Python : Generating cyclic permutations code (An unexpected code error to be clarified)

Question

I didn't manage to correct a code I thought it would work for sure. Any advice to make the code functional is accepted. Expected outputs of the following code is a list containing a cyclic permuation of the list

l = [1,2,3,4] (i.e : [[4, 1, 2, 3],[3, 4, 1, 2],[2, 3, 4, 1],[1, 2, 3, 4]])

Although what I get is : [[2, 3, 4, 1]]

The code :

def cycGen(l):
    L=[]
    while not(l in L) :
        L.append(l)
        for i in range(len(l)):
            if l[i] == len(l) :
                l[i]=1
            else :
                l[i] = 1 + l[i] 
    return L
print(cycGen([1,2,3,4]))

Another variation of the solution is to consider the following code wich seems unfortunatly not working either :

def cycGen(l):
    L=[]
    for k in range(len(l)):
        L.append(l)
        for i in range(len(l)):
            if l[i] == len(l) :
                l[i]=1
            else :
                l[i] = 1 + l[i]   
    return L

Help me with your generous knowlege sharing please.

Possible duplicate https://stackoverflow.com/questions/5313900/generating-cyclic-permutations-reduced-latin-squares-in-python/48376124#48376124 — pylang, Jan 22 '18 at 06:57

score 4 · Answer 1 · edited May 23 '17 at 12:01

4

You can use collections.deque:

from collections import deque
a = [1, 2, 3, 4]

d = deque(a)
for _ in range(len(a)):
    d.rotate()
    print(list(d))

Which gives you the output:

[4, 1, 2, 3]
[3, 4, 1, 2]
[2, 3, 4, 1]
[1, 2, 3, 4]

As mentioned in Efficient way to shift a list in python

edited May 23 '17 at 12:01

Community

1
1

answered Jul 17 '14 at 01:25

ceremcem

3,900
4
28
66

1

You can just use `for _ in a:` – Suor Jul 17 '14 at 05:11

score 1 · Answer 2 · answered Jul 17 '14 at 03:49

1

An easy way is just:

In [12]: x = [1,2,3,4]

In [13]: [x[i:]+x[:i] for i in range(len(x))]
Out[13]: [[1, 2, 3, 4], [2, 3, 4, 1], [3, 4, 1, 2], [4, 1, 2, 3]]

answered Jul 17 '14 at 03:49

ssm

5,277
1
24
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score 0 · Accepted Answer · answered Jul 17 '14 at 01:18

0

In your first code sample, the line L.append(l) appends a "reference" (loosely speaking) to the list l to the end of L, rather than appending a copy as you seem to be expecting. Thus, when l is later modified, the reference to it contained in L is modified as well, and so when l in L is tested, l will equal the reference to itself in L, and so the loop will end. The same basic problem causes your second code sample to return multiples of the same list rather than several different lists.

To store a copy of l at the current point in time in L instead, use L.append(l[:]).

answered Jul 17 '14 at 01:18

jwodder

54,758
12
108
124

Can i get a detailed material from your choice to avoid such confusions in future ? – melasiam Jul 17 '14 at 01:44
Is this syntaxe problem existed for all the python versions or is it just concerning the python 3.0 version ? – melasiam Jul 17 '14 at 01:53

score 0 · Answer 4 · answered Jul 17 '14 at 01:35

0

Here is an easy way:

>>> def cycGen(l):
        size = len(l)
        return [[l[(i+j)%size] for i in range(size)] for j in range(size)]

>>> l = [1,2,3,4]
>>> print cycGen(l)
[[1, 2, 3, 4], [2, 3, 4, 1], [3, 4, 1, 2], [4, 1, 2, 3]]

answered Jul 17 '14 at 01:35

mshsayem

17,557
11
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score 0 · Answer 5 · answered Jul 17 '14 at 03:11

You could do this using a generator too:

a = [1, 2, 3, 4]

def next_pos(max):
    i = 0
    while True:
        for n in xrange(max):
            yield n + i
        i += 1

pos = next_pos(len(a))

b = []
for i in xrange(len(a)):
    n = []
    for j in xrange(len(a)):
        m = pos.next()
        if m >= len(a):
            m -= len(a)
        n.append(a[m])
    b.append(n)

print b

output:

[[1, 2, 3, 4], [2, 3, 4, 1], [3, 4, 1, 2], [4, 1, 2, 3]]

Python : Generating cyclic permutations code (An unexpected code error to be clarified)

5 Answers5