返回与特定分支遍历匹配的对象。

huangapple
go评论99阅读模式
英文:

Return object that matches specific branch traversal

问题

这个问题起源于这里。

给定此数据：

  1. {
  2.     "tabs-3": {
  3.         "Collection A": {
  4.             "Level 2": {
  5.                 "Data A": {
  6.                     "tab3graph25": {
  7.                         "30/04": 21750,
  8.                         "31/03": 19428,
  9.                         "29/05": 20955
  10.                     }
  11.                 }
  12.             }
  13.         },
  14.         "Collection B": {
  15.             "Level 2": {
  16.                 "Data A": {
  17.                     "tab3graph33": {
  18.                         "30/04": 56863,
  19.                         "31/03": 62298,
  20.                         "29/05": 56044
  21.                     }
  22.                 }
  23.             }
  24.         },
  25.         "Collection C": {
  26.             "Level 2": {
  27.                 "Data A": {
  28.                     "tab3graph40": {
  29.                         "30/04": 56044,
  30.                         "31/03": 62298,
  31.                         "29/05": 56863
  32.                     }
  33.                 }
  34.             }
  35.         }
  36.     }
  37. }

函数的形式是这样的，它将以对象和搜索列表作为输入，例如，var searchList = [Collection B, Level 2, Data A]，然后返回tab3graph33:{...values...}。

初始函数可参考如下：

  1. function searchObjectForValues(obj, searchList) {
  2.     var matchingObjects = {};
  4.     function recursiveSearch(currentObj, path) {
  5.         if (typeof currentObj === 'object' && currentObj !== null) {
  6.             for (var key in currentObj) {
  7.                 if (currentObj.hasOwnProperty(key)) {
  8.                     var value = currentObj[key];
  9.                     var currentPath = path.concat(key);
  11.                     if (searchList.includes(key)) {
  12.                         matchingObjects[key] = currentObj[key];
  13.                     }
  15.                     recursiveSearch(value, currentPath);
  16.                 }
  17.             }
  18.         }
  19.     }
  21.     recursiveSearch(obj, []);
  23.     return matchingObjects;
  24. }
英文:

This question started here

The conversation exceeded the original scope, so asking the expanded question here.

Given this data:

  1. {
  2. "tabs-3": {
  3.     "Collection A": {
  4.         "Level 2": {
  5.             "Data A": {
  6.                 "tab3graph25": {
  7.                     "30/04": 21750,
  8.                     "31/03": 19428,
  9.                     "29/05": 20955
  10.                 }
  11.             }
  12.         }
  13.     }
  14.     "Collection B": {
  15.         "Level 2": {
  16.             "Data A": {
  17.                 "tab3graph33": {
  18.                     "30/04": 56863,
  19.                     "31/03": 62298,
  20.                     "29/05": 56044
  21.                 }
  22.             }
  23.         }
  24.     }
  25.     "Collection C": {
  26.         "Level 2": {
  27.             "Data A": {
  28.                 "tab3graph40": {
  29.                     "30/04": 56044,
  30.                     "31/03": 62298,
  31.                     "29/05": 56863
  32.                 }
  33.             }
  34.         }
  35.     }
  36. }

}

What would a function look like that would take as it's input the holding object to search, in this case, "tabs-3", and a key set to look for, e.g., var searchList = [Collection B, Level 2, Data A] and then return "tab3graph33:{...values...}

The original starting function I have is here for reference ,

  1. function searchObjectForValues(obj, searchList) {
  2.   var matchingObjects = {};
  4.   function recursiveSearch(currentObj, path) {
  5.     if (typeof currentObj === 'object' && currentObj !== null) {
  6.       for (var key in currentObj) {
  7.         if (currentObj.hasOwnProperty(key)) {
  8.           var value = currentObj[key];
  9.           var currentPath = path.concat(key);
  11.           if (searchList.includes(key)) {
  12.             matchingObjects[key] = currentObj[key];
  13.           }
  15.           recursiveSearch(value, currentPath);
  16.         }
  17.       }
  18.     }
  19.   }
  21.   recursiveSearch(obj, []);
  23.   return matchingObjects;
  24. }

Although the answers in the original question are much more elegant than mine.

