我想了解从另一个数组的所有元素中过滤一个数组的最佳方法。我尝试了过滤功能，但它不来我如何给它的值，我想删除。喜欢的东西:
var array = [1,2,3,4];
var anotherOne = [2,4];
var filteredArray = array.filter(myCallback);
// filteredArray should now be [1,3]
function myCallBack(){
return element ! filteredArray;
//which clearly can't work since we don't have the reference <,<
}
如果过滤器函数没有用处，您将如何实现它? 编辑:我检查了可能的重复问题，这可能对那些容易理解javascript的人有用。如果答案勾选“好”，事情就简单多了。
对过滤功能最好的描述是https://developer.mozilla.org/pl/docs/Web/JavaScript/Referencje/Obiekty/Array/filter
你应该简单地条件函数:
function conditionFun(element, index, array) {
return element >= 10;
}
filtered = [12, 5, 8, 130, 44].filter(conditionFun);
在变量值被赋值之前，您不能访问它
你可以设置过滤器函数来遍历“过滤器数组”。
var arr = [1, 2, 3 ,4 ,5, 6, 7];
var filter = [4, 5, 6];
var filtered = arr.filter(
function(val) {
for (var i = 0; i < filter.length; i++) {
if (val == filter[i]) {
return false;
}
}
return true;
}
);
var array = [1,2,3,4];
var anotherOne = [2,4];
var filteredArray = array.filter(myCallBack);
function myCallBack(el){
return anotherOne.indexOf(el) < 0;
}
在回调中，检查数组的每个值是否在另一个数组中
https://jsfiddle.net/0tsyc1sx/
如果使用lodash.js，请使用_.difference
filteredArray = _.difference(array, anotherOne);
Demo
如果你有一个对象数组:
var array = [{id :1, name :"test1"},{id :2, name :"test2"},{id :3, name :"test3"},{id :4, name :"test4"}];
var anotherOne = [{id :2, name :"test2"}, {id :4, name :"test4"}];
var filteredArray = array.filter(function(array_el){
return anotherOne.filter(function(anotherOne_el){
return anotherOne_el.id == array_el.id;
}).length == 0
});
对象的演示数组
用lodash演示不同的对象数组
您可以使用过滤器，然后为过滤器函数使用过滤数组的约简，当它找到匹配时检查并返回true，然后在返回(!)时反转。filter函数对数组中的每个元素调用一次。在你的文章中，你没有对函数中的任何元素进行比较。
Var a1 = [1,2,3,4]， A2 = [2,3]; Var filter = a1.filter(函数(x) { 返回! a2。Reduce(函数(y, z) { 返回x == y || x == z || y == true; }) })； document . write(过滤);
可以使用filter()函数的this参数来避免将过滤器数组存储在全局变量中。
Var filtered = [1,2,3,4].filter( 函数(e) { 返回this.indexOf(e) < 0; }， (2、4) ）; console.log(过滤);
Var arr1= [1,2,3,4]; var arr2 =(2、4) 函数费尔(价值){ 返回value !=arr2[0] && value !=arr2[1] ｝ . getelementbyid (p)。innerHTML = arr1.filter (fil) <！DOCTYPE html > < html > < >头 > < /头 身体< > < p id = p > < / p >
函数的arr (arr1 arr2) { 函数filt(价值){ 返回arr2.indexOf(value) == -1; ｝ 返回arr1.filter (filt) ｝ . getelementbyid (p)。innerHTML = arr([1,2,3,4]，[2,4]) < p id = p > < / p >
我会这样做;
Var arr1 = [1,2,3,4]， Arr2 = [2,4]， Res = arr1。Filter (item => !arr2.includes(item)); console.log (res);
来自另一个包含对象属性的数组的更灵活的过滤数组
function filterFn(array, diffArray, prop, propDiff) { diffArray = !propDiff ? diffArray : diffArray.map(d => d[propDiff]) this.fn = f => diffArray.indexOf(f) === -1 if (prop) { return array.map(r => r[prop]).filter(this.fn) } else { return array.filter(this.fn) } } //You can use it like this; var arr = []; for (var i = 0; i < 10; i++) { var obj = {} obj.index = i obj.value = Math.pow(2, i) arr.push(obj) } var arr2 = [1, 2, 3, 4, 5] var sec = [{t:2}, {t:99}, {t:256}, {t:4096}] var log = console.log.bind(console) var filtered = filterFn(arr, sec, 'value', 't') var filtered2 = filterFn(arr2, sec, null, 't') log(filtered, filtered2)
你可以写一个泛型的filterByIndex()函数，并在TS中使用类型推断来省去回调函数的麻烦:
假设你有一个数组[1,2,3,4]，你想用[2,4]数组中指定的下标来过滤()。
var filtered = [1,2,3,4,].filter(byIndex(element => element, [2,4]))
