我想切片pandas value_counts():
GeoPoint geoPoint = new GeoPoint(mLatitude, mLongitude);
IMapController mapController = mMapView.getController();
mapController.setZoom(5);
mapController.setCenter(geoPoint);
Overlay overlay = new Overlay(getActivity()) {
ItemizedIconOverlay<OverlayItem> items = null;
@Override
protected void draw(Canvas c, MapView osmv, boolean shadow) {
}
@Override
public boolean onSingleTapConfirmed(MotionEvent e, MapView mapView) {
Projection proj = mapView.getProjection();
GeoPoint loc = (GeoPoint) proj.fromPixels((int)e.getX(), (int)e.getY());
double longitude = loc.getLongitude();
double latitude = loc.getLatitude();
mLatitude = latitude;
mLongitude = longitude;
ArrayList<OverlayItem> markers = new ArrayList<>();
OverlayItem item = new OverlayItem("", "", new GeoPoint(latitude, longitude));
item.setMarker(ContextCompat.getDrawable(getActivity(), R.drawable.ic_maps_marker_large));
markers.add(item);
if (items == null) {
items = new ItemizedIconOverlay<>(getActivity(), markers, null);
mMapView.getOverlays().add(items);
mMapView.invalidate();
} else {
mMapView.getOverlays().remove(items);
mMapView.invalidate();
items = new ItemizedIconOverlay<>(getActivity(), markers, null);
mMapView.getOverlays().add(items);
}
return true;
}
};
mMapView.getOverlays().add(overlay);
但是我收到了一个错误:
>sur_perimetre[col].value_counts()
44341006.0 610
14231009.0 441
12131001.0 382
12222009.0 364
12142001.0 354
与ix相同:
> sur_perimetre[col].value_counts()[:5]
KeyError: 5.0
你会如何处理?
也许:
> sur_perimetre[col].value_counts().ix[:5]
KeyError: 5.0
答案 0 :(得分:3)
方法1:
您需要观察value_counts()返回一个Series对象。您可以像处理任何其他系列一样处理它并获取值。您甚至可以构建一个新的数据帧。
In [1]: import pandas as pd
In [2]: df = pd.DataFrame([1,2,3,3,4,5], columns=['C1'])
In [3]: vc = df.C1.value_counts()
In [4]: type(vc)
Out[4]: pandas.core.series.Series
In [5]: vc.values
Out[5]: array([2, 1, 1, 1, 1])
In [6]: vc.values[:2]
Out[6]: array([2, 1])
In [7]: vc.index.values
Out[7]: array([3, 5, 4, 2, 1])
In [8]: df2 = pd.DataFrame({'value':vc.index, 'count':vc.values})
In [8]: df2
Out[8]:
count value
0 2 3
1 1 5
2 1 4
3 1 2
4 1 1
方法2:
然后,我试图重新生成你提到的错误。但是,在DF中使用单个列时,我没有像你提到的那样用相同的符号表示任何错误。
In [1]: import pandas as pd
In [2]: df = pd.DataFrame([1,2,3,3,4,5], columns=['C1'])
In [3]: df['C1'].value_counts()[:3]
Out[3]:
3 2
5 1
4 1
Name: C1, dtype: int64
In [4]: df.C1.value_counts()[:5]
Out[4]:
3 2
5 1
4 1
2 1
1 1
Name: C1, dtype: int64
In [5]: pd.__version__
Out[5]: u'0.17.1'
希望它有所帮助!