Rearrange array element based on the number sequence and represent by array id

Question

Suppose, I have multiple arrays in one array with the number from 0 to n in multiple order. For example,

x = [[0,2,3,5],[1,4]]

Here we have two arrays in x. There could be more than two.

I want to get rearrange all the array elements based on their number sequence. However, they will represent their array ID. The result should be like this

y = [0,1,0,0,1,0]

That means 0,2,3,5 is in array id 0. So, they will show the id in their respective sequence. Same for 1 and 4. Can anyone help me to solve this? [N.B. There could be more than two arrays. So, it will be highly appreciated if the code work for different array numbers]

Sahadat Hossain · Accepted Answer · 2021-01-21 18:56:37Z

2

You can do this by using a dictionary

x = [[0,2,3,5],[1,4]]
lst = {}
for i in range(len(x)):
    for j in range(len(x[i])):
        lst[x[i][j]] = i
print(lst)

You can also do this by using list, list.insert(idx, value) means value is inserted to the list at the idxth index. Here, we are traversing through all the values of x and the value x[i][j] is in the i th number array.

x = [[0,2,3,5],[1,4]]
lst = []
for i in range(len(x)):
    for j in range(len(x[i])):
        lst.insert(x[i][j], i)
print(lst)

Output: [0, 1, 0, 0, 1, 0]

edited Jan 21, 2021 at 18:56

answered Jan 21, 2021 at 18:38

Sahadat Hossain

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3 Comments

Sahadat Hossain Over a year ago

@Lzz0 i hope you got it

Sash Sinha Over a year ago

Using insert on a list is an O(n) operation since all the values that come after the insert index need to be moved down in memory every time, so it would be bad for big lists. Therefore I think you should put your dictionary solution as the first one since it will be more efficient for the OP.

Sahadat Hossain Over a year ago

@ShashSinha thanks for your comment, updated.

Marc · Accepted Answer · 2021-01-21 20:00:24Z

1

You might also consider using np.argsort for rearranging your array values and create the index-array with list comprehension:

x = [[0,2,3,5],[1,4]]
order = np.concatenate(x).argsort()
np.concatenate([ [i]*len(e) for i,e in enumerate(x) ])[order]

array([0, 1, 0, 0, 1, 0])

edited Jan 21, 2021 at 20:00

answered Jan 21, 2021 at 19:54

Marc

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