byIndex函数需要元素函数和数组，如下所示:
byIndex = (getter: (e:number) => number, arr: number[]) => (x: number) => {
var i = getter(x);
return arr.indexOf(i);
}
结果就是这样
filtered = [1,3]
下面的例子使用new Set()创建一个只有唯一元素的过滤数组:
数组的基本数据类型:字符串，数字，布尔，空，未定义，符号:
const a = [1, 2, 3, 4];
const b = [3, 4, 5];
const c = Array.from(new Set(a.concat(b)));
以对象为项的数组:
const a = [{id:1}, {id: 2}, {id: 3}, {id: 4}];
const b = [{id: 3}, {id: 4}, {id: 5}];
const stringifyObject = o => JSON.stringify(o);
const parseString = s => JSON.parse(s);
const c = Array.from(new Set(a.concat(b).map(stringifyObject)), parseString);
All the above solutions "work", but are less than optimal for performance and are all approach the problem in the same way which is linearly searching all entries at each point using Array.prototype.indexOf or Array.prototype.includes. A far faster solution (far faster even than a binary search for most cases) would be to sort the arrays and skip ahead as you go along as seen below. However, one downside is that this requires all entries in the array to be numbers or strings. Also however, binary search may in some rare cases be faster than the progressive linear search. These cases arise from the fact that my progressive linear search has a complexity of O(2n1+n2) (only O(n1+n2) in the faster C/C++ version) (where n1 is the searched array and n2 is the filter array), whereas the binary search has a complexity of O(n1ceil(log2n2)) (ceil = round up -- to the ceiling), and, lastly, the indexOf search has a highly variable complexity between O(n1) and O(n1n2), averaging out to O(n1ceil(n2÷2)). Thus, indexOf will only be the fastest, on average, in the cases of (n1,n2) equaling {1,2}, {1,3}, or {x,1|x∈N}. However, this is still not a perfect representation of modern hardware. IndexOf is natively optimized to the fullest extent imaginable in most modern browsers, making it very subject to the laws of branch prediction. Thus, if we make the same assumption on indexOf as we do with progressive linear and binary search -- that the array is presorted -- then, according to the stats listed in the link, we can expect roughly a 6x speed up for IndexOf, shifting its complexity to between O(n1÷6) and O(n1n2), averaging out to O(n1ceil(n27÷12)). Finally, take note that the below solution will never work with objects because objects in JavaScript cannot be compared by pointers in JavaScript.
function sortAnyArray(a,b) { return a>b ? 1 : (a===b ? 0 : -1); }
function sortIntArray(a,b) { return (a|0) - (b|0) |0; }
function fastFilter(array, handle) {
var out=[], value=0;
for (var i=0, len=array.length|0; i < len; i=i+1|0)
if (handle(value = array[i]))
out.push( value );
return out;
}
const Math_clz32 = Math.clz32 || (function(log, LN2){
return function(x) {
return 31 - log(x >>> 0) / LN2 | 0; // the "| 0" acts like math.floor
};
})(Math.log, Math.LN2);
/* USAGE:
filterArrayByAnotherArray(
[1,3,5],
[2,3,4]
) yields [1, 5], and it can work with strings too
*/
function filterArrayByAnotherArray(searchArray, filterArray) {
if (
// NOTE: This does not check the whole array. But, if you know
// that there are only strings or numbers (not a mix of
// both) in the array, then this is a safe assumption.
