Javascript 以最小的复杂性对数组中的元素进行分组的最佳方法
我有一个JSON数组,如下所示:Javascript 以最小的复杂性对数组中的元素进行分组的最佳方法,javascript,performance,algorithm,Javascript,Performance,Algorithm,我有一个JSON数组,如下所示: var map_results = [{"Type":"Flat","Price":100.9}, {"Type":"Room","Price":23.5}, {"Type":"Flat","Price":67.5}, {"Type":"Flat","Price":100.9} {"Type":"Plot","P
var map_results = [{"Type":"Flat","Price":100.9},
{"Type":"Room","Price":23.5},
{"Type":"Flat","Price":67.5},
{"Type":"Flat","Price":100.9}
{"Type":"Plot","Price":89.8}]
var expected_output = [{"Type":"Flat", "Data":[{"Price":100.9, "Total":2},
{"Price":67.5, "Total":1}] },
{"Type":"Room","Data":[{"Price":23.5,"Total":1}]},
{"Type":"Plot","Data":[{"Price":89.8, "Total:1"}]}]
var reduce = function (map_results) {
var results = [];
for (var i in map_results) {
var type_found = 0;
for(var result in results){
if (map_results[i]["Type"] == results[result]["Type"]){
type_found = 1;
var price_found = 0;
for(var data in results[result]["Data"]){
if(map_results[i]["Price"] == results[result]["Data"][data]["Price"]){
price_found = 1;
results[result]["Data"][data]["Total"] +=1;
}
}
if(price_found == 0){
results[result]["Data"].push({"Price":map_results[i]["Price"], "Total":1});
}
}
}
if(type_found == 0){
results.push({"Type":map_results[i]["Type"], "Data":[{"Price":map_results[i]["Price"],"Total":1}]});
}
}
return results;
};
var reduce_func = function (previous, current) {
if(previous.length == 0){
previous.push({Type: current.Type, Data:[{Price:current.Price,Total:1}]});
return previous;
}
var type_found = 0;
for (var one in previous) {
if (current.Type == previous[one].Type){
type_found = 1;
var price_found = 0;
for(var data in previous[one].Data){
if(current.Price == previous[one].Data[data].Price){
price_found = 1;
previous[one].Data[data].Total += 1;
}
}
if(price_found == 0){
previous[one].Data.push({Price:current.Price, Total:1});
}
}
}
if(type_found == 0){
previous.push({Type:current.Type, Data:[{Price : current.Price ,Total:1}]});
}
return previous;
}
map_results.reduce(reduce_func,[]);
这个数组包含大约100000条记录。我希望输出按“类型”和“价格”分组。应该是这样的:
var map_results = [{"Type":"Flat","Price":100.9},
{"Type":"Room","Price":23.5},
{"Type":"Flat","Price":67.5},
{"Type":"Flat","Price":100.9}
{"Type":"Plot","Price":89.8}]
var expected_output = [{"Type":"Flat", "Data":[{"Price":100.9, "Total":2},
{"Price":67.5, "Total":1}] },
{"Type":"Room","Data":[{"Price":23.5,"Total":1}]},
{"Type":"Plot","Data":[{"Price":89.8, "Total:1"}]}]
var reduce = function (map_results) {
var results = [];
for (var i in map_results) {
var type_found = 0;
for(var result in results){
if (map_results[i]["Type"] == results[result]["Type"]){
type_found = 1;
var price_found = 0;
for(var data in results[result]["Data"]){
if(map_results[i]["Price"] == results[result]["Data"][data]["Price"]){
price_found = 1;
results[result]["Data"][data]["Total"] +=1;
}
}
if(price_found == 0){
results[result]["Data"].push({"Price":map_results[i]["Price"], "Total":1});
}
}
}
if(type_found == 0){
results.push({"Type":map_results[i]["Type"], "Data":[{"Price":map_results[i]["Price"],"Total":1}]});
}
}
return results;
};
var reduce_func = function (previous, current) {
if(previous.length == 0){
previous.push({Type: current.Type, Data:[{Price:current.Price,Total:1}]});
return previous;
}
var type_found = 0;
for (var one in previous) {
if (current.Type == previous[one].Type){
