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Get reference to arbitrarily (deep) nested node inside JSON/JS object using value of key


Get reference to arbitrarily (deep) nested node inside JSON/JS object using value of key

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Get reference to arbitrarily (deep) nested node inside JSON/JS object using value of key
Tag : javascript , By : akr
Date : November 28 2020, 12:01 PM

like below fixes the issue I have been looking all night on SO with lots of similar issues but none that directly solves my problem at the moment. So please have a look below. , Here's one possibility, using a for loop in a recursive function:
let data=[{id:777,name:"Level 1_section_1",children:[{id:778,name:"Level 2a",children:[]},{id:783,name:"Level 2b",children:[]}]},{id:786,name:"Level 1_section_2",children:[{id:781,name:"Level 2c",children:[]}]}];

const findNode = (arr, idToFind) => {
  for (const item of arr) {
    if (item.id === idToFind) {
      return item;
    }
    const possibleResult = findNode(item.children, idToFind);
    if (possibleResult) {
      return possibleResult;
    }
  }
};

console.log(findNode(data, 778));

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Arbitrarily Deep Nested Pattern Matching


Tag : development , By : Felix Almeida
Date : March 29 2020, 07:55 AM
Does that help The suggested solution
There seems to be no built-in construct to pattern-test nested heads automatically. We can achieve the goal by writing a function which would, for any given (sub)expression of the form f[___]...[___], efficiently determine f (which, with a slight abuse of terminology, we may call a symbolic head for the expression). Here is the code:
ClearAll[shead];
SetAttributes[shead, HoldAllComplete];
shead[expr_] := Scan[Return, Unevaluated[expr], {-1}, Heads -> True];
In[105]:= Cases[{f[1], g[f[1]], f[1, 2, 3][1], f[1][2][3][4]}, x_ /; shead[x] === f]

Out[105]= {f[1], f[1, 2, 3][1], f[1][2][3][4]}
Clear[headPattern];
headPattern[head_] := _?(Function[Null, shead[#] === head, HoldFirst]);

In[108]:= Cases[{f[1], g[f[1]], f[1, 2, 3][1], f[1][2][3][4]}, headPattern[f]]

Out[108]= {f[1], f[1, 2, 3][1], f[1][2][3][4]}
In[110]:= m = n = 0;
g[x_] := n++;
h[x_] := m++;
{Cases[Hold[f[g[1]][h[2]]], x_ /; shead[x] === f :> Hold[x], Infinity], {m, n}}

Out[113]= {{Hold[f[g[1]][h[2]]]}, {0, 0}}
sheadEval[expr_] := Scan[Return, expr, {-1}, Heads -> True]
In[114]:= {Cases[Hold[f[g[1]][h[2]]], x_ /; sheadEval[x] === f :> Hold[x], Infinity], {m, n}}

Out[114]= {{Hold[f[g[1]][h[2]]]}, {2, 1}}

How to render an arbitrarily deep nested list?


Tag : javascript , By : n800s
Date : March 29 2020, 07:55 AM
help you fix your problem Use a recursive template, e.g. like this:
var tree = {
  subItems: [
    { 
      name: "A", 
      subItems: [ { name:"AA", subItems: [] }, { name:"AB", subItems: [] }, { name:"AC", subItems: [] } ] 
    },
    { 
      name: "B", 
      subItems: [ { name:"BA", subItems: [] }, { name:"BB", subItems: [{name:"BB1 (etc)", subItems: []}] } ] 
    }
  ]
};

ko.applyBindings(tree);
<script src="https://cdnjs.cloudflare.com/ajax/libs/knockout/3.2.0/knockout-min.js"></script>

<script type="text/html" id="myTemplate">
    <ul data-bind="foreach: $data">
        <li>
            <label data-bind="text: name"></label>
            <div data-bind="template: { name: 'myTemplate', data: subItems }"></div>
        </li>
    </ul>
</script>

<div data-bind="template: { name: 'myTemplate', data: $root.subItems }"></div>

How to delete something in a JavaScript object that can be nested arbitrarily deep?


Tag : javascript , By : Noah
Date : March 29 2020, 07:55 AM
wish helps you I have a JavaScript object that manages all of my app's data. It looks something like this: , This'll do it.
remove = function(address, tree) {
    if (address.length === 1) { 
        delete tree.children[address[0]]; 
    } else {
        remove(address.slice(1), tree.children[address[0]]);
    }
}

Coping with arbitrarily deep nested lists


Tag : list , By : Brian Drum
Date : March 29 2020, 07:55 AM

Flattening arbitrarily deep nested lists


Tag : python , By : semicolonth
Date : January 02 2021, 06:48 AM
Any of those help You can use a recursive generator to yield elements from nested lists:
from typing import Collection

def check_nested(obj):
    for sub_obj in obj:
        # tuples, lists, dicts, and sets are all Collections
        if isinstance(sub_obj, Collection):
            yield from check_nested(sub_obj)
        else:
            yield sub_obj


l = [[[[[1, 2]]]]]
list(check_nested(l))
[1, 2]

# This will work for other formats
l = [[[[[1, 2]]]], [[3, 4]]]

list(check_nested(l))
[1, 2, 3, 4]
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