// Always use `==` with `typeof` because browsers can optimize
// the `==` into `===` (ONLY IN THIS CIRCUMSTANCE)
typeof searchArray[0] == "number" &&
typeof filterArray[0] == "number" &&
(searchArray[0]|0) === searchArray[0] &&
(filterArray[0]|0) === filterArray[0]
) {filterArray
// if all entries in both arrays are integers
searchArray.sort(sortIntArray);
filterArray.sort(sortIntArray);
} else {
searchArray.sort(sortAnyArray);
filterArray.sort(sortAnyArray);
}
var searchArrayLen = searchArray.length, filterArrayLen = filterArray.length;
var progressiveLinearComplexity = ((searchArrayLen<<1) + filterArrayLen)>>>0
var binarySearchComplexity= (searchArrayLen * (32-Math_clz32(filterArrayLen-1)))>>>0;
// After computing the complexity, we can predict which algorithm will be the fastest
var i = 0;
if (progressiveLinearComplexity < binarySearchComplexity) {
// Progressive Linear Search
return fastFilter(searchArray, function(currentValue){
while (filterArray[i] < currentValue) i=i+1|0;
// +undefined = NaN, which is always false for <, avoiding an infinite loop
return filterArray[i] !== currentValue;
});
} else {
// Binary Search
return fastFilter(
searchArray,
fastestBinarySearch(filterArray)
);
}
}
// see https://stackoverflow.com/a/44981570/5601591 for implementation
// details about this binary search algorithm
function fastestBinarySearch(array){
var initLen = (array.length|0) - 1 |0;
const compGoto = Math_clz32(initLen) & 31;
return function(sValue) {
var len = initLen |0;
switch (compGoto) {
case 0:
if (len & 0x80000000) {
const nCB = len & 0x80000000;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 1:
if (len & 0x40000000) {
const nCB = len & 0xc0000000;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 2:
if (len & 0x20000000) {
const nCB = len & 0xe0000000;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 3:
if (len & 0x10000000) {
const nCB = len & 0xf0000000;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 4:
if (len & 0x8000000) {
const nCB = len & 0xf8000000;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 5:
if (len & 0x4000000) {
const nCB = len & 0xfc000000;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 6:
if (len & 0x2000000) {
const nCB = len & 0xfe000000;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 7:
if (len & 0x1000000) {
const nCB = len & 0xff000000;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 8:
if (len & 0x800000) {
const nCB = len & 0xff800000;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 9:
if (len & 0x400000) {
const nCB = len & 0xffc00000;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 10:
if (len & 0x200000) {
const nCB = len & 0xffe00000;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 11:
if (len & 0x100000) {
const nCB = len & 0xfff00000;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 12:
if (len & 0x80000) {
const nCB = len & 0xfff80000;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 13:
if (len & 0x40000) {
const nCB = len & 0xfffc0000;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 14:
if (len & 0x20000) {
const nCB = len & 0xfffe0000;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 15:
if (len & 0x10000) {
const nCB = len & 0xffff0000;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 16:
if (len & 0x8000) {
const nCB = len & 0xffff8000;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 17:
if (len & 0x4000) {
const nCB = len & 0xffffc000;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 18:
if (len & 0x2000) {
const nCB = len & 0xffffe000;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 19:
if (len & 0x1000) {
const nCB = len & 0xfffff000;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 20:
if (len & 0x800) {
const nCB = len & 0xfffff800;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 21:
if (len & 0x400) {
const nCB = len & 0xfffffc00;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 22:
if (len & 0x200) {
const nCB = len & 0xfffffe00;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 23:
if (len & 0x100) {
const nCB = len & 0xffffff00;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 24:
if (len & 0x80) {
const nCB = len & 0xffffff80;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 25:
if (len & 0x40) {
const nCB = len & 0xffffffc0;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 26:
if (len & 0x20) {
const nCB = len & 0xffffffe0;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 27:
if (len & 0x10) {
const nCB = len & 0xfffffff0;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 28:
if (len & 0x8) {
const nCB = len & 0xfffffff8;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 29:
if (len & 0x4) {
const nCB = len & 0xfffffffc;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 30:
if (len & 0x2) {
const nCB = len & 0xfffffffe;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 31:
if (len & 0x1) {
const nCB = len & 0xffffffff;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
}
// MODIFICATION: Instead of returning the index, this binary search
// instead returns whether something was found or not.
