Question

我有一个带有目录和频率（目录的清洗频率）的parsed_JSON。

如果所有子目录频率都设置为“ not”，我想删除该部分。

例如：

[ { label: 'Offices',
     rows: [ { freq: '1w'},{ freq: '2w'},{ freq: 'not'} ]  },
  { label: 'Kitchen and community areas – Extras',
    rows: [ { freq: 'not'},{ freq: 'not'},{ freq: 'not'} ] 
},
]

在这种情况下，应删除标有“厨房和社区区域–其他”的部分。

我通过以下代码实现了这一点：

  const mapped_sections      = _.map(parsed_json, section => ({
        label   : section.label,
        rows    : _.map(section.rows, row => _.merge({}, default_row, row)),
    }));

    const sections = _.forEach(mapped_sections, (section, i) => {
        let not_length_count = 0;
        _.forEach(section, (rows) => {
            _.forEach(rows, (row) => {
                if (row.freq === "not") {
                    not_length_count += 1;
                }
            });
            if (not_length_count === rows.length) {
                mapped_sections.splice(i, 1);
            }
        });
    });

但是我想使用filter()之类的ES6方法并仅通过mapped_sections进行映射来重构它

我一直在尝试，但是被卡在这里：

const sections      = _.map(parsed_json, (section, i) => {
        const test = ((section.rows.filter(item => item.freq === "not"))
                && (section.rows.filter(item => item.freq === "not").length === section.rows.length)
            ? section.rows.slice(i, 1)
            : section.rows
        );
        return (
            section.label,
            _.map(test, row => _.merge({}, default_row, row))
        );
    });

任何帮助将不胜感激。谢谢！

Answer 1

您可以使用every函数在元素行上运行 not ! ，如下所示：

const myList = [
  {
    label: 'Offices',
    rows: [{ freq: '1w'},{ freq: '2w'},{ freq: 'not'}]
  },
  {
    label: 'Kitchen and community areas – Extras',
    rows: [{ freq: 'not'},{ freq: 'not'},{ freq: 'not'}] 
  },
]

const result = myList.filter(el => !el.rows.every(r => r.freq === 'not'))

console.log(result)

所有freq个not的项目都被过滤掉。

Answer 2

让我知道它是否有效

let arr = [ { label: 'Offices',
     rows: [ { freq: '1w'},{ freq: '2w'},{ freq: 'not'} ]  },
  { label: 'Kitchen and community areas – Extras',
    rows: [ { freq: 'not'},{ freq: 'not'},{ freq: 'not'} ] 
},
]

arr.filter(ar => {
    let rowCount = 0;
    ar.rows.forEach(row => {
        row.freq === 'not' ? rowCount++: rowCount;
    })
    return rowCount !== ar.rows.length
}) 
console.log(arr);

Answer 3

结合_.find和_.some

let data = [{
    label: 'Offices',
    rows: [{
      freq: '1w'
    }, {
      freq: '2w'
    }, {
      freq: 'not'
    }]
  },
  {
    label: 'Kitchen and community areas – Extras',
    rows: [{
      freq: 'not'
    }, {
      freq: 'not'
    }, {
      freq: 'not'
    }]
  }
];

const result = _.filter(data, item => _.some(item.rows, (x) =>  x.freq !== 'not'))

console.log({
  result
});

<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.15/lodash.min.js"></script>

根据过滤值删除

3 个答案: