remove consequtive repeating values python

Question

I have a list for which I need to remove continuously repeating values.

For example: input:

Xin = [1,1,1,1,1,2,2,2,3,2,2,2,2,2,3,3,1]

output:

Yout = [1,2,3,2,3,1]

(Zero is not a legal value in this case).

My current code is this:

Yout = np.zeros(len(Xin))          #initialize a list

for i in range(len(Xin)):          
    if Xin[i] != Xin[i-1]:     
        You[j] = Xin[i]
        j = j + 1

Yout = np.ma.masked_equal(Yout,0) #mask all the zeros
Yout = Yout.compressed()          #Remove all the masks

Someone had answered this for 'R', but I need this for Python.

your current code seems to miss the initialization of `j` and some imports - hence it produces errors. — Patrick Artner, Jun 10 '20 at 05:57

score 0 · Accepted Answer · answered Jun 10 '20 at 05:55

0

import numpy as np

a = np.array([1,1,1,1,1,2,2,2,3,2,2,2,2,2,3,3,1])
print(a)
# [1 1 1 1 1 2 2 2 3 2 2 2 2 2 3 3 1]

b = a[a != np.r_[a[1:], None]]
print(b)
# [1 2 3 2 3 1]

answered Jun 10 '20 at 05:55

0xF4D3C0D3

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This works great. – Malcolm Rest Jun 10 '20 at 06:04
Can you please explain the syntax ? – Malcolm Rest Jun 10 '20 at 06:05
it uses `advanced indexing in numpy`. if you compare two numpy arrays, they're element-wise compared. `np.r_[a[1:], None]` precedes one element over just `a`. so `a != np.r_[a[1:], None]` will have True at which consecutive elements are different. – 0xF4D3C0D3 Jun 10 '20 at 06:17
Thanks. and how do I mark the index in which the element has changed ? – Malcolm Rest Jun 10 '20 at 06:32
@MalcolmRest you can use `np.where(a != np.r_[a[1:], None])[0]` – 0xF4D3C0D3 Jun 10 '20 at 07:01

remove consequtive repeating values python

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