Build groups in array based on index of item (Python)

Question

I try to create a function which allows me to group an array into groups.

arr = [1,2,3,4,5,6,7,8,9,10,11]

group1 = [1,1,1,1,1,1,2,2,2,2,2] # build 2 groups
group2 = [1,1,1,1,2,2,2,2,3,3,3] # build 3 groups


newarr = []
for index, item in enumerate(arr):
    if index + 1 < len(arr) / 2:
        newarr.append(1)
    else:
        newarr.append(2)
        
print(newarr)
[1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2]

This works somewhat, but I am not able to scale that function in any way, since I would have to add if and else clauses if I want 3, 4, x groups. Is there a function to dynamically create my desired results?

I know in R is a seq function, but I was not able to find something similar in Python.

How do you define groups? should it have defined length, what should the occurrence of numbers be? if i see the occurrence of last number in group1 and group2, its length is one less than others.. — Yash
– Yash, Commented Sep 16, 2020 at 7:25
group1 and group2 are the expected outputs, arr is the input. @Yash Shah: Yes, that is the kind of a problem here. When you can not devide the array into equal length, I would like to assign to the first group — Data Mastery
– Data Mastery, Commented Sep 16, 2020 at 7:34

Yash · Accepted Answer · 2020-09-16 07:59:42Z

1

Are you looking for something like this?

arr = [1,2,3,4,5,6,7,8,9,10,11]
groups=2 #variable
newarr = []
for x in arr[:groups]:
    newarr+=[x]*(int(len(arr)/groups)+1)
newarr=newarr[:len(arr)]
print(newarr)

answered Sep 16, 2020 at 7:59

Yash

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Comments

Christian Sloper · Accepted Answer · 2020-09-16 07:19:14Z

Not the prettiest but faster than the pretty version below (it assumes that groups in group1/group2 are sequencial):

arr = [1,2,3,4,5,6,7,8,9,10,11]

group1 = [1,1,1,1,1,1,2,2,2,2,2] # build 2 groups
group2 = [1,1,1,1,2,2,2,2,3,3,3] # build 3 groups

groups = []
current_group = group1[0]
group = []
for i,v in zip(group1, arr):
    if i==current_group:
        group.append(v)
    else:
        groups.append(group)
        group = []
        group.append(v)
        current_group = i
groups.append(group)

yields (for group 1):

[[1, 2, 3, 4, 5, 6], [7, 8, 9, 10, 11]]

yields (for group 2):

[[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11]]

You can do it pretty, but its looping the table more, on the other hand it doesn't assume the groups to be sequencial:

[[ v for j,v in zip(group2,arr) if i == j] for i in set(group2)]

for same result.

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