Split an array based on another array-CodePudding

*I don't know what's the proper title should be. But I will explain my problem clearly.

In this situation, I have two arrays

oldIDs = [1, 2]
newIDs = [2, 3, 4]

question:
how to split "newIDs" into [[2], [3,4]] ?

explanation:
2 is an old id (It is existed in oldIDs)
3 & 4 are new ids (which are not existed in oldIDs)

Other examples:

oldIDs = [1]
newIDs = [1, 2]
// the result should be [[1], [2]]

oldIDs = [1, 3, 5]
newIDs = [3, 4, 7, 8]
// the result should be [[3], [4, 7, 8]]

This is what I'm doing with Go.

func splitIDs(oldIDs []int64, newIDs []int64) (arr1 []int64, arr2 []int64) {
    for _, newID := range newIDs {
        if contains(oldIDs, newID) {
            arr1 = append(arr1, newID)
        }
    }
    for _, oldID := range oldIDs {
        for _, newID := range newIDs {
            if contains(arr1, newID) {
                continue
            }
            if contains(arr2, newID) {
                continue
            }
            if oldID != newID {
                arr2 = append(arr2, newID)
            }
        }
    }
    return
}

func contains(a []int64, x int64) bool {
    for _, n := range a {
        if x == n {
            return true
        }
    }
    return false
}

This can also be divided into two arrays, not necessarily two dimensional array.

I really appreciate someone who can solve my problem or improve my solution.

CodePudding user response：

You can get the arrays intersection and the arrays difference and combine them into one array as follows:

const oldIDs = [1, 3, 5]
const newIDs = [3, 4, 7, 8]

const splitArray = (arr1, arr2) => {
  const result = []
  //Getting the arrays intersection
  result[0] = arr2.filter((item) => {
    return arr1.includes(item)
  })
  //Getting the arrays difference
  result[1] = arr2.filter((item) => {
    return !arr1.includes(item)
  })
  return result
}

console.log(splitArray(oldIDs, newIDs))

CodePudding user response：

Below is a program in python.
This first read through old id, creates a dictionary out of it. and read through the new ids looking for common number, to split.

oldIDs = [1, 3, 5]
newIDs = [3, 4, 7, 8]

def split(oldIDs,newIDs):
    o_dict={}
    for i in oldIDs:
        o_dict[i]=True
    for i,num in enumerate(newIDs):
        if num in o_dict:
            return [newIDs[:i 1],newIDs[i 1:]]

print(split(oldIDs,newIDs))

CodePudding user response：

It turns out there is a library (not surprisingly) for this same purpose. It is split_into available in more_itertools

All you need to do is:

res = list(more_itertools.split_into(newIDs, oldIDs   [None]*(len(newIDs) - len(oldIDs))))

res = list(filter(lambda x:len(x)>0, res))

Otherwise if you want to write your own loop ( which I prefer ):

i = 0
j = 0
l = []
while i < len(newIDs):
    l.extend([newIDs[i:i oldIDs[j]]])
    i  = oldIDs[j]
    if j < len(oldIDs)-1:
        j  = 1

What is this doing is, keep checking if you have seen all elements in newIDs but also keeping a check on the index in your oldIDs so that you keep using the oldIDs if you fall short.