如何动态声明一组过滤条件而无需指定过滤器数量?
例如,如果我有一组数据,如下所示:
var data = [
{ item: { type: 'wood', size: 10 } },
{ item: { type: 'wood', size: 8 } },
{ item: { type: 'metal', size: 8 } }
]
我知道我可以使用JS .filter()
获取包含type
wood
和size
8
的所有项目:
function filterItems() {
return data.filter(function(val) {
return val['item'].type == 'wood' &&
val['item'].size == 8;
}
}
但是,如果我想过滤使用未知数量的过滤条件的商品,并.filter()
返回符合这些条件的所有data
项,该怎么办?
答案 0 :(得分:8)
您可以将一系列条件传递给filterItems()
函数。试试这个:
function filterItems(filters) {
return data.filter(function(val) {
for(var i = 0; i < filters.length; i++)
if(val['item'][filters[i][0]] != filters[i][1])
return false;
return true;
}
}
filterItems([['type', 'wood'], ['size', 8], ['someother', 'value']]);
同样的想法可以应用于各种格式,例如使用对象而不是数组来提高可读性。
答案 1 :(得分:1)
我只是对 Amit 对任何数据结构和嵌套属性支持的回答进行了一行重构
from typing import Dict, Any
from pydantic import BaseModel
class Model(BaseModel):
__root__: Dict[str, Any]
def __iter__(self):
return iter(self.__root__)
def __getattr__(self, item):
return self.__root__[item]
m = Model.parse_obj({'key1': 'val1', 'key2': 'val2'})
assert m.key1 == "val1"
答案 2 :(得分:0)
function isSingle(filter) {
return (filter && 'o' in filter && 'm' in filter && 'v' in filter);
}
function isComposite(filter) {
return (filter && 'lo' in filter);
}
function createBody(filter) {
if (isComposite(filter)) {
var bdy = "";
if (filter.v.length > 1) {
var o = filter.lo;
return "(" + createBody(filter.v.shift()) + " " + o + " " + createBody({ lo: filter.lo, v: filter.v }) + ")";
} else if (filter.v.length == 1) {
return createBody(filter.v.shift());
}
return bdy;
} else if (isSingle(filter)) {
var o = filter.o;
if (typeof filter.v == "string") filter.v = "'" + filter.v + "'"
return "item." + filter.m + " " + o + " " + filter.v;
}
}
var createFunc = function (filter) {
var body = createBody(filter);
var f = new Function("item", " return " + body + ";");
return f;
}
function applyFilter(input, filter) {
if (filter == undefined) {
return input;
}
var fun = createFunc(filter);
var output = input.filter(fun);
return output;
};
//m:member,o:operator,v:value.
var filterQuery1 = { m: "item.type", o: "==", v: "metal" };//simpe query
var filterQuery2 = { m: "item.size", o: ">", v: 8 };
var filterQuery3 = {
lo: "&&", v: [
{ m: "item.type", o: "==", v: "metal" },
{ m: "item.size", o: "<", v: 9 }]
}; //composite query
var data = [
{ item: { type: 'wood', size: 10 } },
{ item: { type: 'wood', size: 8 } },
{ item: { type: 'metal', size: 8 } }
]
var result = applyFilter(data, filterQuery1);// or filterQuery2,filterQuery3
console.log(result);
答案 3 :(得分:0)
Amit的回答很好,但我想补充一下。就我而言,我需要返回所有参数,否则返回none / false。这是阿米特(Amit)的编辑代码
function filterItems(filters) {
return data.filter(function(val) {
let result = true;
for(var i = 0; i < filters.length; i++)
if(val['item'][filters[i][0]] != filters[i][1])
result = false;
return result;
}
}
filterItems([['type', 'wood'], ['size', 8], ['someother', 'value']]);