我有两个对象。一个是单词数组,另一个是第一个数组中两个单词之间的关系:
nodes = ["apple","cherry","pear","strawberry"]
data = [{word1 : "apple", word2 : "cherry" , weight : "0.1"},
{word1 : "apple", word2 : "strawberry" , weight : "0.2"},
{word1 : "cherry", word2 : "pear" , weight : "0.3"},
{word1 : "cherry", word2 : "strawberry" , weight : "0.4"},
{word1 : "strawberry", word2 : "pear" , weight : "0.5"}]
我想创建一个具有数据结构的新对象,而不是单词,而是带有索引。像这样:
links = [{source : "0", target : "1" , value : "0.1"},
{source : "0", target : "3" , value : "0.2"},
{source : "1", target : "2" , value : "0.3"},
{source : "1", target : "3" , value : "0.4"},
{source : "3", target : "2" , value : "0.5"}]
到目前为止,我有这个:
for (var i = 0 ; i < nodes.length; i++) {
test[i] = nodes.findIndex(d => d == nodes[i]);
}
这给了我这个新数组,对应于nodes
的索引:
test = [0,1,2,3]
那么我现在该如何将这些索引分配给links
?
答案 0 :(得分:3)
尝试:
var nodes = ["apple","cherry","pear","strawberry"];
var data = [{word1 : "apple", word2 : "cherry" , weight : "0.1"},
{word1 : "apple", word2 : "strawberry" , weight : "0.2"},
{word1 : "cherry", word2 : "pear" , weight : "0.3"},
{word1 : "cherry", word2 : "strawberry" , weight : "0.4"},
{word1 : "strawberry", word2 : "pear" , weight : "0.5"}];
var links = data.map((item) => {
return {
source: nodes.indexOf(item.word1),
target: nodes.indexOf(item.word2),
value: item.weight
};
});
console.log(links);
console.log(data);
答案 1 :(得分:1)
您可以在第二个数组(而不是第一个)和Array#map
上使用Object.assign
,以避免原始数据发生变异。
为了减少时间复杂度,可以创建Map
,这样可以在恒定时间内获取索引。该地图可以通过this
传递给地图方法。通过检查每个属性的匹配项,您可以将其设置为通用属性,因此算法本身中没有硬编码的word1
或word2
:
const nodes = ["apple","cherry","pear","strawberry"]
const data = [
{word1 : "apple", word2 : "cherry" , weight : "0.1"},
{word1 : "apple", word2 : "strawberry" , weight : "0.2"},
{word1 : "cherry", word2 : "pear" , weight : "0.3"},
{word1 : "cherry", word2 : "strawberry" , weight : "0.4"},
{word1 : "strawberry", word2 : "pear" , weight : "0.5"}
];
const result = data.map(function(o) {
return Object.assign(...Object.keys(o).map(key => ({
[key]: this.has(o[key]) ? this.get(o[key]) : o[key]
})));
}, new Map(nodes.map( (s,i) => [s,i] )));
console.log(result);
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答案 2 :(得分:0)
使用Array#map()
创建新阵列。获取indexOf(word1)
和indexOf(word2)
来替换当前值
const nodes = ["apple","cherry","pear","strawberry"];
const data = [{word1 : "apple", word2 : "cherry" , weight : "0.1"},
{word1 : "apple", word2 : "strawberry" , weight : "0.2"},
{word1 : "cherry", word2 : "pear" , weight : "0.3"},
{word1 : "cherry", word2 : "strawberry" , weight : "0.4"},
{word1 : "strawberry", word2 : "pear" , weight : "0.5"}];
const updatedData = data.map((item) => ({
source: nodes.indexOf(item.word1),
target: nodes.indexOf(item.word2),
weight: item.weight
}));
console.log(updatedData);
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答案 3 :(得分:0)
这样可行:
nodes = ["apple", "cherry", "pear", "strawberry"]
data = [{
word1: "apple",
word2: "cherry",
weight: "0.1"
},
{
word1: "apple",
word2: "strawberry",
weight: "0.2"
},
{
word1: "cherry",
word2: "pear",
weight: "0.3"
},
{
word1: "cherry",
word2: "strawberry",
weight: "0.4"
},
{
word1: "strawberry",
word2: "pear",
weight: "0.5"
}
]
var links = [];
for (var i = 0; i < data.length; i++) {
links.push({
source: nodes.indexOf(data[i].word1),
target: nodes.indexOf(data[i].word2),
value: data[i].weight
})
}
console.log(links);