减去数组 - javascript
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Subtract arrays - javascript
提问by bsaglamtimur
I have arrays like;
我有像这样的数组；
var john = { name: "John Smith", age: 23 };
var mary = { name: "Mary Key", age: 18 };
var bob = { name: "Bob-small", age: 6 };
var people = [john, mary, bob];
var john2 = { name: "John Smith", age: 23 };
var people2 = [john2];
What I would like to do is subtract people2 from people and get result;
我想做的是从 people 中减去 people2 并得到结果；
[mary, bob];
How can I achieve this? TIA
我怎样才能做到这一点？TIA
回答by Aadit M Shah
The first thing you need is a way to compare two people for mathematical equivalence. We cannot use the ===operator because it tests two objects for identity equivalence, not mathematical equivalence. For example:
您需要的第一件事是比较两个人的数学等价性的方法。我们不能使用===运算符，因为它测试两个对象的身份等价，而不是数学等价。例如：
var john = {name: "John Smith", age: 23};
var john2 = {name: "John Smith", age: 23};
console.log(john === john2); // false
Hence we create a function to compare two people:
因此，我们创建了一个函数来比较两个人：
console.log(similarPeople(john, john2));
function similarPeople(a, b) {
return a.name === b.name &&
a.age === b.age;
}
The reason the function is named similarPeopleand not samePeopleis because two different people may have the same name and may be of the same age.
函数被命名similarPeople而不是命名的原因samePeople是因为两个不同的人可能有相同的名字并且可能年龄相同。
The difference of two sets, A and B, is defined as the set of all those elements of A which are not in B.
两个集合 A 和 B 的差被定义为 A 中所有不在 B 中的元素的集合。
At worst computing the set difference of two sets, A and B, of sizes mand nrespectively would take O(m * n)time. Not very efficient:
在最坏的情况计算的两组，A和B，尺寸的差集m和n分别将采取O(m * n)时间。效率不高：
function difference(a, b, eq) {
if (arguments.length < 3) eq = function (a, b) {
return a === b;
};
var m = a.length;
var n = b.length;
var c = [];
loop: for (var i = 0; i < m; i++) {
var x = a[i];
for (var j = 0; j < n; j++)
if (eq(b[j], x))
continue loop;
c.push(x);
}
return c;
}
Now you could get the difference of two lists of people as follows:
现在你可以得到两个人列表的差异，如下所示：
var people = [john, mary, bob];
var people2 = [john2];
console.log(difference(people, people2, similarPeople)); // [mary, bob]
See the demo: http://jsfiddle.net/F7RDs/
看演示：http: //jsfiddle.net/F7RDs/
Fortunately there's a faster way to compute the set difference for large sets: indices. Let's create a function to create an index of a list of people:
幸运的是，有一种更快的方法来计算大集合的集合差：索引。让我们创建一个函数来创建人员列表的索引：
function indexFrom(people) {
var length = people.length, index = {};
for (var i = 0; i < length; i++) {
var person = people[i];
var name = person.name;
if (index.hasOwnProperty(name)) var subindex = index[name];
else var subindex = index[name] = {};
subindex[person.age] = {};
}
return index;
}
Now we create an index for the second list of people:
现在我们为第二个人员列表创建一个索引：
var index2 = indexFrom(people2);
We also need a function to test whether a person is in a list using its index:
我们还需要一个函数来使用它的索引来测试一个人是否在列表中：
function hasPerson(person, index) {
var name = person.name;
return index.hasOwnProperty(name) &&
index[name].hasOwnProperty(person.age);
}
Finally we can create a more efficient implementation of differenceas follows:
最后，我们可以创建一个更有效的实现，difference如下所示：
function difference(a, index, has) {
var m = a.length, c = [];
for (var i = 0; i < m; i++) {
var x = a[i];
if (!has(x, index))
c.push(x);
}
return c;
}
You use it as follows:
您可以按如下方式使用它：
console.log(difference(people, index2, hasPerson)); // [mary, bob]
The advantage is that creating an index takes O(n)time and calculating the difference takes O(m)time. Hence in total it only takes O(m + n)time instead of O(m * n)time. In addition you can cache the index for future use.
优点是创建索引需要O(n)时间，计算差异需要O(m)时间。因此，总的来说它只需要O(m + n)时间而不是O(m * n)时间。此外，您可以缓存索引以备将来使用。
See the demo: http://jsfiddle.net/F7RDs/1/
看演示：http: //jsfiddle.net/F7RDs/1/
Hope this helped.
希望这有帮助。
回答by Tibos
Here is an easy solution:
这是一个简单的解决方案：
var diff = people.filter(function(item) {
return !people2.some(function(test){
return test.name === item.name && test.age === item.age;
});
});
Make sure the function passed to people2.somecorrectly checks that the two objects are equal, since ==would fail, as you have references to different objects with identical properties.
确保传递给的函数people2.some正确检查两个对象是否相等，因为==会失败，因为您引用了具有相同属性的不同对象。
回答by user2804021
So, here goes the final running code. Copy , paste it and check for the output.It is first of all separating the strings (comma-separation). Then it is comparing them individually . Check this link: http://jsfiddle.net/DXRZ4/
所以，这是最终的运行代码。复制、粘贴并检查输出。首先是分隔字符串（逗号分隔）。然后是单独比较它们。检查此链接：http: //jsfiddle.net/DXRZ4/
<html>
<head>
<script>
var a = new Array(); var i=1; var people = new Array();
function fun() {
var str = "john,john,bob",
l = str.split(",");
for(i=0; i<3; i++) {
a[i] = l[i];
// this will show the comma separated strings
document.writeln(l[i]);
}
for(i=0; i<3; i++) {
t=a[i];
j=a[i+1];
if((a[i]==a[i+1]) || (a[i]==a[i+2])) {
// it will store the position of string which are same
p=i;
}
}
for(i=0; i<3; i++) {
if(p!=i){
document.writeln("Subtracted Strings are:-");
document.writeln(l[i]);
}
}
}
</script>
<body>
<input type="button" name="b1" onClick="fun()">
</body>
</html>
