如何从 forEach 循环中的数组中删除元素？

Question

I am trying to remove an element in an array in a forEach loop, but am having trouble with the standard solutions I've seen.

This is what I'm currently trying:

review.forEach(function(p){
   if(p === '\u2022 \u2022 \u2022'){
      console.log('YippeeeE!!!!!!!!!!!!!!!!')
      review.splice(p, 1);
   }
});

I know it's getting into the if because I'm seeing YippeeeeeE!!!!!!!!!!!!! in the console.

MY PROBLEM: I know that my for loop and if logic are sound, but my attempt to remove the current element from the array is failing.

UPDATE:

Tried out Xotic750's answer, and the element is still not being removed:

Here is the function in my code:

review.forEach(function (item, index, object) {
    if (item === '\u2022 \u2022 \u2022') {
       console.log('YippeeeE!!!!!!!!!!!!!!!!')
       object.splice(index, 1);
    }
    console.log('[' + item + ']');
});

Here is the output where the array is still not removed:

[Scott McNeil]
[reviewed 4 months ago]
[ Mitsubishi is AMAZING!!!]
YippeeeE!!!!!!!!!!!!!!!!
[• • •]

So obviously it is going into the if statement as directed, but it's also obvious that the [• • •] is still there.

Answer 1

It looks like you are trying to do this?

Iterate and mutate an array using Array.prototype.splice

 var pre = document.getElementById('out'); function log(result) { pre.appendChild(document.createTextNode(result + '\\n')); } var review = ['a', 'b', 'c', 'b', 'a']; review.forEach(function(item, index, object) { if (item === 'a') { object.splice(index, 1); } }); log(review);

 <pre id="out"></pre>

Which works fine for simple case where you do not have 2 of the same values as adjacent array items, other wise you have this problem.

 var pre = document.getElementById('out'); function log(result) { pre.appendChild(document.createTextNode(result + '\\n')); } var review = ['a', 'a', 'b', 'c', 'b', 'a', 'a']; review.forEach(function(item, index, object) { if (item === 'a') { object.splice(index, 1); } }); log(review);

 <pre id="out"></pre>

So what can we do about this problem when iterating and mutating an array? Well the usual solution is to work in reverse. Using ES3 while but you could use for sugar if preferred

 var pre = document.getElementById('out'); function log(result) { pre.appendChild(document.createTextNode(result + '\\n')); } var review = ['a' ,'a', 'b', 'c', 'b', 'a', 'a'], index = review.length - 1; while (index >= 0) { if (review[index] === 'a') { review.splice(index, 1); } index -= 1; } log(review);

 <pre id="out"></pre>

Ok, but you wanted to use ES5 iteration methods. Well and option would be to use Array.prototype.filter but this does not mutate the original array but creates a new one, so while you can get the correct answer it is not what you appear to have specified.

We could also use ES5 Array.prototype.reduceRight , not for its reducing property by rather its iteration property, ie iterate in reverse.

 var pre = document.getElementById('out'); function log(result) { pre.appendChild(document.createTextNode(result + '\\n')); } var review = ['a', 'a', 'b', 'c', 'b', 'a', 'a']; review.reduceRight(function(acc, item, index, object) { if (item === 'a') { object.splice(index, 1); } }, []); log(review);

 <pre id="out"></pre>

Or we could use ES5 Array.protoype.indexOf like so.

 var pre = document.getElementById('out'); function log(result) { pre.appendChild(document.createTextNode(result + '\\n')); } var review = ['a', 'a', 'b', 'c', 'b', 'a', 'a'], index = review.indexOf('a'); while (index !== -1) { review.splice(index, 1); index = review.indexOf('a'); } log(review);

 <pre id="out"></pre>

But you specifically want to use ES5 Array.prototype.forEach , so what can we do? Well we need to use Array.prototype.slice to make a shallow copy of the array and Array.prototype.reverse so we can work in reverse to mutate the original array.

 var pre = document.getElementById('out'); function log(result) { pre.appendChild(document.createTextNode(result + '\\n')); } var review = ['a', 'a', 'b', 'c', 'b', 'a', 'a']; review.slice().reverse().forEach(function(item, index, object) { if (item === 'a') { review.splice(object.length - 1 - index, 1); } }); log(review);

 <pre id="out"></pre>

Finally ES6 offers us some further alternatives, where we do not need to make shallow copies and reverse them. Notably we can use Generators and Iterators . However support is fairly low at present.

 var pre = document.getElementById('out'); function log(result) { pre.appendChild(document.createTextNode(result + '\\n')); } function* reverseKeys(arr) { var key = arr.length - 1; while (key >= 0) { yield key; key -= 1; } } var review = ['a', 'a', 'b', 'c', 'b', 'a', 'a']; for (var index of reverseKeys(review)) { if (review[index] === 'a') { review.splice(index, 1); } } log(review);

 <pre id="out"></pre>

Something to note in all of the above is that, if you were stripping NaN from the array then comparing with equals is not going to work because in Javascript NaN === NaN is false. But we are going to ignore that in the solutions as it it yet another unspecified edge case.

So there we have it, a more complete answer with solutions that still have edge cases. The very first code example is still correct but as stated, it is not without issues.

Answer 2

Use Array.prototype.filter instead of forEach :

var pre = document.getElementById('out');

function log(result) {
  pre.appendChild(document.createTextNode(result + '\n'));
}

var review = ['a', 'b', 'c', 'b', 'a', 'e'];
review = review.filter(item => item !== 'a');
log(review);

Answer 3

Although Xotic750's answer provides several good points and possible solutions, sometimes simple is better .

