Cycling values of a list

up vote
6
down vote

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I'm new to coding and am trying to write a simple code that will take a list, say [1,2,3] and cycle the elements n number of times. So if n=1, I should get A=[3,1,2]. If n=2, I should get A=[2,3,1].The code I have written is:

n=1
j=0
A = [1,2,3]
B = [None]*len(A)

while j<=n:
 for i in range(0,len(A)):
 B[i] = A[-1+i]
 j=j+1
print(B)

The problem is that no matter what the value of n is I get the same answer which is only cycled once. I think the problem is that the loop is cycling through the same B every time, so I need to store the new B as something else and then repeat the loop with new B. But I can't figure out how to do that. Any tips would be appreciated

edited 29 mins ago

martineau

62.7k887167

asked 2 hours ago

DeathbyGreen

333

New contributor

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up vote
6
down vote

favorite

n=1
j=0
A = [1,2,3]
B = [None]*len(A)

while j<=n:
 for i in range(0,len(A)):
 B[i] = A[-1+i]
 j=j+1
print(B)

edited 29 mins ago

martineau

62.7k887167

asked 2 hours ago

DeathbyGreen

333

New contributor

add a commentÂ |Â

up vote
6
down vote

favorite

n=1
j=0
A = [1,2,3]
B = [None]*len(A)

while j<=n:
 for i in range(0,len(A)):
 B[i] = A[-1+i]
 j=j+1
print(B)

edited 29 mins ago

martineau

62.7k887167

asked 2 hours ago

DeathbyGreen

333

New contributor

n=1
j=0
A = [1,2,3]
B = [None]*len(A)

while j<=n:
 for i in range(0,len(A)):
 B[i] = A[-1+i]
 j=j+1
print(B)

python python-3.x list for-loop

edited 29 mins ago

martineau

62.7k887167

asked 2 hours ago

DeathbyGreen

333

New contributor

edited 29 mins ago

martineau

62.7k887167

asked 2 hours ago

DeathbyGreen

333

New contributor

edited 29 mins ago

martineau

62.7k887167

edited 29 mins ago

martineau

62.7k887167

edited 29 mins ago

martineau

62.7k887167

asked 2 hours ago

DeathbyGreen

333

New contributor

asked 2 hours ago

DeathbyGreen

333

asked 2 hours ago

DeathbyGreen

333

New contributor

DeathbyGreen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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6 Answers
6

active

oldest

votes

up vote
4
down vote

accepted

I think you're overcomplicating it. Consider changing it to something like the following:

n = 1
A = [1,2,3]
B = A.copy()

for _ in range(n):
 # Cycle through by concatenating the last element and all other elements together 
 B = [B[-1]]+B[0:-1]

print(B)

In case of n=1, you get [3, 1, 2], and n=2 gives you [2, 3, 1]

Note that what you are trying to do is implemented in numpy.roll (I suppose you're asking about the process, not the result, but just in case)

import numpy as np

>>> np.roll(A,1)
array([3, 1, 2])
>>> np.roll(A,2)
array([2, 3, 1])

answered 2 hours ago

sacul

21k31434

add a commentÂ |Â

up vote
2
down vote

A simpler function for this is:

def roll(L, n):
 n %= len(L)
 return L[-n:] + L[:-n]

A = [1,2,3]
roll(A, 1) # [3, 1, 2]
roll(A, 2) # [2, 3, 1]
roll(A, 3) # [1, 2, 3]
roll(A, 4) # [3, 1, 2]

Taking the modulus (n %= len(L)) avoids the need to keep cycling through. We then just concatenate an appropriate-sized slice off the end of the list to the beginning of it.

edited 52 mins ago

answered 58 mins ago

Stuart

5,07811122

add a commentÂ |Â

up vote
0
down vote

@sacul's answer works, but you were close! You missed updating A for the next iteration of the while loop after you created your new B.

n=1
j=0
A = [1,2,3]
B = [None]*len(A)

while j<=n:
 for i in range(0,len(A)):
 B[i] = A[-1+i]
 A = B[:] # update A so that next time B works on the new A 
 print('A is now ', A) # to debug
 j=j+1

print(B)

