Rxjs 类型“Observable”上不存在属性“filter”。谷歌地图

我有一个出租车预订应用程序，使用离子地理定位和Firebase Firestore。我的代码与firestore包不兼容，因此我必须将rxjs更新到版本6。它提出了错误

'Property 'filter' does not exist on type 'Observable<Geoposition>'. '

mypackage.json文件

"@ionic-native/core": "~4.15.0",
"@ionic-native/geolocation": "^4.17.0",
"@ionic-native/splash-screen": "~4.15.0",
"@ionic-native/status-bar": "~4.15.0",
"@ionic/storage": "2.2.0",
"angularfire2": "^5.1.0",
"cordova-plugin-geolocation": "^4.0.1",
"firebase": "^5.5.7",
"ionic-angular": "3.9.2",
"ionic-native": "^2.9.0",
"ionicons": "3.0.0",
"rxjs": "^6.3.3",
"rxjs-compat": "^6.3.3",

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我的代码怎么了？

感谢@JB Nizet和@martin的贡献。我记得在rxJs的最新版本中，链接txjs操作符（如过滤器和映射）时必须使用.pipe操作符。这是我的更新代码

const subscription = this.geo.watchPosition()
                          .pipe(filter((p) => p.coords !== undefined)) //Filter Out Errors
                          .subscribe(position => {
                          let latLng = new google.maps.LatLng(position.coords.latitude, position.coords.longitude);
                          this.map.setCenter(latLng);
                          this.lat = position.coords.latitude;
                          this.long = position.coords.longitude;
                          console.log(this.long + ' ' + this.lat);
});

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使用可管道运算符：，@JBNizet我该如何使用它？如前所述，在我链接的两个页面中有示例和所有内容。@JBNizet对于noob问题表示抱歉，已修复。

const subscription = this.geo.watchPosition()
                          .pipe(filter((p) => p.coords !== undefined)) //Filter Out Errors
                          .subscribe(position => {
                          let latLng = new google.maps.LatLng(position.coords.latitude, position.coords.longitude);
                          this.map.setCenter(latLng);
                          this.lat = position.coords.latitude;
                          this.long = position.coords.longitude;
                          console.log(this.long + ' ' + this.lat);
});