答案1

得分: 1

我们可以为此编写一个相当简单的递归函数。通常，实用程序库中已经包含了这样一个函数，如果您已经在使用类似 Underscore 或 Ramda 的东西的话。

  1. const getPath = ([p, ...ps]) => (o) =>
  2.   p == undefined ? o : getPath(ps)(o && o[p])
  4. const data = {
  5.   "tabs-3": {
  6.     "Collection A": {
  7.       "Level 2": {
  8.         "Data A": {
  9.           tab3graph25: {
  10.             "30/04": 21750,
  11.             "31/03": 19428,
  12.             "29/05": 20955
  13.           }
  14.         }
  15.       },
  16.       "Collection B": {
  17.         "Level 2": {
  18.           "Data A": {
  19.             tab3graph33: {
  20.               "30/04": 56863,
  21.               "31/03": 62298,
  22.               "29/05": 56044
  23.             }
  24.           }
  25.         }
  26.       },
  27.       "Collection C": {
  28.         "Level 2": {
  29.           "Data A": {
  30.             tab3graph40: {
  31.               "30/04": 56044,
  32.               "31/03": 62298,
  33.               "29/05": 56863
  34.             }
  35.           }
  36.         }
  37.       }
  38.     }
  39.   }
  40. }
  42. const tabs3 = data['tabs-3']
  43. console.log('Partial path:', getPath(['Collection B', 'Level 2', 'Data A'])(tabs3))
  45. // 或者
  46. console.log('Full-path:', getPath(['tabs-3', 'Collection B', 'Level 2', 'Data A'])(data))

第一个 console.log 显示了您可能希望使用的方式，其中引用了数据中 'tabs-3' 属性中的对象。第二个示例展示了我更喜欢的方式，将 'tabs-3' 作为属性列表中的一个属性。

英文:

We can write a fairly simple recursive function for this. Utility libraries usually include one, too, if you're already using something like Underscore or Ramda.

<!-- begin snippet: js hide: false console: true babel: false -->

<!-- language: lang-js -->

  1. const getPath = (
    ) =&gt; (o) =&gt;
  2.   p == undefined ? o : getPath(ps) (o &amp;&amp; o
    )
  4. const data = {&quot;tabs-3&quot;: {&quot;Collection A&quot;: {&quot;Level 2&quot;: {&quot;Data A&quot;: {tab3graph25: {&quot;30/04&quot;: 21750, &quot;31/03&quot;: 19428, &quot;29/05&quot;: 20955}}}}, &quot;Collection B&quot;: {&quot;Level 2&quot;: {&quot;Data A&quot;: {tab3graph33: {&quot;30/04&quot;: 56863, &quot;31/03&quot;: 62298, &quot;29/05&quot;: 56044}}}}, &quot;Collection C&quot;: {&quot;Level 2&quot;: {&quot;Data A&quot;: {tab3graph40: {&quot;30/04&quot;: 56044, &quot;31/03&quot;: 62298, &quot;29/05&quot;: 56863}}}}}}
  7. const tabs3 = data[&#39;tabs-3&#39;]
  8. console .log (&#39;Partial path:&#39;, getPath ([&#39;Collection B&#39;, &#39;Level 2&#39;, &#39;Data A&#39;])(tabs3))
  10. // or
  11. console .log (&#39;Full-path:&#39;, getPath ([&#39;tabs-3&#39;, &#39;Collection B&#39;, &#39;Level 2&#39;, &#39;Data A&#39;])(data))

<!-- language: lang-css -->

  1. .as-console-wrapper {max-height: 100% !important; top: 0}

<!-- end snippet -->

The first console.log shows how you would use this the way you seem to want, with a reference to the object in the &#39;tabs-3&#39; property of your data. The second one shows my preferred way, using &#39;tabs-3&#39; as one more property in the list.

答案2

得分: 1

以下是您请求的翻译部分：

这是一个有趣的get函数，它接受*和**通配符。给定这个data -

  1. const data = {
  2.   a: {
  3.     b: {
  4.       x: {y:1, z:2},
  5.       y: {z:3},
  6.     },
  7.     c: {
  8.       x: {y:4},
  9.       y: {z:5},
  10.     },
  11.     d: {x:6},
  12.   },
  13.   b: {c: {z:7}}
  14. }

我们可以将get写成一个递归生成器 -

  1. function *get(t, path) {
  2.   function *loop(t, path, match, wild) {
  3.     if (path.length === 0)
  4.       yield [match, t]
  5.     else if (path[0] === &quot;**&quot;)
  6.       yield *loop(t, path.slice(1), match, true)
  7.     else if (Object(t) === t)
  8.       for (const k of Object.keys(t))
  9.         if (path[0] === &quot;*&quot; || path[0] === k)
  10.           yield *loop(t[k], path.slice(1), [...match, k], false)
  11.         else if (wild)
  12.           yield *loop(t[k], path, [...match, k], true)
  13.   }
  14.   yield *loop(t, path, [], false)
  15. }

