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I have a list as

{1,2,3,4,5,6,7,8,9}

I want to extract a new list as

{{1,2},{2,3},{3,4},{4,5},{5,6},{6,7},{7,8},{8,9},{9,1}}

I've tried to search Partition, but it didn't give me what I want. Or I've missed out something. Can you please suggest me a way to do that?

a = {1, 2, 3, 4, 5, 6, 7, 8, 9};

The command

Partition[a, 2]

just partitions the input into as many nonoverlapping pairs as possible:

{{1, 2}, {3, 4}, {5, 6}, {7, 8}}

The three-argument version creates pairs with an offset of `1:

Partition[a, 2, 1]

{{1, 2}, {2, 3}, {3, 4}, {4, 5}, {5, 6}, {6, 7}, {7, 8}, {8, 9}}

That is already very close to what you want!

In order to pair also the last element with the first, use the four argument version of Partition to make it cycle

Partition[a, 2, 1, {1, 1}]

{{1, 2}, {2, 3}, {3, 4}, {4, 5}, {5, 6}, {6, 7}, {7, 8}, {8, 9}, {9,
1}}

whose shorter version is

Partition[a, 2, 1, 1]

In the usage notes of Partition (?Partition), this is somewhat hidden. The notes state (highlighting by me):

Partition[list,n,d,{Subscript[k, L],Subscript[k, R]}] specifies that
the first element of list should appear at position Subscript[k, L] in
the first sublist, and the last element of list should appear at or
after position Subscript[k, R] in the last sublist. If additional
elements are needed, Partition fills them in by treating list as
cyclic.

For the specific case you can also use

lst = Range[9];

Transpose[{lst, RotateLeft[lst]}]

{{1, 2}, {2, 3}, {3, 4}, {4, 5}, {5, 6}, {6, 7}, {7, 8}, {8, 9}, {9, 1}}

which is the same result Partition[lst, 2, 1, 1] gives.

I propose you another option that works similarly to Partition[list,n,d,{Subscript[k, L],Subscript[k, R]}] but making use of other list-manipulation functions that might be useful in other contexts.

a = {1, 2, 3, 4, 5, 6, 7, 8, 9};

Partition[Riffle[a, RotateLeft[a, 1]], 2]

Explanation of overhangs

This answer is in response to the comment

I did, but didn't understand it. Could you please elaborate a bit more
about each parameter?

As J.M. and Henrik have noted, this problem can be solved with the fourth argument of Partition, the so-called overhangs. I will explain in detail how overhangs work.

l = Range[3]

{1, 2, 3}

Partition[l, 2, 1, {-1, -1}]

{{3, 1}, {1, 2}, {2, 3}}

To understand why the result above is what it is, imagine that you have a window of size two that you slide across the list. Just as if you were computing a moving average over the current element and the one before it. The overhangs specify where this window starts.

{3, 1} is the first sublist of the result. The first overhang in {-1, -1} refers to this sublist. It says that the first element in the original list should be the last element (-1) of the first sublist.

{2, 3} is the last sublist of the result. The second overhang in {-1, -1} refers to this sublist. It says that the first element in the original list should be the last element (-1) of the last sublist.

The first overhang specifies something about the first sublist in the result, and the second overhang specifies something about the last sublist in the result.

Let's now specify that the first sublist should not wrap around to the end, while the last sublist should. This is the opposite of what we did before. Since the first sublist should not wrap around, we want the first element of the first sublist to be the first element of the original list, i.e. 1. Since the last element of the last sublist should be the first element of the original list, we want it to be 1.

Partition[l, 2, 1, {1, 1}]

{{1, 2}, {2, 3}, {3, 1}}

Using this type of reasoning, you should be able to understand there as well:

Partition[l, 2, 1, {-1, 1}]

{{3, 1}, {1, 2}, {2, 3}, {3, 1}}

Partition[l, 2, 1, {1, -1}]

{{1, 2}, {2, 3}}

This reasoning also works for larger windows. Here's an example with a window size of 3, where we specify that the third element of the first sublist should be the first element of the original list.

Partition[Range[5], 3, 1, {3, -1}]

{{4, 5, 1}, {5, 1, 2}, {1, 2, 3}, {2, 3, 4}, {3, 4, 5}}

Returning to the idea of the sliding window, we can also think of the overhang as specifying that, in this last example, the sliding window should start two steps to the left of the first element. But what is to the left of the first element? As we have seen, the list is padded cyclically by default. To the left of the first element will then be the last elements of the list.

Other paddings can be set with the fifth argument of Partition:

Partition[Range[5], 3, 1, {3, -1}, 0]

{{0, 0, 1}, {0, 1, 2}, {1, 2, 3}, {2, 3, 4}, {3, 4, 5}}

With this setting, the elements to the left of the first element are all zero, and the same goes for the elements to the right of the last element:

Partition[l, 2, 1, {-1, 1}, 0]

{{0, 1}, {1, 2}, {2, 3}, {3, 0}}