This results in the following print statements:

A is now [3, 1, 2] 
A is now [2, 3, 1] 
[2, 3, 1] # this is what finally B is

answered 1 hour ago

slider

2,210925

But this is the wrong outcome for n = 1... set while j<n instead.
â€“Â Stuart
29 mins ago

@Stuart What do you mean "wrong"? j = 1 can mean 2 iterations; it's just a matter of definition. This is OP's original code only modified slightly to explain where they went wrong.
â€“Â slider
17 mins ago

Wrong in the sense that it does not provide the output specified by the OP for the given A and n.
â€“Â Stuart
14 mins ago

add a commentÂ |Â

up vote
0
down vote

An itertools based solution

from itertools import cycle, islice

def roll(x, n):
 start = len(x) - n
 return islice(cycle(x), start, start + len(x))

If we cycle through the values in x (we'd actually only do this at most twice). Then slice it so that it begins at the element we want and includes the correct number of elements.

It's probably a little more esoteric than some of the other solutions, but worth throwing into the bag, given there are so many alternatives.

itertools.islice returns a generator. If you wanted to print the result you'd need to convert it to list.

answered 19 mins ago

Paul Rooney

12.1k62642

add a commentÂ |Â

up vote
0
down vote

See @sacul's answer for the problem with your code. But list is not the most appropriate structure for such a requirement as each shift has O(n) complexity.

deque ("double-ended qeueue") from the collections module provides this feature via its rotate method. This method works in-place and has O(k) complexity, where k is argument representing the number of rotations. Here's an example:

from collections import deque

d = deque([1,2,3])
d.rotate(2)

print(d)

deque([2, 3, 1])

answered 16 mins ago

jpp

69.2k174085

add a commentÂ |Â

up vote
-1
down vote

Not going to in depth since someone already answered your question.. But an important property to note here is also that

A = [1, 2, 3] #Assuming that the length of the matrix you want to rotate is of length 3

If you want to cycle/rotate by 3, then you want to actually cycle/rotate by `len(A)%3` = 0
If you want to cycle/rotate by 6, then you want to actually cycle/rotate by `len(A)%6` = 0
If you want to cycle/rotate by 4, then you want to actually cycle/rotate by `len(A)%3` = 1
If you want to cycle/rotate by a number lesser than the length itself.. then just rotate it by that number!

so if you want to cycle by 2, then well.. rotate it by two..

answered 1 hour ago

Abhishek

947219

Okay.. Why am I getting this downvote?
â€“Â Abhishek
4 mins ago

add a commentÂ |Â

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6 Answers
6

active

oldest

votes

6 Answers
6

active

oldest

votes

up vote
4
down vote

accepted

I think you're overcomplicating it. Consider changing it to something like the following:

n = 1
A = [1,2,3]
B = A.copy()

for _ in range(n):
 # Cycle through by concatenating the last element and all other elements together 
 B = [B[-1]]+B[0:-1]

print(B)

In case of n=1, you get [3, 1, 2], and n=2 gives you [2, 3, 1]

Note that what you are trying to do is implemented in numpy.roll (I suppose you're asking about the process, not the result, but just in case)

import numpy as np

>>> np.roll(A,1)
array([3, 1, 2])
>>> np.roll(A,2)
array([2, 3, 1])

answered 2 hours ago

sacul

21k31434

add a commentÂ |Â

up vote
4
down vote

accepted

I think you're overcomplicating it. Consider changing it to something like the following:

n = 1
A = [1,2,3]
B = A.copy()

for _ in range(n):
 # Cycle through by concatenating the last element and all other elements together 
 B = [B[-1]]+B[0:-1]

print(B)

In case of n=1, you get [3, 1, 2], and n=2 gives you [2, 3, 1]