You know the array being iterated on is being mutated in the iteration itself (ie removing an item => index changes), thus the simplest logic is to go backwards in an old fashioned for (à la C language):

 let arr = ['a', 'a', 'b', 'c', 'b', 'a', 'a']; for (let i = arr.length - 1; i >= 0; i--) { if (arr[i] === 'a') { arr.splice(i, 1); } } document.body.append(arr.join());

If you really think about it, a forEach is just syntactic sugar for a for loop... So if it's not helping you, just please stop breaking your head against it.

Answer 4

You could also use indexOf instead to do this

var i = review.indexOf('\u2022 \u2022 \u2022');
if (i !== -1) review.splice(i,1);

Answer 5

I understood that you want to remove from the array using a condition and have another array that has items removed from the array. Is right?

How about this?

 var review = ['a', 'b', 'c', 'ab', 'bc']; var filtered = []; for(var i=0; i < review.length;) { if(review[i].charAt(0) == 'a') { filtered.push(review.splice(i,1)[0]); }else{ i++; } } console.log("review", review); console.log("filtered", filtered);

Hope this help...

By the way, I compared 'for-loop' to 'forEach'.

If remove in case a string contains 'f', a result is different.

 var review = ["of", "concat", "copyWithin", "entries", "every", "fill", "filter", "find", "findIndex", "flatMap", "flatten", "forEach", "includes", "indexOf", "join", "keys", "lastIndexOf", "map", "pop", "push", "reduce", "reduceRight", "reverse", "shift", "slice", "some", "sort", "splice", "toLocaleString", "toSource", "toString", "unshift", "values"]; var filtered = []; for(var i=0; i < review.length;) { if( review[i].includes('f')) { filtered.push(review.splice(i,1)[0]); }else { i++; } } console.log("review", review); console.log("filtered", filtered); /** * review [ "concat", "copyWithin", "entries", "every", "includes", "join", "keys", "map", "pop", "push", "reduce", "reduceRight", "reverse", "slice", "some", "sort", "splice", "toLocaleString", "toSource", "toString", "values"] */ console.log("========================================================"); review = ["of", "concat", "copyWithin", "entries", "every", "fill", "filter", "find", "findIndex", "flatMap", "flatten", "forEach", "includes", "indexOf", "join", "keys", "lastIndexOf", "map", "pop", "push", "reduce", "reduceRight", "reverse", "shift", "slice", "some", "sort", "splice", "toLocaleString", "toSource", "toString", "unshift", "values"]; filtered = []; review.forEach(function(item,i, object) { if( item.includes('f')) { filtered.push(object.splice(i,1)[0]); } }); console.log("-----------------------------------------"); console.log("review", review); console.log("filtered", filtered); /** * review [ "concat", "copyWithin", "entries", "every", "filter", "findIndex", "flatten", "includes", "join", "keys", "map", "pop", "push", "reduce", "reduceRight", "reverse", "slice", "some", "sort", "splice", "toLocaleString", "toSource", "toString", "values"] */

And remove by each iteration, also a result is different.

 var review = ["of", "concat", "copyWithin", "entries", "every", "fill", "filter", "find", "findIndex", "flatMap", "flatten", "forEach", "includes", "indexOf", "join", "keys", "lastIndexOf", "map", "pop", "push", "reduce", "reduceRight", "reverse", "shift", "slice", "some", "sort", "splice", "toLocaleString", "toSource", "toString", "unshift", "values"]; var filtered = []; for(var i=0; i < review.length;) { filtered.push(review.splice(i,1)[0]); } console.log("review", review); console.log("filtered", filtered); console.log("========================================================"); review = ["of", "concat", "copyWithin", "entries", "every", "fill", "filter", "find", "findIndex", "flatMap", "flatten", "forEach", "includes", "indexOf", "join", "keys", "lastIndexOf", "map", "pop", "push", "reduce", "reduceRight", "reverse", "shift", "slice", "some", "sort", "splice", "toLocaleString", "toSource", "toString", "unshift", "values"]; filtered = []; review.forEach(function(item,i, object) { filtered.push(object.splice(i,1)[0]); }); console.log("-----------------------------------------"); console.log("review", review); console.log("filtered", filtered);

Answer 6

Here is how you should do it:

review.forEach(function(p,index,object){
   if(review[index] === '\u2022 \u2022 \u2022'){
      console.log('YippeeeE!!!!!!!!!!!!!!!!')
      review.splice(index, 1);
   }
});

Answer 7

The following will give you all the elements which is not equal to your special characters!

review = jQuery.grep( review, function ( value ) {
    return ( value !== '\u2022 \u2022 \u2022' );
} );

Answer 8

I had problems with the following code. My solution follows after. First, the problem: Let's assume you want to trim strings, and discard the string with a 'c' in it:

 var review = [' a ', ' b ', ' c ', ' d ', ' e ']; review.forEach(function(item, index) { review[index] = item.trim(); console.log("After trimming, the item is ", review[index]); if (review[index] === 'c') { review.splice(index, 1); } }); console.log("Review is now: "); console.log(review);

If you run the above piece of code, you will see that ' d ' is never trimmed. It remains in the review array, with spaces before and after it.

That's because you are messing with the review array that the foreach is built on. It's a lot better to make a copy of the array, and push elements into it that are OK to keep, and skip the ones that are not. Then, output the new FinalArray, like this:

 var review = [' a ', ' b ', ' c ', ' d ', ' e ']; var finalArray = []; review.forEach(function(item, index) { // Add it to the new array first. If it turns out to be bad, remove it before going onto // the next iteration. finalArray.push(item.trim()); console.log("After trimming, the item is ", item.trim()); if (item.trim() == 'c') { // skip c finalArray.pop(); // This way, removing it doesn't affect the original array. } }); console.log("FinalArray is now: "); console.log(finalArray);

As you can see, this worked perfectly: