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LESSON 234 OF 671 · SETS

Removing From Sets in JavaScript

Lesson Read the full explanation

TL ; DR:

delete(x) takes x out of a set. A for...of loop visits every value once.

let colors = new Set(["red", "blue", "green"]);

colors.delete("blue");

colors.delete("pink");          // not in it, nothing happens

for (let color of colors) {
    console.log(color);         // red, then green
}

Full lesson:

s.delete(x) removes x from the set s. It returns true if x was there. If it wasn't, nothing happens and it returns false.

for (let x of s) visits each value of the set once.

A JavaScript set loops in the order the values were added. Most languages don't promise any order for a set. So a set is best for work where the order doesn't matter. A total, a count and a check are fine.

A set has no indexes. A for loop with i and s[i] doesn't work on it. Use for...of.

Your task

In a word game, the set found holds the words a player found. The words "qqq" and "zzz" aren't allowed. Remove both from found.

The player never found "zzz", so removing it changes nothing.

Then add up the points. A word scores one point per letter. Loop over found and add each word's length to score.

The code already prints how many words are left, and the points.

Expected output:

4 words
14 points

STEP 1 OF 3

Guided

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JAVASCRIPT · SETS

Removing From Sets in JavaScript, explained

Removing a value

s.delete(x) takes x out of the set s. It returns true if x was there. If it wasn't, nothing happens: there is no error, and the call returns false.

let colors = new Set(["red", "blue", "green"]);
console.log(colors.delete("blue"));   // true
console.log(colors.delete("pink"));   // false, it was never there

true
false

That makes delete safe to call without checking has first.

Looping over a set

for (let x of s) visits each value of the set once.

for (let color of colors) {
    console.log(color);
}

red
green

A JavaScript set loops in the order the values were added. Most languages don't promise any order for a set, so a set is best for work where the order doesn't matter: a total, a count or a check.

Traps

No indexes. A for loop with i and s[i] doesn't work on a set. With the set of four words from the exercise below, found[0], found[1] and so on are all undefined. Use for...of.
The wrong method name. Arrays use splice or pop to remove values. A set uses delete.
Type matters. A set compares values exactly. For new Set([1, 2, 3]), delete("1") returns false and leaves the size at 3, because the text "1" is not the number 1. Then delete(1) returns true and the size drops to 2.
Re-adding goes to the end. If you delete a value and add it again, it counts as newly added, so it is visited last. After deleting 3 from [3, 1, 2] and adding it back, the loop visits 1, 2, 3.
Counting with length. Use size.

The exercise, step by step

In a word game, the set found holds the words a player found. The words "qqq" and "zzz" aren't allowed, so both must be removed. The player never found "zzz", so that removal changes nothing. Then the score is one point per letter, so the task is to add each word's length to score.

Call delete on found with "qqq", then with "zzz", one call per line.
Below let score = 0;, write a for...of loop over found, with the loop variable word.
Inside the loop, add word.length to score.

let words = ["cat", "tree", "sun", "qqq", "moon"];
let found = new Set(words);
found.delete("qqq");
found.delete("zzz");
console.log(found.size + " words");
let score = 0;
for (let word of found) {
    score += word.length;
}
console.log(score + " points");

The output is:

4 words
14 points

The four words left are cat, tree, sun and moon, with 3 + 4 + 3 + 4 = 14 letters. Because a total doesn't depend on the order, the loop gives 14 however the words are visited.

Using the answer from delete

Because delete says whether the value was there, you can count real removals. Here two words are removed from the exercise's set, and only "qqq" was present:

let found = new Set(["cat", "tree", "sun", "qqq", "moon"]);
let removed = 0;
for (const w of ["qqq", "zzz"]) {
    if (found.delete(w)) {
        removed += 1;
    }
}
console.log(removed, found.size);

1 4

To empty a whole set at once, call found.clear(); its size is then 0.

Deleting while looping

JavaScript lets you call delete on the set you are looping over: the loop skips values that are already gone and carries on. This code empties a set and ends with a size of 0:

let live = new Set([1, 2, 3]);
for (const x of live) {
    live.delete(x);
}
console.log(live.size);

Key takeaways

s.delete(x) removes x and returns whether it was there; a missing value is not an error.
for (let x of s) visits every value once, in the order they were added.
Sets have no indexes, so loop with for...of.
Totals, counts and checks give the same answer in any order.

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