Javascript 使用 Lodash 删除数组中的元素
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Removing elements in an array using Lodash
提问by bard
I have this array:
我有这个数组：
var fruits = ['Apple', 'Banana', 'Orange', 'Celery'];
And I use Lodash's removelike so:
我remove像这样使用Lodash ：
_.remove(fruits, function (fruit) {
return fruit === 'Apple' || 'Banana' || 'Orange';
})
The result is ['Apple', 'Banana', 'Orange', 'Celery'], while I expected it to be ['Apple', 'Banana', 'Orange']. Why is this so?
结果是['Apple', 'Banana', 'Orange', 'Celery']，而我预期它是['Apple', 'Banana', 'Orange']。为什么会这样？
回答by Amadan
Because when fruitis "Celery", you are testing:
因为 when fruitis "Celery"，您正在测试：
"Celery" === 'Apple' || 'Banana' || 'Orange'
which evaluates to
其评估为
false || true || true
which is true.
这是true.
You can't use that syntax. Either do it the long way around:
您不能使用该语法。要么做很长的路：
_.remove(fruits, function (fruit) {
return fruit === 'Apple' || fruit === 'Banana' || fruit === 'Orange'
});
or test for array membership:
或测试数组成员资格：
_.remove(fruits, function (fruit) {
return _.indexOf(['Apple', 'Banana', 'Orange'], fruit) !== -1
});
This is not limited to JavaScript, and is in fact a common mistake (e.g. this question)
这不仅限于 JavaScript，而且实际上是一个常见的错误（例如这个问题）
回答by gustavogelf
You can use the method _.pullfrom lodash 2.0 and up
您可以使用_.pulllodash 2.0 及更高版本的方法
var fruits = ['Apple', 'Banana', 'Orange', 'Celery'];
_.pull(fruits, 'Apple', 'Banana', 'Orange'); // ['Celery']
document.write(fruits);<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.6.1/lodash.js"></script>回答by Mouscellaneous
If you want to remove a set of items from another set, there are set operations intended specifically for that. Lodash has https://lodash.com/docs/4.17.2#differencewhich takes two array parameters A and B and will return another array which contains all of the elements of A which are not in B.
如果您想从另一个集合中删除一组项目，则有专门用于该集合的操作。Lodash 有https://lodash.com/docs/4.17.2#difference，它接受两个数组参数 A 和 B，并将返回另一个数组，其中包含 A 中不在 B 中的所有元素。
In your case, you could write
在你的情况下，你可以写
const fruits = ['Apple', 'Banana', 'Orange', 'Celery'];
const filteredFruits = _.difference(fruits, ['Apple', 'Banana', 'Orange']);
which will result in ['Celery'].
这将导致['Celery'].
回答by ncksllvn
The problem isn't with Lo-Dash; your problem is with your conditional within your callback function. This:
问题不在于 Lo-Dash；您的问题在于您的回调函数中的条件。这个：
return fruit === 'Apple' || 'Banana' || 'Orange';
Is notcorrect. You need to actually compare fruitwith each string:
是不正确的。您需要实际fruit与每个字符串进行比较：
return fruit === 'Apple' || fruit === 'Banana' || fruit === 'Orange';
Or, you can use another Lo-Dash function to make it a little more compact:
或者，您可以使用另一个 Lo-Dash 函数使其更紧凑：
_.remove(fruits, function (fruit) {
return _.contains(['Apple', 'Banana', 'Orange'], fruit);
})
Note:In the latest versions of Lo-Dash the _.containsfunction is deprecated. Please use _.includes
注意：在 Lo-Dash 的最新版本中，该_.contains功能已被弃用。请用_.includes
回答by Sampson
Use an array of values that you'd like to compare against, and check for a returned index greater than -1. This indicates the value evaluated was found in the collection.
使用您想要比较的值数组，并检查返回的索引是否大于 -1。这表明在集合中找到了评估的值。
_.remove( fruits, function ( fruit ) {
return _.indexOf( [ "Apple", "Banana", "Orange" ], fruit ) >= 0;
});
Alternatively you could use lo-dash's _.containsmethodto get a boolean response.
或者，您可以使用lo-dash 的_.contains方法来获得布尔响应。
The problem with the approach you took was that you weren't comparing fruitagainst each one of those strings; instead, the only comparison taking place was fruitagainst "Apple", after that you were coercing strings all on their own.
您采用的方法的问题在于您没有与fruit这些字符串中的每一个进行比较；相反，唯一发生的比较是fruit针对"Apple"，之后你就自己强制字符串了。
Non-empty strings coerce to true(!!"Banana"), and as such are truthy. Therefore, the following condition will always short-circuit at "Banana" (unless fruitstrictly equals "Apple"), returning true:
非空字符串强制为true( !!"Banana")，因此为真。因此，以下条件将始终在“香蕉”处短路（除非fruit严格等于"Apple"），返回true：
return fruit === "Apple" || 'Banana' || "Orange";
