Reordering an Ordered List

Reordering an ordered list

Use the 'orrible old trick made famous(?) by old BASIC coders - set your orders to be 100, 200, 300, 400 etc. and then you can pick an order 'inbetween' when you need to. This could get messy - and if you're anticipating a lot of reordering then I'd recommend that you have a scheduled task to 'reorder' the order values every now and then for the entire table.

Reorder an element in an ordered list with a single update

A value m = (upper+lower)/2 is always in the middle of two values.

For further reading on numbering I refer to a lecture:
http://www.cs.uni-paderborn.de/fachgebiete/ag-boettcher/lehre/ss2012/dbis-2/download.html

Reordering data and making a re-ordered list in r

This is as easy as using unique. I coerce to character so as not to get confused between levels and actual values.

 unique(as.character(dfordered$spp))

To get here from the original data, use table

table(df$spp)

##  A B C 
##  2 4 3 

You can sort this

sppCount <- table(df$spp)

names(sort(sppCount, decreasing = TRUE))

# [1] "B" "C" "A"

python reordering of list elements with conversion to set

Thats because sets are based on the Hash Table data structure, where records are stored in buckets using hash keys, and the order of items (when printed or converted to list) depends on the hashes, but internally the order doest matter. so it doesnt really bother to change the order, it just adds the item and creates a hash index for it. when you print the set it is probably printed according to the lexographical order of hashes, or something like that.

As you can see from the following, the list when created from a set, takes the same order of hashes of these items.

>>> s=set([5,4,3,7,6,2,1,0])
>>> s
{0, 1, 2, 3, 4, 5, 6, 7}
>>> list(s)
[0, 1, 2, 3, 4, 5, 6, 7]

How can I reorder a list?

You can do it like this

mylist = ['a', 'b', 'c', 'd', 'e']
myorder = [3, 2, 0, 1, 4]
mylist = [mylist[i] for i in myorder]
print(mylist)         # prints: ['d', 'c', 'a', 'b', 'e']

javascript - reorder a list without creating new list according to another list

• Use Array spread syntax or reduce your array of indexes into a DocumentFragment
• Finally, .append() the iterable to the same parent UL element.

const orderList = [3, 1, 2, 0];

const ul = document.querySelector('#list');
const li = ul.querySelectorAll('li'); 
ul.append(...orderList.map(i => li[i]));

<ul id="list">
  <li>0</li>
  <li>1</li>
  <li>2</li>
  <li>3</li>
</ul>

Reorder R list by unconventionally ordered vector?

It might help to take a look at ?"[". Specifically, the difference between [[ and [ confused me for a while. Here, the key thing to note is that you can use [ to subset to multiple elements, while [[ can only be used for one element.

So as @lmo points out, what you want to do here is MyList[pattern]. In case you didn't realize, this can work because names(MyList) matches up with the elements in pattern -- you can index/ subset not only by numerical index (e.g., 1:3), but also by name (e.g., c("A","B","C")).

TLDR:

NewList <- MyList[pattern]

---

Related Topics