Creating a Python list with given indexes for each repeating element

Question

First list : contains the list indexes of corresponding category name

Second list : contains the category names as string

Intervals=[[Indexes_Cat1],[Indexes_Cat2],[Indexes_Cat3], ...]

Category_Names=["cat1","cat2","cat3",...]

Desired Output:

list=["cat1", "cat1","cat2","cat3","cat3"]

where indexes of any element in output list is placed using Intervals list.

Ex1:

Intervals=[[0,4], [2,3] , [1,5]]
Category_Names=["a","b","c"]

Ex: Output1

["a","c","b","b","a","c"]

Edit: More Run Cases

Ex2:

Intervals=[[0,1], [2,3] , [4,5]]
Category_Names=["a","b","c"]

Ex: Output2

["a","a","b","b","c","c"]

Ex3:

Intervals=[[3,4], [1,5] , [0,2]]
Category_Names=["a","b","c"]

Ex: Output3

["c","b","c","a","a","b"]

My solution:

Create any empty array of size n.

Run a for loop for each category.

output=[""]*n
for i in range(len(Category_Names)):
    for index in Intervals[I]:
       output[index]=Categories[i]

Is there a better solution, or a more pythonic way? Thanks

Do you have an example that others can actually run please?

Paddy3118
– Paddy3118

2019-03-10 11:24:52 +00:00
Commented Mar 10, 2019 at 11:24 — Paddy3118
– Paddy3118, Commented Mar 10, 2019 at 11:24
@Paddy3118 I added 2 more cases

Gokberk Yar
– Gokberk Yar

2019-03-10 11:29:43 +00:00
Commented Mar 10, 2019 at 11:29 — Gokberk Yar
– Gokberk Yar, Commented Mar 10, 2019 at 11:29

Paddy3118 · Accepted Answer · 2019-03-10 12:09:44Z

2

def categorise(Intervals=[[0,4], [2,3] , [1,5]],
               Category_Names=["a","b","c"]):
    flattened = sum(Intervals, [])
    answer = [None] * (max(flattened) + 1)
    for indices, name in zip(Intervals, Category_Names):
        for i in indices:
            answer[i] = name
    return answer

assert categorise() == ['a', 'c', 'b', 'b', 'a', 'c']
assert categorise([[3,4], [1,5] , [0,2]], 
                  ["a","b","c"]) == ['c', 'b', 'c', 'a', 'a', 'b']

Note that in this code you will get None values in the answer if the "intervals" don't cover all integers from zero to the max interval number. It is assumed that the input is compatable.

answered Mar 10, 2019 at 12:09

Paddy3118

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Comments

jgmontoya · Accepted Answer · 2019-03-10 11:32:07Z

2

I am not sure if there is a way to avoid the nested loop (I can't think of any right now) so it seems your solution is good.

A way you could do it a bit better is to construct the output array with one of the categories:

output = [Category_Names[0]]*n

and then start the iteration skipping that category:

for i in range(1, len(Category_Names)):

If you know there is a category that appears more than the others then you should use that as the one initializing the array.

I hope this helps!

answered Mar 10, 2019 at 11:32

jgmontoya

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4 Comments

Gokberk Yar Over a year ago

I actually thought your solution but couldn't make sure creating an empty array or with default values runs faster. Thanks ,Appreciated

jgmontoya Over a year ago

Well, in your solution you were already creating the n-size array but with empty strings (output=[""]*n) so using one of the categories saves some time as you don't need to rewrite those ones. I am glad that my answer was helpful.

jgmontoya Over a year ago

BTW does the n come as an input or do you need to calculate it?

Gokberk Yar Over a year ago

@jpmontoya n is known, it is known (a O(1) look up). Thanks

Patrick Artner · Accepted Answer · 2019-03-10 11:36:28Z

1

You can reduce the amount of strings created and use enumerate to avoid range(len(..)) for indexing.

Intervals=[[0,4], [2,3] , [1,5]]
Category_Names=["a","b","c"]

n = max(x for a in Intervals for x in a) + 1

# do not construct strings that get replaced anyhow    
output=[None] * n

for i,name in enumerate(Category_Names):
    for index in Intervals[i]:
       output[index]=name

print(output)

Output:

["a","c","b","b","a","c"]

edited Mar 10, 2019 at 11:36

answered Mar 10, 2019 at 11:25

Patrick Artner

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