搜索以a开头，然后跟一个通配符，然后是x的路径 -

  1. for (const [path, value] of get(data, [&quot;a&quot;,&quot;*&quot;,&quot;x&quot;]))
  2.   console.log(
  3.     JSON.stringify(path),
  4.     JSON.stringify(value)
  5.   )
  1. [&quot;a&quot;,&quot;b&quot;,&quot;x&quot;] {&quot;y&quot;:1,&quot;z&quot;:2}
  2. [&quot;a&quot;,&quot;c&quot;,&quot;x&quot;] {&quot;y&quot;:4}
  3. [&quot;a&quot;,&quot;d&quot;,&quot;x&quot;] 6

搜索以z结尾的任何路径 -

  1. for (const [path, value] of get(data, [&quot;**&quot;,&quot;z&quot;]))
  2.   console.log(
  3.     JSON.stringify(path),
  4.     JSON.stringify(value)
  5.   )
  1. [&quot;a&quot;,&quot;b&quot;,&quot;x&quot;,&quot;z&quot;] 2
  2. [&quot;a&quot;,&quot;b&quot;,&quot;y&quot;,&quot;z&quot;] 3
  3. [&quot;a&quot;,&quot;c&quot;,&quot;y&quot;,&quot;z&quot;] 5
  4. [&quot;b&quot;,&quot;c&quot;,&quot;z&quot;] 7

获取完整路径a，c，x，y -

  1. for (const [path, value] of get(data, [&quot;a&quot;,&quot;c&quot;,&quot;x&quot;,&quot;y&quot;]))
  2.   console.log(
  3.     JSON.stringify(path),
  4.     JSON.stringify(value)
  5.   )
  1. [&quot;a&quot;,&quot;c&quot;,&quot;x&quot;,&quot;y&quot;] 4

创建复杂的模式 -

  1. for (const [path, value] of get(data, [&quot;*&quot;,&quot;b&quot;,&quot;**&quot;,&quot;y&quot;]))
  2.   console.log(
  3.     JSON.stringify(path),
  4.     JSON.stringify(value)
  5.   )
  1. [&quot;a&quot;,&quot;b&quot;,&quot;x&quot;,&quot;y&quot;] 1
  2. [&quot;a&quot;,&quot;b&quot;,&quot;y&quot;] {&quot;z&quot;:3}

获取不存在的路径 -

  1. for (const [path, value] of get(data, [&quot;x&quot;,&quot;y&quot;,&quot;z&quot;]))
  2.   console.log(
  3.     JSON.stringify(path),
  4.     JSON.stringify(value)
  5.   )
  1. // &lt;empty result&gt;
英文:

Here's a fun get function which accepts * and ** wildcards. Given this data -

  1. const data = {
  2.   a: {
  3.     b: {
  4.       x: {y:1, z:2},
  5.       y: {z:3},
  6.     },
  7.     c: {
  8.       x: {y:4},
  9.       y: {z:5},
  10.     },
  11.     d: {x:6},
  12.   },
  13.   b: {c: {z:7}}
  14. }

We can write get as a recursive generator -

  1. function *get(t, path) {
  2.   function *loop(t, path, match, wild) {
  3.     if (path.length === 0)
  4.       yield [match, t]
  5.     else if (path[0] === &quot;**&quot;)
  6.       yield *loop(t, path.slice(1), match, true)
  7.     else if (Object(t) === t)
  8.       for (const k of Object.keys(t))
  9.         if (path[0] === &quot;*&quot; || path[0] === k)
  10.           yield *loop(t[k], path.slice(1), [...match, k], false)
  11.         else if (wild)
  12.           yield *loop(t[k], path, [...match, k], true)
  13.   }
  14.   yield *loop(t, path, [], false)
  15. }

Search for a path starting with a, followed by one wildcard, followed by x -

  1. for (const [path, value] of get(data, [&quot;a&quot;,&quot;*&quot;,&quot;x&quot;]))
  2.   console.log(
  3.     JSON.stringify(path),
  4.     JSON.stringify(value)
  5.   )
  1. [&quot;a&quot;,&quot;b&quot;,&quot;x&quot;] {&quot;y&quot;:1,&quot;z&quot;:2}
  2. [&quot;a&quot;,&quot;c&quot;,&quot;x&quot;] {&quot;y&quot;:4}
  3. [&quot;a&quot;,&quot;d&quot;,&quot;x&quot;] 6

Search for any path ending in z -

  1. for (const [path, value] of get(data, [&quot;**&quot;,&quot;z&quot;]))
  2.   console.log(
  3.     JSON.stringify(path),
  4.     JSON.stringify(value)
  5.   )
  1. [&quot;a&quot;,&quot;b&quot;,&quot;x&quot;,&quot;z&quot;] 2
  2. [&quot;a&quot;,&quot;b&quot;,&quot;y&quot;,&quot;z&quot;] 3
  3. [&quot;a&quot;,&quot;c&quot;,&quot;y&quot;,&quot;z&quot;] 5
  4. [&quot;b&quot;,&quot;c&quot;,&quot;z&quot;] 7

