英文:
Check if all values in an array are equal but ignore null values
问题
我有一个字符串和空值的列表。现在我想检查列表中的所有字符串是否相同，应该忽略空值。例如：
- `[null, "a", "a", null, "a"]` 应该返回 `true`，因为所有字符串都相等
- `[null, "a", "a", null, "b"]` 应该返回 `false`，因为有 "b" 和 "a"
在SO上有其他与查找重复项或相等值有关的问题，但似乎没有一个考虑到`null`值。
我已经考虑过用以下方法实现：
- 一个循环，在第一个字符串被保存后，将在后续迭代中与其进行比较
- 一个`filter`后跟一个`every`
不过我想知道是否有更好、更直观的方法来做这件事。
英文:
I have a list of strings and nulls. Now I want to check if all strings in the list are the same and nulls should be ignored. So for instance
[null, "a", "a", null, "a"]should evaluate totrue, because all strings are equal[null, "a", "a", null, "b"]should evaluate tofalse, because there is "b" and "a"
There are other questions on SO that are concerned with finding duplicates or equal values, but none of them seem to consider null values.
I already thought about implementing this with
- a loop where I would save the first string and then compare to it in the later iterations
- a
filterfollowed byevery
However I was wondering if there is a better more intuitive way to do this.
答案1
得分: 3
你可以这样做：
set = new Set(yourArray)
set.delete(null)
if (set.size === 1)
// unique
英文:
You could do something like
set = new Set(yourArray)
set.delete(null)
if (set.size === 1)
// unique
答案2
得分: 1
- 一个循环，在其中我会保存第一个字符串，然后在后续迭代中与其进行比较。
- 一个筛选，然后是每个。
你可以通过一个空变量，然后接着使用 every 来实现它：
let j; // 第一个非空值
const result = array.every((v, i, a) => v === null || v === a[j ??= i]) && j != null;
编辑： 好的，让我们来分解这个代码：
every((v, i, a) => v === null || ...)检查每个元素是null或者...- 因此，只有在
v不是null时才执行v === a[j ??= i] - j ??= i 基本上是
j === null || j === undefined ? (j = i) : j的简写，所以j将适应第一个v !== null的i - 但是我们的循环不区分
array.every((v, i, a) => v === null)和array.every((v, i, a) => v === a[j ??= i])，所以我们还想要检查v === a[j ??= i]至少被执行了一次，通过检查索引j不再为null。
英文:
> - a loop where I would save the first string and then compare to it in the later iterations
> - a filter followed by every
you can do it with an empty variable followed by every:
let j; // first non null value
const result = array.every((v,i,a) => v === null || v === a[j??=i]) && j != null;
Edit: OK, let's take this apart:
every((v,i,a) => v === null || ...)check that every element is eithernullor ...- so
v === a[j??=i]is only executed whenvis notnull - j ??= i is basically a shorthand for
j === null || j === undefined ? (j = i) : jsojwill adapt the firstiwherev !== null - but our loop doesn't differentiate between
array.every((v,i,a) => v === null)andarray.every((v,i,a) => v === a[j??=i])so we also want to check thatv === a[j??=i]was executed at least once by checking that the indexjis not null anymore.
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