Note that what you are trying to do is implemented in numpy.roll (I suppose you're asking about the process, not the result, but just in case)

import numpy as np

>>> np.roll(A,1)
array([3, 1, 2])
>>> np.roll(A,2)
array([2, 3, 1])

answered 2 hours ago

sacul

21k31434

add a commentÂ |Â

up vote
4
down vote

accepted

I think you're overcomplicating it. Consider changing it to something like the following:

n = 1
A = [1,2,3]
B = A.copy()

for _ in range(n):
 # Cycle through by concatenating the last element and all other elements together 
 B = [B[-1]]+B[0:-1]

print(B)

In case of n=1, you get [3, 1, 2], and n=2 gives you [2, 3, 1]

Note that what you are trying to do is implemented in numpy.roll (I suppose you're asking about the process, not the result, but just in case)

import numpy as np

>>> np.roll(A,1)
array([3, 1, 2])
>>> np.roll(A,2)
array([2, 3, 1])

answered 2 hours ago

sacul

21k31434

I think you're overcomplicating it. Consider changing it to something like the following:

n = 1
A = [1,2,3]
B = A.copy()

for _ in range(n):
 # Cycle through by concatenating the last element and all other elements together 
 B = [B[-1]]+B[0:-1]

print(B)

In case of n=1, you get [3, 1, 2], and n=2 gives you [2, 3, 1]

Note that what you are trying to do is implemented in numpy.roll (I suppose you're asking about the process, not the result, but just in case)

import numpy as np

>>> np.roll(A,1)
array([3, 1, 2])
>>> np.roll(A,2)
array([2, 3, 1])

answered 2 hours ago

sacul

21k31434

answered 2 hours ago

sacul

21k31434

answered 2 hours ago

sacul

21k31434

answered 2 hours ago

sacul

21k31434

add a commentÂ |Â

up vote
2
down vote

A simpler function for this is:

def roll(L, n):
 n %= len(L)
 return L[-n:] + L[:-n]

A = [1,2,3]
roll(A, 1) # [3, 1, 2]
roll(A, 2) # [2, 3, 1]
roll(A, 3) # [1, 2, 3]
roll(A, 4) # [3, 1, 2]

Taking the modulus (n %= len(L)) avoids the need to keep cycling through. We then just concatenate an appropriate-sized slice off the end of the list to the beginning of it.

edited 52 mins ago

answered 58 mins ago

Stuart

5,07811122

add a commentÂ |Â

up vote
2
down vote

A simpler function for this is:

def roll(L, n):
 n %= len(L)
 return L[-n:] + L[:-n]

A = [1,2,3]
roll(A, 1) # [3, 1, 2]
roll(A, 2) # [2, 3, 1]
roll(A, 3) # [1, 2, 3]
roll(A, 4) # [3, 1, 2]

Taking the modulus (n %= len(L)) avoids the need to keep cycling through. We then just concatenate an appropriate-sized slice off the end of the list to the beginning of it.

edited 52 mins ago

answered 58 mins ago

Stuart

5,07811122

add a commentÂ |Â

up vote
2
down vote

A simpler function for this is:

def roll(L, n):
 n %= len(L)
 return L[-n:] + L[:-n]

A = [1,2,3]
roll(A, 1) # [3, 1, 2]
roll(A, 2) # [2, 3, 1]
roll(A, 3) # [1, 2, 3]
roll(A, 4) # [3, 1, 2]

Taking the modulus (n %= len(L)) avoids the need to keep cycling through. We then just concatenate an appropriate-sized slice off the end of the list to the beginning of it.

edited 52 mins ago

answered 58 mins ago

Stuart

5,07811122

A simpler function for this is:

def roll(L, n):
 n %= len(L)
 return L[-n:] + L[:-n]

A = [1,2,3]
roll(A, 1) # [3, 1, 2]
roll(A, 2) # [2, 3, 1]
roll(A, 3) # [1, 2, 3]
roll(A, 4) # [3, 1, 2]

Taking the modulus (n %= len(L)) avoids the need to keep cycling through. We then just concatenate an appropriate-sized slice off the end of the list to the beginning of it.

