Question

我正在使用lodash进行一些过滤。

我具有以下功能：

filterByCardinalPoint = (key, value) => {
    const { passengersData } = this.props;
    console.log('passengersData', passengersData);
    return filter(passengersData, [key, value]);
  };

  callFilter = () => {
    passengersGoingNorth = this.filterByCardinalPoint('cardinalpoint', 'N')
      .length;
    passengersGoingSouth = this.filterByCardinalPoint('cardinalpoint', 'S')
      .length;
    passengersGoingWest = this.filterByCardinalPoint('cardinalpoint', 'W')
      .length;
    passengersGoingEast = this.filterByCardinalPoint('cardinalpoint', 'E')
      .length;
  };

我正在render方法中调用callFilter。该功能被调用约8次。我该如何优化它才能调用一次？

Answer 1

由于您需要各个方向的人数，因此可以将_.countBy()与_.property()一起使用。分解结果对象以将数字分配给变量。

示例：

const { countBy, property } = _;

let passengersGoingNorth, passengersGoingSouth, passengersGoingWest, passengersGoingEast;

const passengersData = [{ cardinalpoint: 'N' }, { cardinalpoint: 'N' }, { cardinalpoint: 'S' }, { cardinalpoint: 'N' }, { cardinalpoint: 'E' }];

({
  N: passengersGoingNorth = 0,
  S: passengersGoingSouth = 0,
  W: passengersGoingWest = 0,
  E: passengersGoingEast = 0
} = countBy(passengersData, property('cardinalpoint')));

console.log(passengersGoingNorth, passengersGoingSouth, passengersGoingWest, passengersGoingEast); // 3, 1, 0, 1

<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.11/lodash.min.js"></script>

Answer 2

那也许太老套了，但是一个简单的for循环呢：

 for(const { cardinalpoint } of passengersData) {
   if(cardinalpoint === "N")
    passengersGoingNorth++;
   if(cardinalpoint === "S")
    passengersGoingSouth++;
   if(cardinalpoint === "E")
    passengersGoingEast++;
   if(cardinalpoint === "W")
    passengersGoingWest++;
 }

如果这太重复了，只需将方向用作查找键：

 const directions = { N: 0, S: 0, W: 0, E: 0 };
 for(const passenger of passengersData)
   directions[ passenger.cardinalpoint ]++;

然后，您可以将乘客以directions.S的身分带到南方。

当我需要同时过滤多个事物时，如何调用一次filter？

2 个答案: