Repeat index by giving list

Question

Not sure if this kind of question is exist or not. If so, I will delete the post later.

I tried to build a list with repeating index which will follow the list here:

Mylist = [3, 3, 6, 6, 6, 8, 8]

My output should be:

expect = [0, 0, 2, 2, 2, 5, 5]

The index will change when the value in the giving list changes.

I tried the code like this:

z_pts = [0] * (len(z_rows)) 
i_cur = -1
for k in range(len(z_rows)):  
    if z_rows[k] != i_cur:                
       i_cur = z_rows[k]         
       z_pts[i_cur] = k

However, I only could get result like this:

res = [0, 0, 0, 0, 2, 5, 0]

Asked By: Chuen

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Source

Answer 1

Not the cleanest way to do it but with only built-ins I got the answer with this:

Mylist = [3, 3, 6, 6, 6, 8, 8]
expect = [0]

for i, n in enumerate(Mylist[1:], 1):
    expect.append(expect[i-1] if n == Mylist[i-1] else i)

print(expect)

Output:

[0, 0, 2, 2, 2, 5, 5]

Answered By: B Remmelzwaal

Answer 2

Try:

lst = [3, 3, 6, 6, 6, 8, 8]

tmp = 0, -1
out = [tmp[0] if tmp[1] == t[1] else (tmp := t)[0] for t in enumerate(lst)]

print(out)

Prints:

[0, 0, 2, 2, 2, 5, 5]

Answered By: Andrej Kesely

Answer 3

This can be done very simply using a list comprehension.

my_list = [3, 3, 6, 6, 6, 8, 8]

new_list = [my_list.index(i) for i in my_list]

print(new_list) # [0, 0, 2, 2, 2, 5, 5]

Answered By: Mous

Repeat index by giving list

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