type_found = 1;
var price_found = 0;
for(var data in previous[one].Data){
if(current.Price == previous[one].Data[data].Price){
price_found = 1;
previous[one].Data[data].Total += 1;
}
}
if(price_found == 0){
previous[one].Data.push({Price:current.Price, Total:1});
}
}
}
if(type_found == 0){
previous.push({Type:current.Type, Data:[{Price : current.Price ,Total:1}]});
}
return previous;
}
map_results.reduce(reduce_func,[]);
这必须在纯javascript中完成,我不能使用像undersore.js这样的库。我试图解决这个问题,但它有3个嵌套for循环,这使得复杂性为n^4。对于这个问题,有什么更好的解决方案
我的函数如下所示:
var map_results = [{"Type":"Flat","Price":100.9},
{"Type":"Room","Price":23.5},
{"Type":"Flat","Price":67.5},
{"Type":"Flat","Price":100.9}
{"Type":"Plot","Price":89.8}]
var expected_output = [{"Type":"Flat", "Data":[{"Price":100.9, "Total":2},
{"Price":67.5, "Total":1}] },
{"Type":"Room","Data":[{"Price":23.5,"Total":1}]},
{"Type":"Plot","Data":[{"Price":89.8, "Total:1"}]}]
var reduce = function (map_results) {
var results = [];
for (var i in map_results) {
var type_found = 0;
for(var result in results){
if (map_results[i]["Type"] == results[result]["Type"]){
type_found = 1;
var price_found = 0;
for(var data in results[result]["Data"]){
if(map_results[i]["Price"] == results[result]["Data"][data]["Price"]){
price_found = 1;
results[result]["Data"][data]["Total"] +=1;
}
}
if(price_found == 0){
results[result]["Data"].push({"Price":map_results[i]["Price"], "Total":1});
}
}
}
if(type_found == 0){
results.push({"Type":map_results[i]["Type"], "Data":[{"Price":map_results[i]["Price"],"Total":1}]});
}
}
return results;
};
var reduce_func = function (previous, current) {
if(previous.length == 0){
previous.push({Type: current.Type, Data:[{Price:current.Price,Total:1}]});
return previous;
}
var type_found = 0;
for (var one in previous) {
if (current.Type == previous[one].Type){
type_found = 1;
var price_found = 0;
for(var data in previous[one].Data){
if(current.Price == previous[one].Data[data].Price){
price_found = 1;
previous[one].Data[data].Total += 1;
}
}
if(price_found == 0){
previous[one].Data.push({Price:current.Price, Total:1});
}
}
}
if(type_found == 0){
previous.push({Type:current.Type, Data:[{Price : current.Price ,Total:1}]});
}
return previous;
}
map_results.reduce(reduce_func,[]);
由于
Type
s和Price
s在分组后是唯一的,我认为像{“Flat”:{“100.9”:2,“67.5”:1},{“Room”:{“23.5”:1}}
这样的结构更容易处理。因此,可以通过以下方式进行分组:
var output = {};
map_results.map(function(el, i) {
output[el["Type"]] = output[el["Type"]] || [];
output[el["Type"]][el["Price"] = (output[el["Type"]][el["Price"]+1) || 1;
});
如果无法处理此结构,则可以执行另一个到结构的映射。
当您对数组进行一次迭代时,其复杂性应为n
找一把能用的小提琴
编辑:所以将所有内容重新映射到您的结构。重新映射的顺序远小于第一次映射,因为分组已经完成
var expected_output = [];
for(type in output) {
var prices = [];
for(price in output[type]) {
prices.push({"Price": price, "Total": output[type][price]);
}
expected_output.push({"Type": type, "Data": prices});
}
我有一个简短的函数来处理请求功能的第一部分:它将
map\u结果
映射到所需的格式:
var map_results = [{"Type":"Flat","Price":100.9},
{"Type":"Room","Price":23.5},
{"Type":"Flat","Price":67.5},
{"Type":"Flat","Price":100.9},
{"Type":"Plot","Price":89.8}]
var expected_output = map_results.reduce(function(obj, current){
if(!obj[current.Type]){
obj[current.Type] = {'Type':current.Type, 'Data':[]};
}
obj[current.Type].Data.push({'Price':current.Price, 'Total':1});
return obj;
},{})
那么,恐怕计算总数需要这段代码:
for(var type in expected_output){
var d = {};
