JavaScript - 比较两个数组 [英] JavaScript - Compare two arrays
问题描述
我见过很多关于如何在JavaScript中比较两个数组的帖子，如果你找到匹配的话就做点什么。
我想知道如何写相反的东西。
我有两个数组，当我找不到匹配项时我想做点什么。
I have seen many posts about how to compare two arrays in JavaScript and do something if you find a match. I wanted to know how to write the opposite. I have two arrays and I want to do something when I don't find a match.
var a = [1,2,3,5];
var b = [4,7,5,5];
for(i=0;i<a.length;i++){
for(j=0;j<b.length;j++){
if(b[j]===a[i]){
//do something
}
}
}
//somehow return 4, 7
基本上，我想找到与上述情况相反的情况。如果我找不到比赛，我想表演一个动作。我的其他声明会在哪里？
Basically, I would like to find the reverse of the above. If I dont find a match I would like to perform an action. Where would my else statement go?
loop1:
for(var i=0;i<a.length;i++){
loop2:
for(var j=0;j<b.length;j++){
if(b[j]==a[i]){
console.log("break loop");
break loop2;
}
else{
continue loop1;
}
}
}
这就是我我已经走了，我是在正确的轨道上？
This is as far as I've gotten, am I on the right track?
推荐答案
你想检查数组的任何元素的值是否 a [] 存在于数组 b [] 中，而您想要在不匹配的情况下做点什么。
You want to check if the value of any element of array a[] exists in array b[], and you want to "do something" on the condition that there is "no match".
有许多不同的方法可以考虑不匹配的情况。
There are many different ways you can consider a "no match" condition.
在你的第一个例子中，每当找到匹配时做某事，它将检查 a [] <的每个元素的值/ code> ，并将该值逐一与 b [] 的所有元素进行比较。因此，此比较过程不依赖于匹配的位置。此外，如果 a [] 中特定元素的值存在于 b []内的多个位置 ，或者 b [] 中的特定元素的值存在于 a [] ，然后代码将为同一元素的多个匹配中的每个匹配执行某些操作。
In your first example that "does something" each time it finds a match, it will examine the value of each element of a[], and compare that value to all elements of b[], one by one. So, this compare proceedure is not dependent on the position of a match. Also, if the value of a particular element in a[] exists in multiple places within b[], or if the value of a particular element in b[] exists in multiple places within a[], then the code will "do something" for each of the multiple matches of the same element.
因此，对于您提供的样本数据：
So, in the case of the sample data you provided:
var a = [1,2,3,5];
var b = [4,7,5,5];
当代码找到 <$的第4个元素时，它会做点什么 c $ c> a [] 匹配 b [] 的3rd元素，以及它会再次做某事，当它找到 的第4个元素时，[] 匹配的第4个元素 b [] 即可。我认为这就是你想要它的方式，但是如果你发现 a [] 中的元素，你只想一次做某事在 b [] 中的一个或多个位置，您只需要添加中断;声明，紧跟在做某事声明之后。
the code will "do something" when it finds the "4th" element of a[] matches the "3rd" element of b[], and it will "do something" again, when it finds the "4th" element of a[] matches the "4th" element of b[]. I assume that's how you wanted it, but if you only wanted to "do something" one-time if an element in a[] is found in one "or more" places within b[], then you would only need to add a "break;" statement, immediately following the "do something" statement(s).
所以，对于你的问题，你说的是，你想要反对：做 当你找不到匹配的东西。
So, for your question, you said that instead, you want the "opposite": to "do something" when you don't find a match.
从字面上看，你只需更改语句： if（b [j] === a [i]） ，至： if（b [j]！= a [i]） 即可。现在，这可能不你想要什么，因为不匹配的情况可能会经常发生：
Taking that literally, you would just change the statement: "if(b[j]===a[i])", to: "if(b[j] != a[i])". Now, this is probably not what you want, because the "no match" condition would likely happen far too often:
for(i=0;i<a.length;i++){
for(j=0;j<b.length;j++){
if(b[j] != a[i]){
// do something
}
}
}
另一种可能性是，如果 a []的元素值，你想要做某事 的任何地方找到code> 。为此，您将检查 b [] a [] 的每个元素的值，并将该值与 <$ c的所有元素进行比较$ c> b [] ，一个接一个，如果在检查 b [] 您找不到与 a [] ， 当前元素值的匹配项那么 你会做点什么。
Another possibility, is that you want to "do something" if the value of an element of a[] can't be found anywhere in b[]. For this, you would examine the value of each element of a[], and compare that value to all elements of b[], one by one, and if after checking the value of all elements in b[] you couldn't find a match to the value of the current element of a[], then you would "do something".
在这种情况下，如果你找到了来自 <$的当前值的匹配c $ c> a [] ，在 b [] 中，进一步检查该元素 a [] 不是必需的。您可能希望跳过剩下的部分，然后开始检查 code> 的下一个元素 b [] 即可。为此，您需要使用 continue outerloop; 。
In this case, if you found a match of the current value from a[], in b[], further checking of that element of a[] would not be necessary. You would want skip the rest, and begin checking the next element of a[] to the elements of b[]. For this, you would want to use "continue outerloop;".
outerloop:
for(i=0;i<a.length;i++){
for(j=0;j<b.length;j++){
if(b[j] === a[i]){
continue outerloop;
}
}
// if we get to here, then no value
// of b[] matched the a[] value, so:
// "do something"
}
第三种可能性，如果 元素 a [] 中的任何地方找到。为此，您将检查 b [] a [] 的每个元素的值，并将该值与 <$ c的所有元素进行比较$ c> b [] ，一个接一个，如果在检查 b [] 你找不到与 a [] 当前元素的值匹配，你设置了一个标志并使用 break outerloop; 退出所有循环。然后在最后，如果标志是设置，你可以做某事。
A 3rd possibility, is that you want to "do something", only one time, if any element of a[] can't be found anywhere in b[]. For this, you would examine the value of each element of a[], and compare that value to all elements of b[], one by one, and if after checking the value of all elements in b[] you couldn't find a match to the value of the current element of a[], you set a "flag" and exit all the loops using "break outerloop;". Then at the end, if the flag is "set", you can "do something".
bflag=false;
outerloop:
for(i=0;i<a.length;i++){
for(j=0;j<b.length;j++){
if(b[j] === a[i]){
continue outerloop;
}
}
// if we get to here, then no value
// of b[] matched the a[] value, so,
// set the flag (bflag)
bflag=true;
break outerloop:
}
if(bflag){
// "do something"
}
如果将此代码作为函数调用，可以简化一下：
If this code was called as a function, it could be simplified a bit:
outerloop:
for(i=0;i<a.length;i++){
for(j=0;j<b.length;j++){
if(b[j] === a[i]){
continue outerloop;
}
}
// if we get to here, then no value of b[] matched
// the a[] value, so, do-something and return "failed":
// "do something"
return false;
}
// all elements of a[] were matched to
// elements of b[], so: return "success"
return true;
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