Kristaps Fabians Geikins
9ada92ccde
* fix(fe):automation result isolation didnt select objects * minor bugfix
58 lines
1.9 KiB
TypeScript
58 lines
1.9 KiB
TypeScript
import { intersection } from 'lodash-es'
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export { isNonNullable } from '@speckle/shared'
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/**
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* Checks for inclusion of one array (target) into another (source)
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* @param target the array you want to check that is included in the other one
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* @param source the array you want to check INTO for inclusion of the previous one
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*/
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export const containsAll = <T>(target: T[], source: T[]) =>
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target.every((v) => source.includes(v))
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/**
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* Whether or not arrays are equal with the order being ignored
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*/
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export const arraysEqual = (arr1: unknown[], arr2: unknown[]) => {
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if (arr1.length !== arr2.length) return false
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return intersection(arr1, arr2).length === arr1.length
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}
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const isSet = <V = unknown>(vals: V[] | Set<V>): vals is Set<V> => vals instanceof Set
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const toSet = <V = unknown>(vals: V[] | Set<V>): Set<V> =>
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isSet(vals) ? vals : new Set(vals)
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const length = (vals: unknown[] | Set<unknown>) =>
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isSet(vals) ? vals.size : vals.length
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/**
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* Various performance-improved array utilities:
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*/
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/**
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* Whenever you have to compare two arrays/sets, this function will make sure the biggest one is a Set for quick look ups
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* and so you can iterate over the smallest one
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* @param vals1
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* @param vals2
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*/
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const toOptimizedComparisonArrays = <V = unknown>(
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vals1: V[] | Set<V>,
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vals2: V[] | Set<V>
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) => {
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const biggest: Set<V> = length(vals1) > length(vals2) ? toSet(vals1) : toSet(vals2)
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const smallest: V[] | Set<V> = length(vals1) > length(vals2) ? vals2 : vals1
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return { biggest, smallest: isSet(smallest) ? [...smallest] : smallest }
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}
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/**
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* A fast way to check if two arrays/sets have at least one element in common
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*/
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export const hasIntersection = <V = unknown>(
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vals1: V[] | Set<V>,
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vals2: V[] | Set<V>
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) => {
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const { biggest, smallest } = toOptimizedComparisonArrays(vals1, vals2)
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if (!length(biggest) || !length(smallest)) return false
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return smallest.some((v) => biggest.has(v))
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}
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