edited 52 mins ago

answered 58 mins ago

Stuart

5,07811122

edited 52 mins ago

answered 58 mins ago

Stuart

5,07811122

answered 58 mins ago

Stuart

5,07811122

answered 58 mins ago

Stuart

5,07811122

add a commentÂ |Â

up vote
0
down vote

@sacul's answer works, but you were close! You missed updating A for the next iteration of the while loop after you created your new B.

n=1
j=0
A = [1,2,3]
B = [None]*len(A)

while j<=n:
 for i in range(0,len(A)):
 B[i] = A[-1+i]
 A = B[:] # update A so that next time B works on the new A 
 print('A is now ', A) # to debug
 j=j+1

print(B)

This results in the following print statements:

A is now [3, 1, 2] 
A is now [2, 3, 1] 
[2, 3, 1] # this is what finally B is

answered 1 hour ago

slider

2,210925

But this is the wrong outcome for n = 1... set while j<n instead.
â€“Â Stuart
29 mins ago

@Stuart What do you mean "wrong"? j = 1 can mean 2 iterations; it's just a matter of definition. This is OP's original code only modified slightly to explain where they went wrong.
â€“Â slider
17 mins ago

Wrong in the sense that it does not provide the output specified by the OP for the given A and n.
â€“Â Stuart
14 mins ago

add a commentÂ |Â

up vote
0
down vote

@sacul's answer works, but you were close! You missed updating A for the next iteration of the while loop after you created your new B.

n=1
j=0
A = [1,2,3]
B = [None]*len(A)

while j<=n:
 for i in range(0,len(A)):
 B[i] = A[-1+i]
 A = B[:] # update A so that next time B works on the new A 
 print('A is now ', A) # to debug
 j=j+1

print(B)

This results in the following print statements:

A is now [3, 1, 2] 
A is now [2, 3, 1] 
[2, 3, 1] # this is what finally B is

answered 1 hour ago

slider

2,210925

But this is the wrong outcome for n = 1... set while j<n instead.
â€“Â Stuart
29 mins ago

@Stuart What do you mean "wrong"? j = 1 can mean 2 iterations; it's just a matter of definition. This is OP's original code only modified slightly to explain where they went wrong.
â€“Â slider
17 mins ago

Wrong in the sense that it does not provide the output specified by the OP for the given A and n.
â€“Â Stuart
14 mins ago

add a commentÂ |Â

up vote
0
down vote

@sacul's answer works, but you were close! You missed updating A for the next iteration of the while loop after you created your new B.

n=1
j=0
A = [1,2,3]
B = [None]*len(A)

while j<=n:
 for i in range(0,len(A)):
 B[i] = A[-1+i]
 A = B[:] # update A so that next time B works on the new A 
 print('A is now ', A) # to debug
 j=j+1

print(B)

This results in the following print statements:

A is now [3, 1, 2] 
A is now [2, 3, 1] 
[2, 3, 1] # this is what finally B is

answered 1 hour ago

slider

2,210925

@sacul's answer works, but you were close! You missed updating A for the next iteration of the while loop after you created your new B.

n=1
j=0
A = [1,2,3]
B = [None]*len(A)

while j<=n:
 for i in range(0,len(A)):
 B[i] = A[-1+i]
 A = B[:] # update A so that next time B works on the new A 
 print('A is now ', A) # to debug
 j=j+1

print(B)

This results in the following print statements:

A is now [3, 1, 2] 
A is now [2, 3, 1] 
[2, 3, 1] # this is what finally B is

answered 1 hour ago

slider

2,210925

answered 1 hour ago

slider

2,210925

answered 1 hour ago

slider

2,210925

answered 1 hour ago

slider

2,210925

But this is the wrong outcome for n = 1... set while j<n instead.
â€“Â Stuart
29 mins ago

@Stuart What do you mean "wrong"? j = 1 can mean 2 iterations; it's just a matter of definition. This is OP's original code only modified slightly to explain where they went wrong.
â€“Â slider
17 mins ago

Wrong in the sense that it does not provide the output specified by the OP for the given A and n.
â€“Â Stuart
14 mins ago

add a commentÂ |Â

But this is the wrong outcome for n = 1... set while j<n instead.
â€“Â Stuart
29 mins ago

