{"id":4510,"date":"2022-08-23T04:42:16","date_gmt":"2022-08-22T23:12:16","guid":{"rendered":"https:\/\/jassweb.com\/solved\/solved-array-cross-over-loop-in-javascript\/"},"modified":"2022-08-23T04:42:16","modified_gmt":"2022-08-22T23:12:16","slug":"solved-array-cross-over-loop-in-javascript","status":"publish","type":"post","link":"https:\/\/jassweb.com\/solved\/solved-array-cross-over-loop-in-javascript\/","title":{"rendered":"[Solved] array cross-over loop in javascript"},"content":{"rendered":"

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This is a functional ES6 approach.<\/p>\n<\/p>\n

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let variants = [{\r\n    variantName: \"Size\",\r\n    variantItems: [\r\n      \"XL\",\r\n      \"MD\",\r\n      \"SM\"\r\n    ]\r\n  },\r\n  {\r\n    variantName: \"Color\",\r\n    variantItems: [\r\n      \"Red\",\r\n      \"Blue\"\r\n    ]\r\n  }];\r\n\r\nlet crossJoined = new Array(variants.reduce((product, variant) => (product * variant.variantItems.length), 1))\r\n  .fill(0)\r\n  .reduce(crossJoin => {\r\n    crossJoin.data.push(crossJoin.currentIndexes.map((itemIndex, variantIndex) => `${variants[variantIndex].variantName}: ${variants[variantIndex].variantItems[itemIndex]}`).join(\", \"));\r\n\r\n    let incrementableIndex = variants.length - crossJoin.currentIndexes\r\n      .slice()\r\n      .reverse()\r\n      .findIndex((itemIndex, variantIndex) => variants[variants.length - variantIndex - 1].variantItems.length > itemIndex + 1) - 1;\r\n\r\n    crossJoin.currentIndexes[incrementableIndex]++;\r\n    crossJoin.currentIndexes = crossJoin.currentIndexes.map((value, index) => (index > incrementableIndex\r\n      ? 0\r\n      : value));\r\n    \r\n    return (crossJoin.currentIndexes.length == variants.length\r\n      ? crossJoin\r\n      : crossJoin.data);\r\n  }, {\r\n    data: [],\r\n    currentIndexes: new Array(variants.length).fill(0)\r\n  }).join(\"\\n\");\r\n\r\nconsole.log(crossJoined);<\/code><\/pre>\n<\/div>\n<\/div>\n
First, an array is created with a length of all variantItems<\/code> array lengths multiplied, then it is zero-filled. Next, we reduce<\/code> it to another array.<\/p>\n
crossJoin<\/code> is the aggregator that holds an object of this structure most of the time:<\/p>\n
{\n  data: [ \u2026 ],\n  currentIndexes: [ 1, 3 ] \/\/ Corresponds to variants[0].variantItems[1]\n                           \/\/            and variants[1].variantItems[3]\n}\n<\/code><\/pre>\n
That is \u201cmost of the time\u201d, until the end, where currentIndexes<\/code> won\u2019t be used anymore and only<\/em> its data<\/code> property is returned.<\/p>\n
So in each reduce<\/code> iteration, we first push a new entry to crossJoin.data<\/code>, using crossJoin.currentIndexes<\/code> and the indexes of that array. Removing .join(\", \")<\/code> will result in an entry of the structure<\/p>\n
[ \"Size\", \"XL\" ]\n<\/code><\/pre>\n
Next, we need to increment the numbers in crossJoin.currentIndexes<\/code>:<\/p>\n
[ 0, 0 ]<\/code> should be incremented to [ 0, 1 ]<\/code>, because there is a second color at the index 1<\/code>. [ 0, 1 ]<\/code> should be incremented to [ 1, 0 ]<\/code>, because there is no third color, but a second size, but then we need the first color again, and so on. The last valid index array is [ 2, 1 ]<\/code> which corresponds to the third size and the second color, which is the last combination.<\/p>\n
incrementableIndex<\/code> is the last possible index that can still be incremented. Once incremented, all subsequent indexes have to be 0<\/code>.<\/p>\n
You see those variants.length - <\/code>something<\/em> - 1<\/code> twice, because you need to find the first index from the end<\/em>, so you have to reverse<\/code> (a copy of<\/em>\u200a\u2014\u200ahence the slice<\/code>) the array, then re-interpret the found index as an index from the start again.<\/p>\n
The return<\/code> returns this crossJoin<\/code> object for the next iteration. The condition crossJoin.currentIndexes.length == variants.length<\/code> applies to the very end, where no index can be incremented anymore. It adds a NaN<\/code> to the currentIndexes<\/code> array, so the lengths don\u2019t match, and instead of filtering out or preventing the NaN<\/code>, I just ignored and discarded it altogether.<\/p>\n
The output will be an array of all combinations. You can .join(\"\\n\")<\/code> it to make a string where each combination is separated by a line-break, or you can use .forEach<\/code> to append each combination to a list, for example.<\/p>\n<\/p><\/div>\n
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solved array cross-over loop in javascript <\/p>\n","protected":false},"excerpt":{"rendered":"
[ad_1] This is a functional ES6 approach. let variants = [{ variantName: “Size”, variantItems: [ “XL”, “MD”, “SM” ] }, { variantName: “Color”, variantItems: [ “Red”, “Blue” ] }]; let crossJoined = new Array(variants.reduce((product, variant) => (product * variant.variantItems.length), 1)) .fill(0) .reduce(crossJoin => { crossJoin.data.push(crossJoin.currentIndexes.map((itemIndex, variantIndex) => `${variants[variantIndex].variantName}: ${variants[variantIndex].variantItems[itemIndex]}`).join(“, “)); let incrementableIndex = variants.length – … Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[320],"tags":[361,898,333],"class_list":["post-4510","post","type-post","status-publish","format-standard","hentry","category-solved","tag-arrays","tag-ecmascript-6","tag-javascript"],"yoast_head":"\n[Solved] array cross-over loop in javascript - JassWeb<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/jassweb.com\/solved\/solved-array-cross-over-loop-in-javascript\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"[Solved] array cross-over loop in javascript - JassWeb\" \/>\n<meta property=\"og:description\" content=\"[ad_1] This is a functional ES6 approach. let variants = [{ variantName: "Size", variantItems: [ "XL", "MD", "SM" ] }, { variantName: "Color", variantItems: [ "Red", "Blue" ] }]; let crossJoined = new Array(variants.reduce((product, variant) => (product * variant.variantItems.length), 1)) .fill(0) .reduce(crossJoin => { crossJoin.data.push(crossJoin.currentIndexes.map((itemIndex, variantIndex) => `${variants[variantIndex].variantName}: ${variants[variantIndex].variantItems[itemIndex]}`).join(", ")); let incrementableIndex = variants.length - ... 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