Create two linear lists from a conceptual circular list in python

Question

Consider the following list:

>>> circle = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h']  
>>> list(enumerate(circle))  
[(0, 'a'), (1, 'b'), (2, 'c'), (3, 'd'), (4, 'e'), (5, 'f'), (6, 'g'), (7, 'h')]

If circle is conceptualized as a circular list i.e. circle[0] is connected to circle[7], given a start index and an end index where start != end, I want to construct two lists that represent linear traversal orders in clockwise and counter-clockwise directions.

Depending on the values of start and end, here's what I have come up with:

Case 1: start < end

>>> circle = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h']  
>>> start, end = 1, 6  
>>> clockwise = circle[start:end+1]  
>>> clockwise  
['b', 'c', 'd', 'e', 'f', 'g']  
>>> counter_clockwise = circle[start::-1] + circle[:end-1:-1]  
>>> counter_clockwise  
['b', 'a', 'h', 'g']

Case 2: start > end

>>> circle = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h']  
>>> start, end = 6, 1  
>>> clockwise = circle[start:] + circle[:end+1]  
>>> clockwise  
['g', 'h', 'a', 'b']  
>>> counter_clockwise = circle[start:end-1:-1]  
>>> counter_clockwise  
['g', 'f', 'e', 'd', 'c', 'b']

Is there a pythonic/more-efficient/easier way of constructing these two lists?

I don't have an exact answer for you, but Numpy (numpy.org) has a bunch of useful features in their array implementation. — Xorgon
– Xorgon, Commented May 18, 2017 at 4:51

Taku · Accepted Answer · 2017-05-18 05:45:36Z

3

You can use itertools.cycle:

import itertools

circle = ['a', 'b', 'c', 'd', 'e', 'f', 
  'g', 'h']

def clockwise(start, end):
    endless = itertools.cycle(circle)
    if start > end:
        end = start + (len(circle)-(start
          -end))
    return [next(endless) for i in range
      (end+1)][start:]

def counter_clockwise(start, end):
    endless = itertools.cycle(circle)
    if end > start:
        start = end + (len(circle)-(end
          -start))    
    return [next(endless) for i in range
      (start+1)][end:][::-1]

# start < end:
forward=clockwise(1, 6)
b1=counter_clockwise(1, 6)
#start > end:
f1=clockwise(6, 1)
backward=counter_clockwise(6, 1)

print(forward)
print(b1)
print(f1)
print(backward)

Outputs:

['b', 'c', 'd', 'e', 'f', 'g']
['b', 'a', 'h', 'g']
['g', 'h', 'a', 'b']
['g', 'f', 'e', 'd', 'c', 'b']

edited May 18, 2017 at 5:45

answered May 18, 2017 at 4:52

Taku

34.1k12 gold badges79 silver badges88 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

kuro Over a year ago

But what about the case when start > end?

Joran Beasley · Accepted Answer · 2017-05-18 04:58:35Z

1

import itertools

def slice_it(circle,start,end,step=1):
    if end < start and step > 0:
        end = len(circle)+end
    if step < 0:
        return list(itertools.islice(itertools.cycle(circle),end,start,-1*step))[::-1]
    return list(itertools.islice(itertools.cycle(circle),start,end,step))

circle = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h']

print slice_it(circle,6,1)
print slice_it(circle,6,1,-1)

maybe?

answered May 18, 2017 at 4:58

Joran Beasley

114k13 gold badges167 silver badges187 bronze badges

Comments

Keerthana Prabhakaran · Accepted Answer · 2017-05-18 05:11:30Z

0

You can use lambda expression for creating clockwise and anticlockwise methods and list slicing!

>>> clockwise = lambda circle, start, end: circle[start:] + circle[:end+1] if start > end else circle[start:end+1]
>>> counter_clockwise = lambda circle, start, end : clockwise(circle, end, start)[::-1]

This is somewhat similar to what you've tried, but in a more pythonic way and generalized way!

>>> clockwise(circle,1,6)
['b', 'c', 'd', 'e', 'f', 'g']
>>> counter_clockwise(circle,1,6)
['b', 'a', 'h', 'g']
>>> clockwise(circle,6,1)
['g', 'h', 'a', 'b']
>>> counter_clockwise(circle,6,1)
['g', 'f', 'e', 'd', 'c', 'b']

edited May 18, 2017 at 5:11

answered May 18, 2017 at 5:06

Keerthana Prabhakaran

3,8071 gold badge16 silver badges25 bronze badges

Comments

Chiheb Nexus · Accepted Answer · 2017-05-18 19:00:23Z

You can use deque from collections module to handle some parts of the counter_clockwise conditon like this example (you can modify it or improve it if you want):

from collections import deque

circle = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h']


clockwise = lambda x,start,end: x[start:end+1] if start < end else x[start:] + x[:end+1]

def counter_clockwise(itertable, start, end):
    if end > start:
        a = deque(itertable[:start+1])
        b = deque(itertable[end:])
        a.rotate()
        b.rotate()
        # Or:
        # return list(a.__iadd__(b))
        return list(a) + list(b)

    if end < start:
        return itertable[start:end-1:-1]


print("start > end:")
start, end = 6,1
print(clockwise(circle, start, end))
print(counter_clockwise(circle, start, end))

print("start < end:")
start, end = 1,6
print(clockwise(circle, start, end))
print(counter_clockwise(circle, start, end))

Output:

start > end:
['g', 'h', 'a', 'b']
['g', 'f', 'e', 'd', 'c', 'b']
start < end:
['b', 'c', 'd', 'e', 'f', 'g']
['b', 'a', 'h', 'g']

Collectives™ on Stack Overflow

Create two linear lists from a conceptual circular list in python

4 Answers 4

1 Comment

Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

1 Comment

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related