@Stuart What do you mean "wrong"? j = 1 can mean 2 iterations; it's just a matter of definition. This is OP's original code only modified slightly to explain where they went wrong.
â€“Â slider
17 mins ago

Wrong in the sense that it does not provide the output specified by the OP for the given A and n.
â€“Â Stuart
14 mins ago

But this is the wrong outcome for n = 1... set while j<n instead.
â€“Â Stuart
29 mins ago

@Stuart What do you mean "wrong"? j = 1 can mean 2 iterations; it's just a matter of definition. This is OP's original code only modified slightly to explain where they went wrong.
â€“Â slider
17 mins ago

Wrong in the sense that it does not provide the output specified by the OP for the given A and n.
â€“Â Stuart
14 mins ago

add a commentÂ |Â

up vote
0
down vote

An itertools based solution

from itertools import cycle, islice

def roll(x, n):
 start = len(x) - n
 return islice(cycle(x), start, start + len(x))

If we cycle through the values in x (we'd actually only do this at most twice). Then slice it so that it begins at the element we want and includes the correct number of elements.

It's probably a little more esoteric than some of the other solutions, but worth throwing into the bag, given there are so many alternatives.

itertools.islice returns a generator. If you wanted to print the result you'd need to convert it to list.

answered 19 mins ago

Paul Rooney

12.1k62642

add a commentÂ |Â

up vote
0
down vote

An itertools based solution

from itertools import cycle, islice

def roll(x, n):
 start = len(x) - n
 return islice(cycle(x), start, start + len(x))

If we cycle through the values in x (we'd actually only do this at most twice). Then slice it so that it begins at the element we want and includes the correct number of elements.

It's probably a little more esoteric than some of the other solutions, but worth throwing into the bag, given there are so many alternatives.

itertools.islice returns a generator. If you wanted to print the result you'd need to convert it to list.

answered 19 mins ago

Paul Rooney

12.1k62642

add a commentÂ |Â

up vote
0
down vote

An itertools based solution

from itertools import cycle, islice

def roll(x, n):
 start = len(x) - n
 return islice(cycle(x), start, start + len(x))

If we cycle through the values in x (we'd actually only do this at most twice). Then slice it so that it begins at the element we want and includes the correct number of elements.

It's probably a little more esoteric than some of the other solutions, but worth throwing into the bag, given there are so many alternatives.

itertools.islice returns a generator. If you wanted to print the result you'd need to convert it to list.

answered 19 mins ago

Paul Rooney

12.1k62642

An itertools based solution

from itertools import cycle, islice

def roll(x, n):
 start = len(x) - n
 return islice(cycle(x), start, start + len(x))

If we cycle through the values in x (we'd actually only do this at most twice). Then slice it so that it begins at the element we want and includes the correct number of elements.

It's probably a little more esoteric than some of the other solutions, but worth throwing into the bag, given there are so many alternatives.

itertools.islice returns a generator. If you wanted to print the result you'd need to convert it to list.

answered 19 mins ago

Paul Rooney

12.1k62642

answered 19 mins ago

Paul Rooney

12.1k62642

answered 19 mins ago

Paul Rooney

12.1k62642

answered 19 mins ago

Paul Rooney

12.1k62642

add a commentÂ |Â

up vote
0
down vote

See @sacul's answer for the problem with your code. But list is not the most appropriate structure for such a requirement as each shift has O(n) complexity.

from collections import deque

d = deque([1,2,3])
d.rotate(2)

print(d)

deque([2, 3, 1])

answered 16 mins ago

jpp

69.2k174085

add a commentÂ |Â

up vote
0
down vote

See @sacul's answer for the problem with your code. But list is not the most appropriate structure for such a requirement as each shift has O(n) complexity.

from collections import deque

d = deque([1,2,3])
d.rotate(2)

print(d)

deque([2, 3, 1])

answered 16 mins ago

jpp

69.2k174085

add a commentÂ |Â

up vote
0
down vote

See @sacul's answer for the problem with your code. But list is not the most appropriate structure for such a requirement as each shift has O(n) complexity.

