如何将一个无状态组件相乘，以使列表的每个组件具有独立的状态？

Question

我使用重组库有一个无状态组件 该文件是一个容器：

import { connect } from 'react-redux';
import {
  compose,
  withState,
  withHandlers,
} from 'recompose';
import PropTypes from 'prop-types';
import ColumnPresentation from './ColumnPresentation';
import { setPlayerStep, setWinner } from '../GameState';


export default compose(
  connect(
    state => ({
      playerStep: state.game.playerStep,
      winner: state.game.winner,
    }),
    dispatch => ({
      setPlayerStep: () => dispatch(setPlayerStep()),
      setWinner: winner => dispatch(setWinner(winner)),
    }),
  ),
  withState('count', 'setCount', 0),
  withHandlers({
    incrementCount: props => () => {
      props.setCount(props.count + 1);
    },
  }),
)(ColumnPresentation);

此文件代表一个视图

import React from 'react';
import Cell from '../../../components/Cell';


function columnPresentation({
  arrayFiller,
  incrementCount,
}) {
  return (
    <div>
      <div
        onClick={incrementCount}
        style={{ flexDirection: 'column', display: 'inline-block' }}
        role="button"
        tabIndex={0}
      >
        <p>Hello</p>
      </div>
    </div>
  );
}


export default columnPresentation;

import React from 'react';
import ColumnPresentation from '../column/ColumnPresentationContainer';

function homePresentation() {
  return (
    <div style={styles.container}>
      <ColumnPresentation />
      <ColumnPresentation />
    </div>
  );
}

const styles = {
  container: {
    flex: 1,
    justifyContent: 'center',
    alignItems: 'center',
    backgroundColor: 'white',
  },
};

export default homePresentation;

如何将一个无状态分量相乘，以使列表的每个分量具有独立的状态？ 当我单击其中一个组件时，状态已更改。如何克隆此组件，以便克隆具有独立状态