增加Javascript数组中项目的值

Question

I have Javascript array that looks like this:

 fruitsGroups: [
    "apple0",
    "banana0",
    "pear0",
  ]

How can I increase the number of each item in this array?

 fruitsGroups: [

    "apple0",
    "apple1",
    "apple2",

    "banana0",
    "banana1",
    "banana2",

    "pear0",
    "pear1",
    "pear2"
  ]

Answer 1

I think your looking for something like that?

var fruitsGroups = [
    "apple0",
    "banana0",
    "pear0",
  ];
  console.log(fruitsGroups);

  var newFruits = [];
  $.each(fruitsGroups, function(i, j) {
     var n = parseInt(j.substring(j.length - 1));
     for(var k = 0; k < 3; k++) {
        newFruits.push(j.substring(0, j.length - 1) + (n + k));
     }
  });

  console.log(newFruits);

Answer 2

You could create function that uses reduce() method and returns new array.

 function mult(data, n) { return data.reduce((r, e) => { return r.push(e, ...Array.from(Array(n), (_, i) => { const [text, n] = e.split(/(\\d+)/); return text + (+n + i + 1) })), r }, []); } console.log(mult(["apple0", "banana0", "pear0"], 2)) console.log(mult(["apple4", "banana2", "pear0"], 3)) 

Answer 3

Since we have 2018 already, another approach using Array.map and destructuring:

const groups = [
    "apple0",
    "banana0",
    "pear0",
  ];

[].concat(...groups.map(item => [
    item,
    item.replace(0, 1),
    item.replace(0, 2)
  ]
))

// result: ["apple0", "apple1", "apple2",
//          "banana0", "banana1", "banana2",
//          "pear0", "pear1", "pear2"]

Explanation:

groups.map(item => [item, item.replace(0, 1), item.replace(0, 2)]) takes each array item one by one ( apple0 , then banana0 , …) and replaces it with an array of:

• item – the item itself ( apple0 )
• item.replace(0, 1) – the item with zero replaced by 1 ( apple1 )
• item.replace(0, 2) – the item with zero replaced by 2 ( apple2 )

so the array looks like…

[
  ["apple0", "apple1", "apple2"],
  ["banana0", "banana1", "banana2"],
  ["pear0", "pear1", "pear2"],
]

…and then we need to flatten it, that's the [].concat(... part. It basically takes array items (the three dots, read more about destructuring here ), and merges them into an empty array.

If you want to replace any digit, not just zero, use regular expression:

"apple0".replace(/\d$/, 1)
 // -> "apple1"
"apple9".replace(/\d$/, 1)
 // -> "apple1"

• \\d – any number character
• $ - end of line
• the surrounding slashes tell JS that it's a regular expression, you could use new RegExp("\\d$") instead, too

Answer 4

I have tried this for you, it might help as per my understanding. Naive Approach

 var a = ['apple0','banana0','pearl0'] var fruitGroups = [] for(var i=0; i < a.length; i++){ for(var j = 0; j<a.length; j++){ let fruit = a[i].replace(/\\d/g,'')+j fruitGroups.push(fruit) } } console.log(fruitGroups) 

Answer 5

Map the original array. For each item, create a sub array, and fill it with the current item. Then map the sub array and replace the digit/s of each item with the index. Flatten the sub arrays, by spreading into Array.concat() :

 const fruitsGroups = [ "apple0", "banana0", "pear0", ]; const len = 3; // map the original array, and use concat to flatten the sub arrays const result = [].concat(...fruitsGroups.map((item) => new Array(len) // create a new sub array with the requested size .fill(item) // fill it with the item .map((s, i) => s.replace(/\\d+/, i)) // map the strings in the sub array, and replace the number with the index )); console.log(result);