if (array[len|0] !== sValue) {
return true; // preserve the value at this index
} else {
return false; // eliminate the value at this index
}
};
}
请参阅我的另一篇文章在这里使用二进制搜索算法的更多细节
如果您对文件大小很挑剔(我尊重这一点)，那么您可以牺牲一点性能，以大大减小文件大小并提高可维护性。
function sortAnyArray(a,b) { return a>b ? 1 : (a===b ? 0 : -1); }
function sortIntArray(a,b) { return (a|0) - (b|0) |0; }
function fastFilter(array, handle) {
var out=[], value=0;
for (var i=0, len=array.length|0; i < len; i=i+1|0)
if (handle(value = array[i]))
out.push( value );
return out;
}
/* USAGE:
filterArrayByAnotherArray(
[1,3,5],
[2,3,4]
) yields [1, 5], and it can work with strings too
*/
function filterArrayByAnotherArray(searchArray, filterArray) {
if (
// NOTE: This does not check the whole array. But, if you know
// that there are only strings or numbers (not a mix of
// both) in the array, then this is a safe assumption.
typeof searchArray[0] == "number" &&
typeof filterArray[0] == "number" &&
(searchArray[0]|0) === searchArray[0] &&
(filterArray[0]|0) === filterArray[0]
) {
// if all entries in both arrays are integers
searchArray.sort(sortIntArray);
filterArray.sort(sortIntArray);
} else {
searchArray.sort(sortAnyArray);
filterArray.sort(sortAnyArray);
}
// Progressive Linear Search
var i = 0;
return fastFilter(searchArray, function(currentValue){
while (filterArray[i] < currentValue) i=i+1|0;
// +undefined = NaN, which is always false for <, avoiding an infinite loop
return filterArray[i] !== currentValue;
});
}
To prove the difference in speed, let us examine some JSPerfs. For filtering an array of 16 elements, binary search is roughly 17% faster than indexOf while filterArrayByAnotherArray is roughly 93% faster than indexOf. For filtering an array of 256 elements, binary search is roughly 291% faster than indexOf while filterArrayByAnotherArray is roughly 353% faster than indexOf. For filtering an array of 4096 elements, binary search is roughly 2655% faster than indexOf while filterArrayByAnotherArray is roughly 4627% faster than indexOf.
反向滤波(如与门)
上一节提供了获取数组A和数组B的代码，并删除A中存在于B中的所有元素:
filterArrayByAnotherArray(
[1,3,5],
[2,3,4]
);
// yields [1, 5]
下一节将提供反向过滤的代码，其中我们从A中删除B中不存在的所有元素。这个过程在功能上相当于只保留A和B的公共元素，如and门:
reverseFilterArrayByAnotherArray(
[1,3,5],
[2,3,4]
);
// yields [3]
下面是反向过滤的代码:
function sortAnyArray(a,b) { return a>b ? 1 : (a===b ? 0 : -1); }
function sortIntArray(a,b) { return (a|0) - (b|0) |0; }
function fastFilter(array, handle) {
var out=[], value=0;
for (var i=0, len=array.length|0; i < len; i=i+1|0)
if (handle(value = array[i]))
out.push( value );
return out;
}
const Math_clz32 = Math.clz32 || (function(log, LN2){
return function(x) {
return 31 - log(x >>> 0) / LN2 | 0; // the "| 0" acts like math.floor
};
})(Math.log, Math.LN2);
/* USAGE:
reverseFilterArrayByAnotherArray(
[1,3,5],
[2,3,4]
) yields [3], and it can work with strings too
*/
function reverseFilterArrayByAnotherArray(searchArray, filterArray) {
if (
// NOTE: This does not check the whole array. But, if you know
// that there are only strings or numbers (not a mix of
// both) in the array, then this is a safe assumption.
