如何访问不相似的嵌套对象并遍历它们？

Question

我有一个到 websocket 的连接，它给我这种格式的数据。

{lastUpdateId: 2303,Bids: [[ '12312.81000000', '0.77085400' ],[ '23423.79000000', '0.02179700' ],[ '23423.76000000', '0.00550000' ],[ '23422.70000000', '0.00052800' ],[ '64536.24000000', '0.00902500' ]],}

我想将“bids” arrays 的所有主要值放入一个新的 xBids 数组中，将“bids” arrays 的次要值放入一个新的 yBids 数组中。 所以它看起来像这样。

xBids = ['12312.81000000','23423.79000000','23423.76000000','23422.70000000''64536.24000000'] yBids = ['0.77085400','0.02179700','0.00550000','0.00052800''0.00902500']

非常感谢任何帮助。

Answer 1

使用 function .map()返回一个新数组，其中包含从现有数组派生的值。

 const result = {lastUpdateId: 2303,Bids: [[ '12312.81000000', '0.77085400' ],[ '23423.79000000', '0.02179700' ],[ '23423.76000000', '0.00550000' ],[ '23422.70000000', '0.00052800' ],[ '64536.24000000', '0.00902500' ]]} const xbids = result.Bids.map(bid=>bid[0]); const ybids = result.Bids.map(bid=>bid[1]); console.log('xbids: ', xbids); console.log('ybids: ', ybids);

Answer 2

您可以使用Array#map 。

 const obj = {lastUpdateId: 2303,Bids: [[ '12312.81000000', '0.77085400' ],[ '23423.79000000', '0.02179700' ],[ '23423.76000000', '0.00550000' ],[ '23422.70000000', '0.00052800' ],[ '64536.24000000', '0.00902500' ]],} const arr1 = obj.Bids.map(x => x[0]), arr2 = obj.Bids.map(x => x[1]); console.log(arr1); console.log(arr2);

只需遍历一次数组即可，您可以使用Array#reduce 。

 const obj = {lastUpdateId: 2303,Bids: [[ '12312.81000000', '0.77085400' ],[ '23423.79000000', '0.02179700' ],[ '23423.76000000', '0.00550000' ],[ '23422.70000000', '0.00052800' ],[ '64536.24000000', '0.00902500' ]],}; const [arr1, arr2] = obj.Bids.reduce((acc,[x,y])=>(acc[0].push(x), acc[1].push(y), acc), [[], []]); console.log(arr1); console.log(arr2);

Answer 3

这有效（ES6 语法）：

const testObj = {lastUpdateId: 2303,Bids: [[ '12312.81000000', '0.77085400' ],[ '23423.79000000', '0.02179700' ],[ '23423.76000000', '0.00550000' ],[ '23422.70000000', '0.00052800' ],[ '64536.24000000', '0.00902500' ]],};
const xBids = [];
const yBids = [];
testObj.Bids.map(bid => { xBids.push(bid[0]), yBids.push(bid[1])});

Answer 4

这很简单，但需要两次迭代。

let xBids = response.Bids.map(element=>{
  element[0];
}
)

let yBids = response.Bids.map(element=>{
element[1];
})

单次迭代法

let xBids= [];
let yBids = [];
response.Bids.forEach((element)=>{
xbids.push(element[0]);
yBids.push(element[1])
}
)

Answer 5

使用map() function 创建 2 个新的 arrays，其中包含所需的数据，如下所示。

 const data = { lastUpdateId: 2303, Bids: [ ['12312.81000000', '0.77085400'], ['23423.79000000', '0.02179700'], ['23423.76000000', '0.00550000'], ['23422.70000000', '0.00052800'], ['64536.24000000', '0.00902500'] ] }; const xBids = data.Bids.map(bid => bid[0]); const yBids = data.Bids.map(bid => bid[1]); console.log(xBids); console.log(yBids);

Answer 6

您也可以使用reduce在1个操作中完成此任务：

主意：

• 创建嵌套数组结构。
• 循环投标和每次迭代，将当前值推送到必要的数组。
• 也为下一次迭代返回这个嵌套结构。
• 在完成时，接收这个结构是分开的。

 const data={lastUpdateId: 2303,Bids: [[ '12312.81000000', '0.77085400' ],[ '23423.79000000', '0.02179700' ],[ '23423.76000000', '0.00550000' ],[ '23422.70000000', '0.00052800' ],[ '64536.24000000', '0.00902500' ]],} const [xBids, yBids] = data.Bids.reduce(([xBidsList, yBidsList], [x, y]) => [ [...xBidsList, x], [...yBidsList, y] ], [[], []] ) console.log(xBids, yBids)

为清楚起见的简化版本：

 const data={lastUpdateId: 2303,Bids: [[ '12312.81000000', '0.77085400' ],[ '23423.79000000', '0.02179700' ],[ '23423.76000000', '0.00550000' ],[ '23422.70000000', '0.00052800' ],[ '64536.24000000', '0.00902500' ]],} const xBids = []; const yBids = []; data.Bids.reduce((bidLists, bid) => { const xValues = bidLists[0] const yValues = bidLists[1] xValues.push(bid[0]) yValues.push(bid[1]) return [xValues, yValues] }, [ xBids, yBids ]) console.log(xBids, yBids)