2

I want to find a better, more elegant solution than using 2 iteration to find a value in an object. I looked at functions to search within an array such as find,indexOf but was wondering I could search through the object without using a for or forEach.

Let say I want to find all swimmers. What I got so far. 

members = [
  { name: "Sue", hobby: ["Swimming", "Running", "Dancing"] },
  { name: "Sam", hobby: ["Cars", "Travelling"] },
  { name: "John", hobby: ["Reading", "Swimming"] },
  { name: "Rob", hobby: ["Running", "Coding"] },      
];
function findSwimmers(members, hobby) {      
  let swimmers = [];      
  members.forEach(function (e){        
    e.hobby.forEach(function (element){          
      if (element.toLowerCase() === hobby.toLowerCase()) {
        swimmers.push(e);
      }
    });
  });
  return swimmers;
}
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1
  • 3
    Any method of searching through an array will use a loop somehow... Commented Aug 20, 2018 at 8:35

5 Answers 5

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5

You can use filter in combination with some by passing callback functions as argument.

let members = [
  { name: "Sue", hobby: ["Swimming", "Running", "Dancing"] },
  { name: "Sam", hobby: ["Cars", "Travelling"] },
  { name: "John", hobby: ["Reading", "Swimming"] },
  { name: "Rob", hobby: ["Running", "Coding"] },      
];

function findSwimmers(members, hobby) {      
  return members_swimming = members.filter(({hobby}) => hobby.some(item => item == "Swimming"));
                                
}

console.log(findSwimmers(members, "Swimming"));

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4 Comments

Suggest incorporating the OP's toLowerCase (or similar) rather than hardcoding "Swimming". And possibly even mention String#localeCompare with sensitivity: "base" (where available).
This works great returning the names, but I want to return the object, same as my function would....
I don't know why you added map and assignment. The return statement can be reduced to: return members.filter(member => member.hobby.some(item => item == hobby)); I think that's what the OP wants.
@PWT, just remove the map method.
2

You can use Array.filter() and Array.map() with destructuring:

var members = [
  { name: "Sue", hobby: ["Swimming", "Running", "Dancing"] },
  { name: "Sam", hobby: ["Cars", "Travelling"] },
  { name: "John", hobby: ["Reading", "Swimming"] },
  { name: "Rob", hobby: ["Running", "Coding"] },      
];
var swimmers = members.filter(({hobby}) => hobby.includes('Swimming')).map(({name}) => name);
console.log(swimmers)

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1 Comment

The OP's code allows for a case-insensitive match. The above does not. Also, the OP's code returns the member object, not just their name. map has no place here.
1

Any solution is going to involve loops. The only question is whether the loops are in your code or a function you call.

In your case, filter and some seem like the tools to use:

function findSwimmers(members, hobby) {
  hobby = hobby.toLowerCase();
  return members.filter(member =>
    member.hobby.some(h => h.toLowerCase() === hobby)
  );
}

You might want a different function name. :-)

Live Example:

const members = [
  { name: "Sue", hobby: ["Swimming", "Running", "Dancing"] },
  { name: "Sam", hobby: ["Cars", "Travelling"] },
  { name: "John", hobby: ["Reading", "Swimming"] },
  { name: "Rob", hobby: ["Running", "Coding"] },      
];
function findSwimmers(members, hobby) {
  hobby = hobby.toLowerCase();
  return members.filter(member =>
    member.hobby.some(h => h.toLowerCase() === hobby)
  );
}

console.log(findSwimmers(members, "swimming"));
.as-console-wrapper {
  max-height: 100% !important;
}

Note that your original code didn't stop looping when it found a hobby match (and if there were duplicate entries in the hobbies array, would have added the same member more than once — but presumably there aren't meant to be duplicates in that array).

You might also use String#localeCompare, if available, with the sensitivity: "base" option rather than comparing lower-case strings.

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3 Comments

Thanks, this works great! I understand the native functions will use a loop. I just wanted to use them instead of my own loop. :)
Unrelated: Today I put up a meta question ( meta.stackoverflow.com/questions/372795/… ) regarding good/effective/... "auto" messages when commenting low quality newbie questions. The feedback I received felt pretty much "you are doing the totally wrong thing". Now I am simply wondering how you look at this? Do you have "generic" messages in your quiver, or do you write solely specific comments in such cases?
@GhostCat - I don't have time now to comment on those specific texts, but I strongly disagree with the people saying "downvote and move on." I do have a quiver of crafted messages, but I ensure that whenever I pick one, I tailor it to the specific situation (I can do that before posting). I don't tell them to take the tour if they have the Informed badge. I don't say "Welcome to Stack Overflow" if they've been a member for two years but this is only their third question. My overriding principle is: The comment must be useful, and must be welcoming. (This was...not always my rule.)
0

You can achieve this with a single line:

let swimmers = members.filter(person => person.hobby.includes('Swimming'));
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3 Comments

The OP's code allows for a case-insensitive match. The above does not.
This works great returning the names, but I want to return the object, same as my function would....
This should return you an array of objects with name and hobby (and anything else you want to add in), where the hobby includes swimming, which I think is what you were looking for? However - as has been pointed out, I haven't accounted for any case mismatch, so my solution is not currently fit for purpose.
0

First solution: bit of a hack but hey, no looping

Second solution: using Array.findIndex may reduce the amount of iterations in the inner loop (well, at least gives you some probability of less iterations) because (MDN):

... If such an element is found, findIndex immediately returns the index for that iteration ...

const findMembersWithValueForKey = (members, key, value) => 
  members.filter(member => member[key] instanceof Array
    && `#${member[key].join("#")}#`.toLowerCase()
         .indexOf(`#${value.toLowerCase()}#`) > -1
    || false);
    
const findMembersWithValueForKey2 = (members, key, value) => {
    value = value.toLowerCase();
    return members.filter(member => 
      member[key] instanceof Array
      && member[key].findIndex(v => v.toLowerCase() === value) > -1
      || false);
};
    
const members = [
  { name: "Sue", hobby: ["Swimming", "Running", "Dancing"] },
  { name: "Sam", hobby: ["Cars", "Travelling"] },
  { name: "John", hobby: ["Reading", "Swimming"] },
  { name: "Rob", hobby: ["Running", "Coding"] },      
];

console.log(findMembersWithValueForKey(members, "hobby", "swimming"));

console.log(findMembersWithValueForKey2(members, "hobby", "swimming"));

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1 Comment

filter and indexOf are both looping operations. Amusing approach, though. :-)

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