子,并开始计步 * 1.传入sersor中的数据 * 2.如果检测到了波峯,并且符合时间差以及阈值的条件,则判定为1步 * 3.符合时间差条件,波峯波谷差值大于initialValue,则将该差值纳入阈值的计算中 * */ public void detectorNewStep(float values) { if (gravityOld == 0) { gravityOld = values; } else { if (detectorPeak(values, gravityOld)) { timeOfLastPeak = timeOfThisPeak; timeOfNow = System.currentTimeMillis(); if (timeOfNow - timeOfLastPeak >= TimeInterval && (peakOfWave - valleyOfWave >= ThreadValue)) { timeOfThisPeak = timeOfNow; /* 走路或者跑步的识别 * */ countStep(); } if (timeOfNow - timeOfLastPeak >= TimeInterval && (peakOfWave - valleyOfWave >= InitialValue)) { timeOfThisPeak = timeOfNow; ThreadValue = peakValleyThread(peakOfWave - valleyOfWave); } } } gravityOld = values; } private void countStep() { timeOfLastStep = timeOfThisStep; timeOfThisStep = System.currentTimeMillis(); long diffValue = timeOfThisStep - timeOfLastStep; if (diffValue <= 3000L){ averageTimeOfEveryStep += diffValue; stepCount++; if(stepCount == 9){ averageTimeOfEveryStep = averageTimeOfEveryStep/10; Log.i("result","averageTimeOfEveryStep: "+averageTimeOfEveryStep); if(averageTimeOfEveryStep <= 380){ Log.i("result","跑步"); }else if(averageTimeOfEveryStep >= 450){ Log.i("result","走路"); }else{ Log.i("result","快走"); } }else if(stepCount > 9){ resSomeValue(); } }else{//超时 resSomeValue(); } } private void resSomeValue(){ stepCount = 0; averageTimeOfEveryStep = 0; } /* * 检测波峯 * 以下四个条件判断为波峯: * 1.目前点为下降的趋势:isDirectionUp为false * 2.之前的点为上升的趋势:lastStatus为true * 3.到波峯为止,持续上升大于等于2次 * 4.波峯值大于20 * 记录波谷值 * 1.观察波形图,可以发现在出现步子的地方,波谷的下一个就是波峯,有比较明显的特征以及差值 * 2.所以要记录每次的波谷值,为了和下次的波峯做对比 * */ public boolean detectorPeak(float newValue, float oldValue) { lastStatus = isDirectionUp; if (newValue >= oldValue) { isDirectionUp = true; continueUpCount++; } else { continueUpFormerCount = continueUpCount; continueUpCount = 0; isDirectionUp = false; } if (!isDirectionUp && lastStatus && (continueUpFormerCount >= 2 || oldValue >= 20)) { peakOfWave = oldValue; return true; } else if (!lastStatus && isDirectionUp) { valleyOfWave = oldValue; return false; } else { return false; } } /* * 阈值的计算 * 1.通过波峯波谷的差值计算阈值 * 2.记录4个值,存入tempValue[]数组中 * 3.在将数组传入函数averageValue中计算阈值 * */ public float peakValleyThread(float value) { float tempThread = ThreadValue; if (tempCount < ValueNum) { tempValue[tempCount] = value; tempCount++; } else { tempThread = averageValue(tempValue, ValueNum); for (int i = 1; i < ValueNum; i++) { tempValue[i - 1] = tempValue[i]; } tempValue[ValueNum - 1] = value; } return tempThread; } /* * 梯度化阈值 * 1.计算数组的均值 * 2.通过均值将阈值梯度化在一个范围里 * */ public float averageValue(float value[], int n) { float ave = 0; for (int i = 0; i < n; i++) { ave += value[i]; } ave = ave / ValueNum; if (ave >= 8) ave = (float) 4.3; else if (ave >= 7 && ave < 8) ave = (float) 3.3; else if (ave >= 4 && ave < 7) ave = (float) 2.3; else if (ave >= 3 && ave < 4) ave = (float) 2.0; else { ave = (float) 1.3; } return ave; } }