// Always use `==` with `typeof` because browsers can optimize
// the `==` into `===` (ONLY IN THIS CIRCUMSTANCE)
typeof searchArray[0] == "number" &&
typeof filterArray[0] == "number" &&
(searchArray[0]|0) === searchArray[0] &&
(filterArray[0]|0) === filterArray[0]
) {
// if all entries in both arrays are integers
searchArray.sort(sortIntArray);
filterArray.sort(sortIntArray);
} else {
searchArray.sort(sortAnyArray);
filterArray.sort(sortAnyArray);
}
var searchArrayLen = searchArray.length, filterArrayLen = filterArray.length;
var progressiveLinearComplexity = ((searchArrayLen<<1) + filterArrayLen)>>>0
var binarySearchComplexity= (searchArrayLen * (32-Math_clz32(filterArrayLen-1)))>>>0;
// After computing the complexity, we can predict which algorithm will be the fastest
var i = 0;
if (progressiveLinearComplexity < binarySearchComplexity) {
// Progressive Linear Search
return fastFilter(searchArray, function(currentValue){
while (filterArray[i] < currentValue) i=i+1|0;
// +undefined = NaN, which is always false for <, avoiding an infinite loop
// For reverse filterning, I changed !== to ===
return filterArray[i] === currentValue;
});
} else {
// Binary Search
return fastFilter(
searchArray,
inverseFastestBinarySearch(filterArray)
);
}
}
// see https://stackoverflow.com/a/44981570/5601591 for implementation
// details about this binary search algorithim
function inverseFastestBinarySearch(array){
var initLen = (array.length|0) - 1 |0;
const compGoto = Math_clz32(initLen) & 31;
return function(sValue) {
var len = initLen |0;
switch (compGoto) {
case 0:
if (len & 0x80000000) {
const nCB = len & 0x80000000;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 1:
if (len & 0x40000000) {
const nCB = len & 0xc0000000;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 2:
if (len & 0x20000000) {
const nCB = len & 0xe0000000;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 3:
if (len & 0x10000000) {
const nCB = len & 0xf0000000;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 4:
if (len & 0x8000000) {
const nCB = len & 0xf8000000;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 5:
if (len & 0x4000000) {
const nCB = len & 0xfc000000;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 6:
if (len & 0x2000000) {
const nCB = len & 0xfe000000;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 7:
if (len & 0x1000000) {
const nCB = len & 0xff000000;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 8:
if (len & 0x800000) {
const nCB = len & 0xff800000;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 9:
if (len & 0x400000) {
const nCB = len & 0xffc00000;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 10:
if (len & 0x200000) {
const nCB = len & 0xffe00000;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 11:
if (len & 0x100000) {
const nCB = len & 0xfff00000;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 12:
if (len & 0x80000) {
const nCB = len & 0xfff80000;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 13:
if (len & 0x40000) {
const nCB = len & 0xfffc0000;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 14:
if (len & 0x20000) {
const nCB = len & 0xfffe0000;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 15:
if (len & 0x10000) {
const nCB = len & 0xffff0000;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 16:
if (len & 0x8000) {
const nCB = len & 0xffff8000;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 17:
if (len & 0x4000) {
const nCB = len & 0xffffc000;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 18:
if (len & 0x2000) {
const nCB = len & 0xffffe000;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 19:
if (len & 0x1000) {
const nCB = len & 0xfffff000;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 20:
if (len & 0x800) {
const nCB = len & 0xfffff800;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 21:
if (len & 0x400) {
const nCB = len & 0xfffffc00;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 22:
if (len & 0x200) {
const nCB = len & 0xfffffe00;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 23:
if (len & 0x100) {
const nCB = len & 0xffffff00;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 24:
if (len & 0x80) {
const nCB = len & 0xffffff80;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 25:
if (len & 0x40) {
const nCB = len & 0xffffffc0;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 26:
if (len & 0x20) {
const nCB = len & 0xffffffe0;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 27:
if (len & 0x10) {
const nCB = len & 0xfffffff0;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 28:
if (len & 0x8) {
const nCB = len & 0xfffffff8;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 29:
if (len & 0x4) {
const nCB = len & 0xfffffffc;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 30:
if (len & 0x2) {
const nCB = len & 0xfffffffe;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
case 31:
if (len & 0x1) {
const nCB = len & 0xffffffff;
len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
}
}
// MODIFICATION: Instead of returning the index, this binary search
// instead returns whether something was found or not.
