# Making a shift permutation in python - Python

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#### [ Making a shift permutation in python ]

I need to make a permutation that looks like a sequence of 1-N with a 0 stuffed in place #k.

This method works, but is there anything simpler using builtin functions?

``````def permshift(n,k):
return [0 if x == k else x+(x&lt;k) for x in range(n)]

&gt;&gt;&gt; permshift(7,0)
[0, 1, 2, 3, 4, 5, 6]
&gt;&gt;&gt; permshift(7,1)
[1, 0, 2, 3, 4, 5, 6]
&gt;&gt;&gt; permshift(7,2)
[1, 2, 0, 3, 4, 5, 6]
&gt;&gt;&gt; permshift(7,3)
[1, 2, 3, 0, 4, 5, 6]
&gt;&gt;&gt; permshift(7,4)
[1, 2, 3, 4, 0, 5, 6]
&gt;&gt;&gt; permshift(7,5)
[1, 2, 3, 4, 5, 0, 6]
&gt;&gt;&gt; permshift(7,6)
[1, 2, 3, 4, 5, 6, 0]
``````

Just create a new list, and insert the `0` wherever you need to.

``````def permshift(n, k):
lst = range(1, n)
lst.insert(k, 0)
return lst
``````

More lines, but I think it's simpler:

``````def permshift(n, k):
x = range(1, n)
x.insert(k, 0)
return x
``````

Here's an interesting one that's shorter than the most popularly repeated answer:

``````def permshift(n, k):
x = range(1, n)
return x[:k] + [0] + x[k:]
``````

Hehe - it's funny when so many people come up with the same solution

``````def permshift(n,k):
r = range(1,n)
r.insert(k,0)
return r
``````

For completeness here's a generator expression if you really care about efficiency

``````def permshift(n,k):
for i in xrange(1,n):
yield i
if i == k:
yield 0

print list(permshift(7,2))
``````

Good answers already but here's a one line version for kicks and giggles:

``````permshift = lambda n,k: range(1, k+1) + [0] + range(k+1, n)
``````

``````from itertools import chain