How to remove every other repeated word

#1

Hi there!

I’m doing an exercise which requires me to remove every other repeated word from an array of words.

Basically i’ve gor an array with 188 words called betterWords and another array called overusedWords which contains the ones i’ve got to remove every other time they appear in the betterWords array.

Could someone give me a hint on how to do it??

Here’s the code: (It’s wrong but I can’t figure out how to do it)

Basically I thought I would keep the count of the overusedWords and then remove those in an odd index position from the betterWords array, but obviously that does not work.

let story = 'Last weekend, I took literally the most beautiful bike ride of my life. The route is called "The 9W to Nyack" and it actually stretches all the way from Riverside Park in Manhattan to South Nyack, New Jersey. It\'s really an adventure from beginning to end! It is a 48 mile loop and it basically took me an entire day. I stopped at Riverbank State Park to take some extremely artsy photos. It was a short stop, though, because I had a really long way left to go. After a quick photo op at the very popular Little Red Lighthouse, I began my trek across the George Washington Bridge into New Jersey.  The GW is actually very long - 4,760 feet! I was already very tired by the time I got to the other side.  An hour later, I reached Greenbrook Nature Sanctuary, an extremely beautiful park along the coast of the Hudson.  Something that was very surprising to me was that near the end of the route you actually cross back into New York! At this point, you are very close to the end.';

let overusedWords = ['really', 'very', 'basically'];

let unnecessaryWords = ['extremely', 'literally', 'actually' ];


const storyWords = story.split(' ');
//console.log(storyWords.length);

const betterWords = storyWords.filter(word => {
  if(unnecessaryWords.includes(word) === false) {
    return word;
  }
});
//console.log(betterWords);

let reallyCount = 0;
let veryCount = 0;
let basicallyCount = 0;

const removeEveryOtherWord = betterWords.forEach(word => {

  if(word === overusedWords[0]) {
    reallyCount++;
  } else if (word === overusedWords[1]) {
    veryCount++;
  } else if (word === overusedWords[2]) {
    basicallyCount++;
  }

  for(i = 0; i < reallyCount.length; i++) {
    if(reallyCount[i] % 2 != 0) {
      betterWords.splice(reallyCount[i]);
    } else if (veryCount[i] % 2 != 0) {
      betterWords.splice(veryCount[i]);
    } else if (basicallyCount[i] % 2 != 0) {
      betterWords.splice(basicallyCount[i]);
    }
  }
});
#2

Sorry @ieahleen leen i don’t know how to correct that, same on my post yesterday… If you could explain i won’t do the same mistake

#3

sorry, your code was a weird color, I just added javascript on the same line of the backticks so that the code was using the correct color scheme

1 Like
#4

the overused words should be removed every other time in general, or for each specific word?

ex. “I’m really exited for this, basically I was waiting my whole life.”
the two words are to be considered separatelly or, there are two overused words and the second one should be removed?

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#5

Each word is to be considered separately, althought it depends on me because the intructions for the excercise do not really specify that.

This is what it says:

‘For the overused words, remove it every other time it appears.’

I gues I’d like to consider each word separately.

#6

can you explain what part of the code is doing that, and what logic you are following?

#7

The last function (removeEveryOtherWord) is supposed to do that, but obviously it is wrong… Basically it has to return the betterWords array (which is the story array without the unnecessary words), deleting every other of each of the overusedWords.

I was trying to keep the count of each overusedWord and then delete every other one for EACH one of the overusedWords…

Someone suggested me that it would have a better result if I use .filter() on betterWords and then create an object which stores the count of each of the overusedWords and then, using the object, returning true or false, choose the words i want to delete…

I haven’t worked with objects a lot yet as I only started JS a month ago. I’m googling on how to do it but it’s being a bit difficult to understand the logic and how then I’d use the object to pick the words I want to delete and returning the final filtered array.

I hope I’ve been clear enough :sweat_smile:

#8

So basically I came up with this but then was when someone suggested me to create an object instead of the IFs sequence…

let reallyCount = 0;
let veryCount = 0;
let basicallyCount = 0;

const removeEveryOtherWord = betterWords.filter(word => {
  if(!overusedWords.includes(word)) {
    return true;
  } else {
    if(word === overusedWords[0]) {
      reallyCount++;
      if(reallyCount % 2 != 0) {
        return true;
      } 
    } 
    if(word === overusedWords[1]) {
      veryCount++;
      if(veryCount % 2 != 0) {
        return true;
      } 
    }
    if(word === overusedWords[2]) {
      basicallyCount++;
      if(basicallyCount % 2 != 0) {
        return true;
      } 
    } 
  }
});

Plus I’ve been told that I shouldn’t refer to each of the overusedWords on my code…

#9

DRY principles are a thing, I mean, if it works is a first awesome step, but then, we want also to make it better (DRY is Don’t Reapeat Yourself)
so, you could use indexOf(), in this way you can also remove the first use of includes, you can do let index = ...; if (index < 0) {return true;} else { /* logic using the new index variable */ }
in this way you have one single block of code that does the thing for the three words, without having to repeat it for each of them

you could use an object or an array to keep count, in both ways you can use the same logic for everything

1 Like
#10

Thanks for your quick answer and your help on my last posts… and for explaining everything properly.

I might need to take a bit of time on undertanding that, I’ll let you know once I get it! :smile:

#11

I didn’t want to give you too much, but if you need help, ask again!

1 Like
#12

I think that’s the best way to learn… thinking about it until you realise it by yourself…

I’ve been all day yesterday trying to figure this out! :joy:

First trying to get the function to work, and once i did, trying to understand how to use an object to make it better and don’t repeat myself…

I ended up going to sleep and not get it done…

So i very much appreciate your help! I’ll get back to you

#13

Plus someone suggested to do something like this which i couldn’t understand…

my container = ......

for word from words
  current count = lookup(container, word)
  new count = current count + 1
  put(container, word, new count)
#14

that doesn’t seem any code language, seems to be pseudocode, or a different language that I don’t know, most probably pseudocode

pseudocode is not any programming language, it’s a way to write down quickly the logic

#15

Yes it is, he was just tying to explain with words what should i do… He obviously didn’t wanna tell me the exact code

#16

So… using indexOf on this…

 let index = ...; if (index < 0) {return true;} else { /* logic using the new index variable */ }

  1. let index = indexOf(word)???

  2. The first IF statement (index < 0) will always equal to false but the last word of the array right?

  3. Of what and how do i have to keep count?

  4. How would I use an object to keep count of the overusedWords and then delete the ones i want? How do i insert information from the .filter() loop in the object??

#17

but where are you searching the word? indexOf is an array method, so you need to use it on an array

try googling the indexOf method and see what you can find about it

#18

let index = betterWords.indexOf(word[, count]) ???

#19

I don’t think the secod argument does what you want: it says at which index to start searching, so maybe just delete it

you will need to keep track of the count, not there

1 Like
#20

Ok I came up with this but it doesn’t track each word individually…

let count = 0;

const removeEveryOtherWord = betterWords.filter(word => {
  let index = overusedWords.indexOf(word);
  if(index < 0) {
    return true;
  } else {
    count ++;
    if (count % 2 != 0) {
      return true;
    }
  }
});

How should i do to keep track of each of the overused words and delete every other one of EACH of them?

I’d like to do it so i do not mention each word on my code if you know what i mean…