'Can you use an if/else inside a .filter() / Is there any other way?
TASK:
There is an array of words called overusedWords. These are words overused in this story. You want to let the user of your program know how many times they have used these overused words. There are two ways to achieve this. Try it on your own first. If you need help, consult the hint.
HINT:
1.You can iterate over the betterWords array three separate times (once for each of the words in the overusedWords array). Create a variable that represents the total times that word appears. Add 1 to the variable every time the current word is the same as that word.
2.You can make this simpler by using one if, and two else if statements in the function code block of your iterator. That way, you can gather the counts of all three overused words at one time.
CODE:
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 storyWords = story.split(' ')
console.log(storyWords)
console.log(storyWords.length)
let overusedWords = ['really', 'very', 'basically'];
let unnecessaryWords = ['extremely', 'literally', 'actually' ];
let betterWords = storyWords.filter( storyWords =>
!unnecessaryWords.includes(storyWords.toLowerCase()));
console.log(betterWords)
Solution 1:[1]
You can define your own function to act as the filter. For example if I was given an array of ages and only wanted to return the ages over between 20 and 35 I could do the following:
var ages = [32, 33, 16, 40];
console.log(ages.filter(checkAge));
function checkAge(age) {
if (age > 20 && age < 35) {
return age;
}
}
this returns an output of:
Array [ 32, 33 ]
Solution 2:[2]
What about using Array.prototype.reduce()
and a single if
?
const 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.`;
const storyWords = story.toLowerCase().split(' ');
const overusedWords = ['really', 'very', 'basically'];
const results = storyWords.reduce((acc, word) => {
if (overusedWords.includes(word)) {
acc[word] = (acc[word] || 0) + 1;
}
return acc;
}, {});
console.log(results)
As you can see in this example, can also call String.prototype.toLowerCase()
a single time outside filter()
/reduce()
.
If you still prefer to use filter()
, the example above should be easy to adapt, so I'll leave that for you.
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
Solution | Source |
---|---|
Solution 1 | Nebri |
Solution 2 | Danziger |