// For reverse filterning, I swapped true with false and vice-versa
if (array[len|0] !== sValue) {
return false; // preserve the value at this index
} else {
return true; // eliminate the value at this index
}
};
}
有关反向过滤代码的较慢的较小版本，请参见下面。
function sortAnyArray(a,b) { return a>b ? 1 : (a===b ? 0 : -1); }
function sortIntArray(a,b) { return (a|0) - (b|0) |0; }
function fastFilter(array, handle) {
var out=[], value=0;
for (var i=0, len=array.length|0; i < len; i=i+1|0)
if (handle(value = array[i]))
out.push( value );
return out;
}
/* USAGE:
reverseFilterArrayByAnotherArray(
[1,3,5],
[2,3,4]
) yields [3], and it can work with strings too
*/
function reverseFilterArrayByAnotherArray(searchArray, filterArray) {
if (
// NOTE: This does not check the whole array. But, if you know
// that there are only strings or numbers (not a mix of
// both) in the array, then this is a safe assumption.
typeof searchArray[0] == "number" &&
typeof filterArray[0] == "number" &&
(searchArray[0]|0) === searchArray[0] &&
(filterArray[0]|0) === filterArray[0]
) {
// if all entries in both arrays are integers
searchArray.sort(sortIntArray);
filterArray.sort(sortIntArray);
} else {
searchArray.sort(sortAnyArray);
filterArray.sort(sortAnyArray);
}
// Progressive Linear Search
var i = 0;
return fastFilter(searchArray, function(currentValue){
while (filterArray[i] < currentValue) i=i+1|0;
// +undefined = NaN, which is always false for <, avoiding an infinite loop
// For reverse filter, I changed !== to ===
return filterArray[i] === currentValue;
});
}
/* Here's an example that uses (some) ES6 Javascript semantics to filter an object array by another object array. */ // x = full dataset // y = filter dataset let x = [ {"val": 1, "text": "a"}, {"val": 2, "text": "b"}, {"val": 3, "text": "c"}, {"val": 4, "text": "d"}, {"val": 5, "text": "e"} ], y = [ {"val": 1, "text": "a"}, {"val": 4, "text": "d"} ]; // Use map to get a simple array of "val" values. Ex: [1,4] let yFilter = y.map(itemY => { return itemY.val; }); // Use filter and "not" includes to filter the full dataset by the filter dataset's val. let filteredX = x.filter(itemX => !yFilter.includes(itemX.val)); // Print the result. console.log(filteredX);
OA也可以在ES6中实现，如下所示
ES6:
const filtered = [1, 2, 3, 4].filter(e => {
return this.indexOf(e) < 0;
},[2, 4]);
你的问题有很多答案，但我没有看到任何人使用lambda表达式:
var array = [1,2,3,4];
var anotherOne = [2,4];
var filteredArray = array.filter(x => anotherOne.indexOf(x) < 0);
Jack Giffin的解决方案很好，但不适用于大于2^32的数组。下面是基于Jack的解决方案来过滤数组的重构快速版本，但它适用于64位数组。
const Math_clz32 = Math.clz32 || ((log, LN2) => x => 31 - log(x >>> 0) / LN2 | 0)(Math.log, Math.LN2);
const filterArrayByAnotherArray = (searchArray, filterArray) => {
searchArray.sort((a,b) => a > b);
filterArray.sort((a,b) => a > b);
let searchArrayLen = searchArray.length, filterArrayLen = filterArray.length;
let progressiveLinearComplexity = ((searchArrayLen<<1) + filterArrayLen)>>>0
