我有这个数据框。我的数据框包含ID，时间和值以及缺口（小时）。我正在独立于每个ID进行重新采样。间隔列给出两个不同时间之间的连续时间间隔。我每10分钟进行一次重新采样，并且如果连续的间隙大于0.86 Hr，我想停止重新采样，并返回下一行作为原始行，并在发现相同条件时再次继续重新采样。 我的空缺条件就是这样
ManyToOne样本数据
package com.example.mrfrag.firebattery;
import android.content.BroadcastReceiver;
import android.content.ContentProvider;
import android.content.Context;
import android.content.Intent;
import android.content.IntentFilter;
import android.os.BatteryManager;
import android.support.v7.app.AppCompatActivity;
import android.os.Bundle;
import android.view.View;
import android.widget.Button;
import android.widget.TextView;
import android.widget.Toast;
public class HomeActivity extends AppCompatActivity{
TextView batteryplus,temp,setting,plug,charging,battery;
private BroadcastReceiver mReceiver;
IntentFilter ifilter =new IntentFilter();
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_home);
batteryplus= findViewById(R.id.batteryplus);
temp = findViewById(R.id.temp);
setting= findViewById(R.id.setting);
plug= findViewById(R.id.plug);
charging= findViewById(R.id.charging);
battery= findViewById(R.id.battery);
mReceiver=new PowerConnectionReceiver();
ifilter.addAction(Intent.ACTION_POWER_CONNECTED);
ifilter.addAction(Intent.ACTION_POWER_DISCONNECTED);
ifilter.addAction(Intent.ACTION_BATTERY_CHANGED);
}
@Override
protected void onResume() {
super.onResume();
registerReceiver(mReceiver, ifilter);
}
@Override
protected void onPause() {
super.onPause();
unregisterReceiver(mReceiver);
}
public class PowerConnectionReceiver extends BroadcastReceiver {
public PowerConnectionReceiver(){
}
@Override
public void onReceive(Context context, Intent batteryStatus) {
int status = batteryStatus.getIntExtra(BatteryManager.EXTRA_STATUS,
-1);
boolean isCharging = status == BatteryManager.BATTERY_STATUS_CHARGING
||
status == BatteryManager.BATTERY_STATUS_FULL;
if (isCharging) {
plug.setText("CHARGING");
} else {
plug.setText("NOT CHARGING");
}
int health =
batteryStatus.getIntExtra(BatteryManager.EXTRA_HEALTH,0);
batteryplus.setText(ConvoHealth(health));
String technology =
batteryStatus.getExtras().getString(BatteryManager.EXTRA_TECHNOLOGY);
setting.setText(technology);
float tpr = ((float)
batteryStatus.getIntExtra(BatteryManager.EXTRA_TEMPERATURE,0)) / 10.0f;
temp.setText( String.valueOf(tpr) + "°C / " +
String.valueOf(tpr*1.8+32)+"F"+"\n");
int voltage =
batteryStatus.getIntExtra(BatteryManager.EXTRA_VOLTAGE,0);
charging.setText(voltage+ " mV");
int mah = batteryStatus.getIntExtra(String.valueOf
(BatteryManager.BATTERY_PROPERTY_CAPACITY),0);
battery.setText(mah+ " Mah");
}
}
private String ConvoHealth(int health){
String result;
switch(health){
case BatteryManager.BATTERY_HEALTH_COLD:
result = "COLD";
break;
case BatteryManager.BATTERY_HEALTH_DEAD:
result = "DEAD";
break;
case BatteryManager.BATTERY_HEALTH_GOOD:
result = "GOOD";
break;
case BatteryManager.BATTERY_HEALTH_OVERHEAT:
result = "OVERHEAT";
break;
case BatteryManager.BATTERY_HEALTH_OVER_VOLTAGE:
result = "OVER VOLTAGE";
break;
case BatteryManager.BATTERY_HEALTH_UNKNOWN:
result = "UNKNOWN";
break;
case BatteryManager.BATTERY_HEALTH_UNSPECIFIED_FAILURE:
result = "UNSPECIFIED_FAILURE";
break;
default:
result = "Unknown";
}
return result;
}
}
如您所见，ID 1的间隙超过0.86 Hr，所以我的想法是在该点停止重新采样。 像这样
a (abs(a-b))
b 0
因此，我想继续对此ID,Time,Value,Gaps
1,1523147332607,2,0.3347541666666667
1,1523148537722,5,0.17346666666666666
1,1523149162202,6,1.6252830555555555
1,1523155013221,4,0.33290027777777775
1,1523156211662,7,0.3722580555555556
1,1523157551791,10,0.0
2,1523156211662,5,0.5115911111111111
2,1523158053390,2,0.3405525
2,1523159279379,9,1.3295477777777778
2,1523164065751,3,0.0
进行重新采样，并且当没有更多采样要做时，我想将最后一行作为原始结果返回，即
ID,Time,Value,Gaps
1,1523147332607,2,0.3347541666666667
...................................
1,1523148537722,5,0.17346666666666666
...................................
...................................
1,1523149162202,6,1.6252830555555555
然后，我想从下一行继续重新采样
Time period 1523149162202然后继续
对于每个ID的正常重采样，
1,1523149162202,6,1.6252830555555555
但是如何跟踪每个重采样，以便在遇到某些条件时可以停止重采样，并在该部分重采样的末尾返回原始行。然后，它从下一行再次继续相同的条件。我本来打算使用1,1523155013221,4,0.33290027777777775
1,1523156211662,7,0.3722580555555556
1,1523157551791,10,0.0
，但距离实现这一目标还遥遥无期。
有什么建议么 ？
答案 0 :(得分：2)
一种方法可能是在df中创建一个临时列'ID_res'，在更改ID时或在间隙超过0.86之后的行中递增数字，例如：
df.loc[(df['ID'] != df['ID'].shift())| (df['Gaps'].shift() > 0.86),'ID_res'] = 1
df['ID_res'] = df['ID_res'].cumsum().ffill()
所以您的df如下：
ID Value Gaps ID_res
Time
2018-04-08 00:28:52.607 1 2 0.334754 1.0
2018-04-08 00:48:57.722 1 5 0.173467 1.0
2018-04-08 00:59:22.202 1 6 1.625283 1.0
2018-04-08 02:36:53.221 1 4 0.332900 2.0
2018-04-08 02:56:51.662 1 7 0.372258 2.0
2018-04-08 03:19:11.791 1 10 0.000000 2.0
2018-04-08 02:56:51.662 2 5 0.511591 3.0
2018-04-08 03:27:33.390 2 2 0.340553 3.0
2018-04-08 03:47:59.379 2 9 1.329548 3.0
2018-04-08 05:07:45.751 2 3 0.000000 4.0
现在，您可以使用在“ ID_res”上进行第一个分组依据的方法，同时保留“ ID”和“ Value”列，并在最后删除“ ID_res”列，因为您不再需要它：
df = (df.groupby('ID_res', axis=0)['ID','Value'].resample('10min').mean()
.groupby(level=0).apply(lambda x: x.interpolate())
.reset_index().drop('ID_res',1))
结果如下：
Time ID Value
0 2018-04-08 00:20:00 1.0 2.0
1 2018-04-08 00:30:00 1.0 3.5
2 2018-04-08 00:40:00 1.0 5.0
3 2018-04-08 00:50:00 1.0 6.0
4 2018-04-08 02:30:00 1.0 4.0
5 2018-04-08 02:40:00 1.0 5.5
6 2018-04-08 02:50:00 1.0 7.0
...
在第3行和第4行之间没有重新采样的情况下，原始df的这些值之间的“间隙”超过0.86