初学者 React/JS 问题在这里。我有一系列条目（表单提交），每个条目都包含如下语言数组：0: {…}dream: Array [ "Bengali", "French" ]feel: Array [ "Arabic", "English" ]speak: Array [ "Afrikaans", "Armenian" ]think: Array [ "Albanian" ]1: {…}dream: Array [ "English", "French" ]feel: Array [ "German", "Italian" ]speak: Array [ "Afrikaans", "English" ]think: Array [ "Cantonese" ]我想要每个类别中所有语言的数组。我知道我必须过滤每个条目并保存语言，但我不知道如何检查重复项。目前，我可以过滤以查看使用单一语言的条目setFilter(entries.filter((key) => key.speak.includes("Afrikaans")));但我不知道如何创建所有语言的主列表。
5 回答
萧十郎
TA贡献1815条经验 获得超13个赞
这是你想要的吗 ？您需要减少和过滤 du 避免重复，结果您将得到一个对象
const myArray = [{
dream: [ "Bengali", "French" ],
feel: [ "Arabic", "English" ],
speak: [ "Afrikaans", "Armenian" ],
think: [ "Albanian" ]
}
,{
dream: [ "Bengaslis", "French" ],
feel: [ "Arabsic", "English" ],
speak: [ "Afrikasans", "Armenian" ],
think: [ "Assslbanian" ]
}
];
const result = myArray.reduce((accumulator, currentValue) => {
const dream = accumulator.dream.concat(currentValue.dream);
const think = accumulator.think.concat(currentValue.think);
const feel = accumulator.feel.concat(currentValue.feel);
const speak = accumulator.speak.concat(currentValue.speak);
accumulator.dream = dream.filter((item, pos) => dream.indexOf(item) === pos)
accumulator.feel = feel.filter((item, pos) => feel.indexOf(item) === pos);
accumulator.speak = speak.filter((item, pos) => speak.indexOf(item) === pos);
accumulator.think = think.filter((item, pos) => think.indexOf(item) === pos);
return accumulator;
})
console.log(result)
尚方宝剑之说
TA贡献1788条经验 获得超4个赞
您可以使用Object.valuesandObject.entries来遍历数据对象。
const data = {
0: {
dream: ["Bengali", "French"],
feel: ["Arabic", "English"],
speak: ["Afrikaans", "Armenian"],
think: ["Albanian"],
},
1: {
dream: ["English", "French"],
feel: ["German", "Italian"],
speak: ["Afrikaans", "English"],
think: ["Cantonese"],
}
};
const categoryMap = Object.values(data)
.reduce((concatedArr, item) => concatedArr.concat(Object.entries(item)), [])
.reduce((result, [category, values]) => {
result[category] = result[category] || [];
result[category] = result[category].concat(values);
return result;
}, {});
console.log(categoryMap);
慕桂英4014372
TA贡献1871条经验 获得超13个赞
老实说，我不太理解这个问题。但我试着这样解释，让我知道：
const entries = [
{
dream: [ "Bengali", "French" ],
feel: [ "Arabic", "English"],
speak: [ "Afrikaans", "Armenian" ],
think: [ "Albanian" ],
},{
dream: [ "English", "French" ],
feel: [ "German", "Italian" ],
speak: [ "Afrikaans", "English" ],
think: [ "Cantonese" ],
}
]
const fields = ["dream", "feel", "speak","think"];
const result = {};
for(field of fields){
// create an array containing all the languages of the specific category of the for-loop (there may be duplications)
const rawArray = entries.map(item => item[field]).flat();
// delete duplicates
const arrayWithoutDuplicates = rawArray.filter((item,index) => rawArray.indexOf(item) === index);
result[field] = arrayWithoutDuplicates;
}
console.log(result);
人到中年有点甜
TA贡献1895条经验 获得超7个赞
试试这个 Obj 结构
var completObj=[
{
dream:[ "Bengali", "French" ],
feel:[ "Arabic", "English" ],
speak:[ "Afrikaans", "Armenian" ],
think:[ "Albanian" ]
},
{
dream:[ "English", "French" ],
feel:[ "German", "Italian" ],
speak:[ "Afrikaans", "English" ],
think:[ "Cantonese" ]
}
]
var Afrikaans=completObj.filter((key) => key.speak.includes("Afrikaans"))
console.log('Afrikaans:'+JSON.stringify( Afrikaans))
var uniqueObj={}
completObj.map((value, index) => {
Object.entries(completObj[index]).forEach(([ObjKey, value]) => {
if (!uniqueObj[ObjKey]) {
uniqueObj[ObjKey] = new Set();
}
debugger
value.map(list => uniqueObj[ObjKey].add(list));
// uniqueObj[ObjKey].add(value)
});
});
var dream=Array.from(uniqueObj.dream)
console.log('dream unique Value:'+JSON.stringify( dream))
忽然笑
TA贡献1806条经验 获得超5个赞
给定数据
const data = [
{
dream: ["Bengali", "French"],
feel: ["Arabic", "English"],
speak: ["Afrikaans", "Armenian"],
think: ["Albanian"]
},
{
dream: ["English", "French"],
feel: ["German", "Italian"],
speak: ["Afrikaans", "English"],
think: ["Cantonese"]
}
];
首先将类别语言数组数组转换为类别语言集的简化对象。这些集合用于删除重复项。
const reducedData = data.reduce((categories, current) => {
// Loop over the categories (dream, feel, etc..)
// and add languages to sets
Object.entries(current).forEach(([currentCategory, languages]) => {
if (!categories[currentCategory]) {
// A set will not allow duplicate entires
categories[currentCategory] = new Set();
}
// Add all languages to the set
languages.forEach(language => categories[currentCategory].add(language));
});
return categories;
}, {});
你现在有一个有形状的物体
{
dream: Set("Bengali", "French", "English"),
feel: Set("Arabic", "English", "German", "Italian"),
speak: Set("Afrikaans", "Armenian", "English"),
think: Set("Albanian", "Cantonese"),
}
然后再次将其缩减为类别对象 - 语言数组。这只是将集合转换回数组。
const reducedDataArray = Object.entries(reducedData).reduce(
(categories, [category, languageSet]) => {
// Convert Set back to an array
categories[category] = [...languageSet];
return categories;
},
{}
);
结果对象形状
{
dream: ["Bengali", "French", "English"],
feel: ["Arabic", "English", "German", "Italian"],
speak: ["Afrikaans", "Armenian", "English"],
think: ["Albanian", "Cantonese"],
}
const data = [
{
dream: ["Bengali", "French"],
feel: ["Arabic", "English"],
speak: ["Afrikaans", "Armenian"],
think: ["Albanian"]
},
{
dream: ["English", "French"],
feel: ["German", "Italian"],
speak: ["Afrikaans", "English"],
think: ["Cantonese"]
}
];
const reducedData = data.reduce((categories, current) => {
Object.entries(current).forEach(([currentCategory, languages]) => {
if (!categories[currentCategory]) {
categories[currentCategory] = new Set();
}
languages.forEach(language => categories[currentCategory].add(language));
});
return categories;
}, {});
const reducedDataArray = Object.entries(reducedData).reduce(
(categories, [category, languageSet]) => {
categories[category] = [...languageSet];
return categories;
},
{}
);
console.log(reducedDataArray);
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