from collections import deque

d = deque([1,2,3])
d.rotate(2)

print(d)

deque([2, 3, 1])

answered 16 mins ago

jpp

69.2k174085

See @sacul's answer for the problem with your code. But list is not the most appropriate structure for such a requirement as each shift has O(n) complexity.

from collections import deque

d = deque([1,2,3])
d.rotate(2)

print(d)

deque([2, 3, 1])

answered 16 mins ago

jpp

69.2k174085

answered 16 mins ago

jpp

69.2k174085

answered 16 mins ago

jpp

69.2k174085

answered 16 mins ago

jpp

69.2k174085

add a commentÂ |Â

up vote
-1
down vote

Not going to in depth since someone already answered your question.. But an important property to note here is also that

A = [1, 2, 3] #Assuming that the length of the matrix you want to rotate is of length 3

If you want to cycle/rotate by 3, then you want to actually cycle/rotate by `len(A)%3` = 0
If you want to cycle/rotate by 6, then you want to actually cycle/rotate by `len(A)%6` = 0
If you want to cycle/rotate by 4, then you want to actually cycle/rotate by `len(A)%3` = 1
If you want to cycle/rotate by a number lesser than the length itself.. then just rotate it by that number!

so if you want to cycle by 2, then well.. rotate it by two..

answered 1 hour ago

Abhishek

947219

Okay.. Why am I getting this downvote?
â€“Â Abhishek
4 mins ago

add a commentÂ |Â

up vote
-1
down vote

Not going to in depth since someone already answered your question.. But an important property to note here is also that

A = [1, 2, 3] #Assuming that the length of the matrix you want to rotate is of length 3

If you want to cycle/rotate by 3, then you want to actually cycle/rotate by `len(A)%3` = 0
If you want to cycle/rotate by 6, then you want to actually cycle/rotate by `len(A)%6` = 0
If you want to cycle/rotate by 4, then you want to actually cycle/rotate by `len(A)%3` = 1
If you want to cycle/rotate by a number lesser than the length itself.. then just rotate it by that number!

so if you want to cycle by 2, then well.. rotate it by two..

answered 1 hour ago

Abhishek

947219

Okay.. Why am I getting this downvote?
â€“Â Abhishek
4 mins ago

add a commentÂ |Â

up vote
-1
down vote

Not going to in depth since someone already answered your question.. But an important property to note here is also that

A = [1, 2, 3] #Assuming that the length of the matrix you want to rotate is of length 3

If you want to cycle/rotate by 3, then you want to actually cycle/rotate by `len(A)%3` = 0
If you want to cycle/rotate by 6, then you want to actually cycle/rotate by `len(A)%6` = 0
If you want to cycle/rotate by 4, then you want to actually cycle/rotate by `len(A)%3` = 1
If you want to cycle/rotate by a number lesser than the length itself.. then just rotate it by that number!

so if you want to cycle by 2, then well.. rotate it by two..

answered 1 hour ago

Abhishek

947219

Not going to in depth since someone already answered your question.. But an important property to note here is also that

A = [1, 2, 3] #Assuming that the length of the matrix you want to rotate is of length 3

If you want to cycle/rotate by 3, then you want to actually cycle/rotate by `len(A)%3` = 0
If you want to cycle/rotate by 6, then you want to actually cycle/rotate by `len(A)%6` = 0
If you want to cycle/rotate by 4, then you want to actually cycle/rotate by `len(A)%3` = 1
If you want to cycle/rotate by a number lesser than the length itself.. then just rotate it by that number!

so if you want to cycle by 2, then well.. rotate it by two..

answered 1 hour ago

Abhishek

947219

answered 1 hour ago

Abhishek

947219

answered 1 hour ago

Abhishek

947219

answered 1 hour ago

Abhishek

947219

Okay.. Why am I getting this downvote?
â€“Â Abhishek
4 mins ago

add a commentÂ |Â

Okay.. Why am I getting this downvote?
â€“Â Abhishek
4 mins ago

Okay.. Why am I getting this downvote?
â€“Â Abhishek
4 mins ago

add a commentÂ |Â

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