Get full path a, c, x, y -

  1. for (const [path, value] of get(data, [&quot;a&quot;,&quot;c&quot;,&quot;x&quot;,&quot;y&quot;]))
  2.   console.log(
  3.     JSON.stringify(path),
  4.     JSON.stringify(value)
  5.   )
  1. [&quot;a&quot;,&quot;c&quot;,&quot;x&quot;,&quot;y&quot;] 4

Make complex patterns -

  1. for (const [path, value] of get(data, [&quot;*&quot;,&quot;b&quot;,&quot;**&quot;,&quot;y&quot;]))
  2.   console.log(
  3.     JSON.stringify(path),
  4.     JSON.stringify(value)
  5.   )
  1. [&quot;a&quot;,&quot;b&quot;,&quot;x&quot;,&quot;y&quot;] 1
  2. [&quot;a&quot;,&quot;b&quot;,&quot;y&quot;] {&quot;z&quot;:3}

Get non-existent path -

  1. for (const [path, value] of get(data, [&quot;x&quot;,&quot;y&quot;,&quot;z&quot;]))
  2.   console.log(
  3.     JSON.stringify(path),
  4.     JSON.stringify(value)
  5.   )
  1. // &lt;empty result&gt;

Verify the results below in your own browser -

<!-- begin snippet: js hide: false console: true babel: false -->

<!-- language: lang-js -->

  1. function *get(t, path) {
  2.   function *loop(t, path, match, wild) {
  3.     if (path.length === 0)
  4.       yield [match, t]
  5.     else if (path[0] === &quot;**&quot;)
  6.       yield *loop(t, path.slice(1), match, true)
  7.     else if (Object(t) === t)
  8.       for (const k of Object.keys(t))
  9.         if (path[0] === &quot;*&quot; || path[0] === k)
  10.           yield *loop(t[k], path.slice(1), [...match, k], false)
  11.         else if (wild)
  12.           yield *loop(t[k], path, [...match, k], true)
  13.   }
  14.   yield *loop(t, path, [], false)
  15. }
  17. const data = {
  18.   a: {
  19.     b: {
  20.       x: {y:1, z:2},
  21.       y: {z:3},
  22.     },
  23.     c: {
  24.       x: {y:4},
  25.       y: {z:5},
  26.     },
  27.     d: {x:6},
  28.   },
  29.   b: {c: {z:7}}
  30. }
  32. for (const [path, value] of get(data, [&quot;a&quot;,&quot;*&quot;,&quot;x&quot;]))
  33.   console.log(
  34.     JSON.stringify(path),
  35.     JSON.stringify(value)
  36.   )
  37.   
  38. for (const [path, value] of get(data, [&quot;**&quot;,&quot;z&quot;]))
  39.   console.log(
  40.     JSON.stringify(path),
  41.     JSON.stringify(value)
  42.   )
  43.   
  44. for (const [path, value] of get(data, [&quot;a&quot;,&quot;c&quot;,&quot;x&quot;,&quot;y&quot;]))
  45.   console.log(
  46.     JSON.stringify(path),
  47.     JSON.stringify(value)
  48.   )
  49.   
  50. for (const [path, value] of get(data, [&quot;x&quot;,&quot;y&quot;,&quot;z&quot;]))
  51.   console.log(
  52.     JSON.stringify(path),
  53.     JSON.stringify(value)
  54.   )

<!-- language: lang-css -->

  1. .as-console-wrapper { min-height: 100%; top: 0; }

<!-- end snippet -->

sans generators

Here's the same solution without using generators. The interface is identical however it always returns an array instead of an iterator -

  1. function get(t, path) {
  2.   const loop = (t, path, match, wild) =&gt;
  3.     path.length === 0
  4.       ? [[match, t]]
  5.   : path[0] === &quot;**&quot;
  6.       ? loop(t, path.slice (1), match, true)
  7.   : Object(t) === t
  8.       ? Object.keys(t).flatMap(k =&gt;
  9.           path[0] === &quot;*&quot; || path[0] === k
  10.             ? loop(t[k], path.slice (1), [...match, k], false)
  11.         : wild
  12.             ? loop(t[k], path, [...match, k], true)
  13.         : []
  14.         )
  15.   : []
  16.   return loop(t, path, [], false)
  17. }

huangapple
  • 本文由 发表于 2023年6月8日 23:00:02
  • 转载请务必保留本文链接：https://go.coder-hub.com/76433185.html
  • javascript
  • recursion
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