for(var item in expected_output[type].Data){
d[expected_output[type].Data[item].Price] = (d[expected_output[type].Data[item].Price] || 0) + 1;
}
expected_output[type].Data = [];
for(var i in d){
expected_output[type].Data.push({
'Price':i,
'Total':d[i]
})
}
}
输出:
{
"Flat":{
"Type":"Flat",
"Data":[{"Price":"100.9","Total":2},
{"Price":"67.5","Total":1}]
},
"Room":{
"Type":"Room",
"Data":[{"Price":"23.5","Total":1}]
},
"Plot":{
"Type":"Plot",
"Data":[{"Price":"89.8","Total":1}]
}
}
下面是另一项努力。这里有一个 为了进行性能测试,我还制作了一个包含163840个元素的模型。在Chrome(OSX)上,原始解决方案比此解决方案慢90% 几点注意: 可以根据您的情况进行优化(例如,对对象克隆进行hasOwnProperty检查) 此外,如果您需要最新的总计作为第一个元素,请使用
unshift
,而不是按将obj添加到数组的开头
function groupBy(arr, key, key2) {
var retArr = [];
arr.reduce(function(previousValue, currentValue, index, array){
if(currentValue.hasOwnProperty(key)) {
var kVal = currentValue[key];
if(!previousValue.hasOwnProperty(kVal)) {
previousValue[kVal] = {};
retArr.push(previousValue[kVal]);
previousValue[kVal][key] = kVal;
previousValue[kVal]["Data"] = [];
}
var prevNode = previousValue[kVal];
if(currentValue.hasOwnProperty(key2)) {
var obj = {};
for(var k in currentValue) {
if(currentValue.hasOwnProperty(k) && k!=key)
obj[k] = currentValue[k];
}
obj["Total"] = prevNode["Data"].length + 1;
prevNode["Data"].push(obj);
}
}
return previousValue;
}, {});
return retArr;
}
var map_results = [{"Type":"Flat","Price":100.9},
{"Type":"Room","Price":23.5},
{"Type":"Flat","Price":67.5},
{"Type":"Flat","Price":100.9},
{"Type":"Plot","Price":89.8}];
var expected_output = groupBy(map_results, "Type", "Price");
console.dir(expected_output);
试过这样的方法:
var map_results = [{"Type":"Flat","Price":100.9},
{"Type":"Room","Price":23.5},
{"Type":"Flat","Price":67.5},
{"Type":"Flat","Price":100.9}
{"Type":"Plot","Price":89.8}]
var expected_output = [{"Type":"Flat", "Data":[{"Price":100.9, "Total":2},
{"Price":67.5, "Total":1}] },
{"Type":"Room","Data":[{"Price":23.5,"Total":1}]},
{"Type":"Plot","Data":[{"Price":89.8, "Total:1"}]}]
var reduce = function (map_results) {
var results = [];
for (var i in map_results) {
var type_found = 0;
for(var result in results){
if (map_results[i]["Type"] == results[result]["Type"]){
type_found = 1;
var price_found = 0;
for(var data in results[result]["Data"]){
if(map_results[i]["Price"] == results[result]["Data"][data]["Price"]){
price_found = 1;
results[result]["Data"][data]["Total"] +=1;
}
}
if(price_found == 0){
results[result]["Data"].push({"Price":map_results[i]["Price"], "Total":1});
}
}
}
if(type_found == 0){
results.push({"Type":map_results[i]["Type"], "Data":[{"Price":map_results[i]["Price"],"Total":1}]});
}
}
return results;
};
var reduce_func = function (previous, current) {
if(previous.length == 0){
previous.push({Type: current.Type, Data:[{Price:current.Price,Total:1}]});
return previous;
}
var type_found = 0;
for (var one in previous) {
if (current.Type == previous[one].Type){
type_found = 1;
var price_found = 0;
for(var data in previous[one].Data){
if(current.Price == previous[one].Data[data].Price){
price_found = 1;
previous[one].Data[data].Total += 1;
}
}
if(price_found == 0){
previous[one].Data.push({Price:current.Price, Total:1});
}
}
}
if(type_found == 0){
previous.push({Type:current.Type, Data:[{Price : current.Price ,Total:1}]});
}
return previous;
}
map_results.reduce(reduce_func,[]);
我认为更好的方法是显示您的4嵌套for循环,然后我们可以帮助您优化it@Madedan张贴。。但是Oops3对于循环…这很好,但是我认为我不能改变结构