React Native：完成拖动后获取元素的实际 x，y 坐标（react-native-gesture-handler）

我正在使用这个库：https://github.com/software-mansion/react-native-gesture-handler https://github.com/software-mansion/react-native-gesture-handler

我对如何获取元素的实际 x,y 坐标有点困惑。 我正在解释...

这是初始状态：

拖动后，用event.nativeEvent.absoluteX我越来越38

但我想要0因为元素获得了左侧区域。

我越来越0仅当我将触摸移动到最左侧区域时。像这样，

但在这种情况下，这种解决方案对我不起作用????

这是我的代码：

import React, {Component} from 'react';
import {Animated, View} from 'react-native';

import {PanGestureHandler, State} from 'react-native-gesture-handler';

export default class App extends Component {
  constructor(props) {
    super(props);
    this._translateX = new Animated.Value(0);
    this._translateY = new Animated.Value(0);
    this._lastOffset = {x: 0, y: 0};
    this._onGestureEvent = Animated.event(
      [
        {
          nativeEvent: {
            translationX: this._translateX,
            translationY: this._translateY,
          },
        },
      ],
      {useNativeDriver: true},
    );
  }

  _onHandlerStateChange = event => {
    if (event.nativeEvent.oldState === State.ACTIVE) {
      this._lastOffset.x += event.nativeEvent.translationX;
      this._lastOffset.y += event.nativeEvent.translationY;
      this._translateX.setOffset(this._lastOffset.x);
      this._translateX.setValue(0);
      this._translateY.setOffset(this._lastOffset.y);
      this._translateY.setValue(0);

      console.log(event.nativeEvent.absoluteX);
      console.log(event.nativeEvent.absoluteY);
    }
  };

  render() {
    return (
      <PanGestureHandler
        {...this.props}
        onGestureEvent={this._onGestureEvent}
        onHandlerStateChange={this._onHandlerStateChange}>
        <Animated.View
          style={{
            transform: [
              {translateX: this._translateX},
              {translateY: this._translateY},
            ],
          }}>
          <View style={{backgroundColor: 'green', height: 100, width: 100}} />
        </Animated.View>
      </PanGestureHandler>
    );
  }
}

请帮忙！ 提前致谢。