let binarySearchComplexity = (searchArrayLen * (32-Math_clz32(filterArrayLen-1)))>>>0;
let i = 0;
if (progressiveLinearComplexity < binarySearchComplexity) {
return searchArray.filter(currentValue => {
while (filterArray[i] < currentValue) i=i+1|0;
return filterArray[i] !== currentValue;
});
}
else return searchArray.filter(e => binarySearch(filterArray, e) === null);
}
const binarySearch = (sortedArray, elToFind) => {
let lowIndex = 0;
let highIndex = sortedArray.length - 1;
while (lowIndex <= highIndex) {
let midIndex = Math.floor((lowIndex + highIndex) / 2);
if (sortedArray[midIndex] == elToFind) return midIndex;
else if (sortedArray[midIndex] < elToFind) lowIndex = midIndex + 1;
else highIndex = midIndex - 1;
} return null;
}
下面的代码是根据另一个数组过滤一个数组的最简单方法。两个数组都可以在其中包含对象而不是值。
Let array1 = [1,3,47,1,6,7]; Let array2 = [3,6]; let filteredArray1 = array1。Filter (el => array2.includes(el)); console.log (filteredArray1);
输出:[3,6]
下面是一个例子
让firstArray =[1、2、3、4、5); 让secondArray =(2、3); let filteredArray = firstArray.filter((a) => secondArray.indexOf(a)<0); console.log (filteredArray);//上面的行给出[1,4,5]
如果你需要比较一个对象数组，这在所有情况下都适用:
let arr = [{ id: 1, title: "title1" },{ id: 2, title: "title2" }]
let brr = [{ id: 2, title: "title2" },{ id: 3, title: "title3" }]
const res = arr.filter(f => brr.some(item => item.id === f.id));
console.log(res);
使用对象筛选结果
[{id:1},{id:2},{id:3},{id:4}].filter(v=>!([{id:2},{id:4}].some(e=>e.id === v.id)))
下面是当数组中的项是对象时的操作方法。
其思想是使用map函数在内部数组中查找仅包含键的数组
然后检查这些键的数组是否包含外层数组中的特定元素键。
const existsInBothArrays = array1.filter((element1) =>
array2.map((element2) => element2._searchKey).includes(element1._searchKey),
);
我只是想给你一个额外的解决方案…
const arr1 = [1,2,3,4];
const arr2 = [2,4];
const container = arr2.reduce((res,item) => {
res[item] = true;
return res;
}, {});
const result = arr1.filter(item => !container[item]);
关于上述代码的时间复杂度:O(n)。 而且,……我们需要更多的空间(空间复杂度O(n)) =>权衡…:))
如果你想过滤一个具有一些匹配属性的不同结构的数组，你应该这样做。
let filteredArray = [];
array1.map(array1Item => {
array2.map(array2Item => {
if (array1.property1 === array2.property2) {
filteredArray.push(array1Item);
}
});
这会让你的生活变得轻松!
这完全取决于数组的类型。
对于简单的数组，比如字符串数组。你可以像@Redu和@Hugolpz指出的那样使用下面的代码
var arr1 = [1,2,3,4],
arr2 = [2,4],
res = arr1.filter(item => !arr2.includes(item));
console.log(res);
然后，对于更复杂的数组过滤器，比如从另一个对象数组中过滤一个对象数组，您可以使用下面的代码
function filterArray(arr1, arr2) {
return arr1.filter(item1 => !arr2.some(item2 => item1.id === item2.id));
}
OR
function filterArray(arr1, arr2) {
const set = new Set(arr2.map(item => item.id));
return arr1.reduce((filtered, item) => {
if (!set.has(item.id)) {
filtered.push(item);
}
return filtered;
}, []);
}
这两种方法都是有效的，即使使用大型数组也能很好地工作。但是，如果您有大量数据，则使用性能更好的数据结构(如Map或HashSet)可能会更优。
Map
function filterArray(arr1, arr2) {
const map = new Map(arr2.map(item => [item.id, item]));
return arr1.filter(item => !map.has(item.id));
}
Hashset
function filterArray(arr1, arr2) {
const set = new Set(arr2.map(item => item.id));
return arr1.filter(item => !set.has(item.id));
}
