Multisort() 3 arrays after usort() one of them - php

I'm building an online store for a clothes selling company. When displaying the products on page, the user has to be able to filter the clothes by SIZE. I also show them how many clothes belong to the specific sizes.
For this porpuse I use three arrays: $size_names, $size_ids, $size_counts
The first array can hold numeric and string values together, so it could look like this:
array(1, 2, 3, 'M', 'L', 'S')
What I basically want is to sort the values within this array in the following logic:
sort the numerical elements first, ascending
sort alphabetical elements in the following order: XXS, XS, S, M, L, XL, XXL
I read about usort(), but the problem is that I need to reorder all three arrays (size_names, size_ids, size_counts) by the reordering rule of only size_names.
So I have to sort the first array, than based on the sort, I have to sort the other two.
EDIT
One possible scenario with my three arrays could be like:
$size_names = array(3, 1, M, S)
$size_ids = array(1, 2, 3, 4)
$size_counts = array(10, 8, 3, 2)
So based on the array values I can say to the visitor that there are TWO clothes that have size S, which size has the id 4. (The ID value is not shown to the visitor, only helps me building the sql for filtering.)
I currently use array_multisort:
array_multisort($size_names, $size_ids, $size_counts);
which produces the following result:
$size_names = array(1, 3, M, S)
$size_ids = array(2, 1, 3, 4)
$size_counts = array(8, 10, 3, 2)
This is half way there, because the numeric values are sorted in the desired order, but the alphabetical values are sorted alphabetically, which is not what I want.
The desired order should be:
$size_names = array(1, 3, S, M)
$size_ids = array(2, 1, 4, 3)
$size_counts = array(8, 10, 2, 3)

Replace the alphabetical elements with numbers, do the sorting and then bring the alphabetical elements back. The following is an example:
$size_names = array(3, 1, "M", "S");
$size_ids = array(1, 2, 3, 4);
$size_counts = array(10, 8, 3, 2);
$arr1 = array("XXS", "XS", "S", "M", "L", "XL", "XXL");
$arr2 = range(1001, 1007);
$size_names = str_replace($arr1, $arr2, $size_names);
array_multisort($size_names, $size_ids, $size_counts);
$size_names = str_replace($arr2, $arr1, $size_names);
DEMO

Related

how to get same values if an array have values same with values of anther array values in PHP?

Having two arrays of authRoom and partiRoom with one same value inside them. Want to find that same value if it was matched
Found array_search function that work only with single variable
$authRoom = [8, 7, 1, 22, 13, 18, 10];
$partiRoom= [3, 6, 5, 9, 8];
I want the output to be 8 which is the same value of these two arrays
You can use array_intersect which will give you an array of the same values in both $authRoom and $partiRoom like so:
$authRoom = [8, 7, 1, 22, 13, 18, 10];
$partiRoom = [3, 6, 5, 9, 8];
$res = array_intersect($authRoom, $partiRoom);
print_r($res); // [8]
If you want to get the value 8 outside of the array, you can simply access the first value using index 0:
$res = array_intersect($authRoom, $partiRoom)[0];
echo $res; // 8

Python subprocess.call and a list of arguments [duplicate]

This question's answers are a community effort. Edit existing answers to improve this post. It is not currently accepting new answers or interactions.
How do I concatenate two lists in Python?
Example:
listone = [1, 2, 3]
listtwo = [4, 5, 6]
Expected outcome:
>>> joinedlist
[1, 2, 3, 4, 5, 6]
Use the + operator to combine the lists:
listone = [1, 2, 3]
listtwo = [4, 5, 6]
joinedlist = listone + listtwo
Output:
>>> joinedlist
[1, 2, 3, 4, 5, 6]
Python >= 3.5 alternative: [*l1, *l2]
Another alternative has been introduced via the acceptance of PEP 448 which deserves mentioning.
The PEP, titled Additional Unpacking Generalizations, generally reduced some syntactic restrictions when using the starred * expression in Python; with it, joining two lists (applies to any iterable) can now also be done with:
>>> l1 = [1, 2, 3]
>>> l2 = [4, 5, 6]
>>> joined_list = [*l1, *l2] # unpack both iterables in a list literal
>>> print(joined_list)
[1, 2, 3, 4, 5, 6]
This functionality was defined for Python 3.5, but it hasn't been backported to previous versions in the 3.x family. In unsupported versions a SyntaxError is going to be raised.
As with the other approaches, this too creates as shallow copy of the elements in the corresponding lists.
The upside to this approach is that you really don't need lists in order to perform it; anything that is iterable will do. As stated in the PEP:
This is also useful as a more readable way of summing iterables into a
list, such as my_list + list(my_tuple) + list(my_range) which is now
equivalent to just [*my_list, *my_tuple, *my_range].
So while addition with + would raise a TypeError due to type mismatch:
l = [1, 2, 3]
r = range(4, 7)
res = l + r
The following won't:
res = [*l, *r]
because it will first unpack the contents of the iterables and then simply create a list from the contents.
It's also possible to create a generator that simply iterates over the items in both lists using itertools.chain(). This allows you to chain lists (or any iterable) together for processing without copying the items to a new list:
import itertools
for item in itertools.chain(listone, listtwo):
# Do something with each list item
You could also use the list.extend() method in order to add a list to the end of another one:
listone = [1,2,3]
listtwo = [4,5,6]
listone.extend(listtwo)
If you want to keep the original list intact, you can create a new list object, and extend both lists to it:
mergedlist = []
mergedlist.extend(listone)
mergedlist.extend(listtwo)
How do I concatenate two lists in Python?
As of 3.9, these are the most popular stdlib methods for concatenating two (or more) lists in Python.
Version Restrictions
In-Place?
Generalize to N lists?
a+b
-
No
sum([a, b, c], [])1
list(chain(a,b))2
>=2.3
No
list(chain(a, b, c))
[*a, *b]3
>=3.5
No
[*a, *b, *c]
a += b
-
Yes
No
a.extend(b)
-
Yes
No
Footnotes
This is a slick solution because of its succinctness. But sum performs concatenation in a pairwise fashion, which means this is a
quadratic operation as memory has to be allocated for each step. DO
NOT USE if your lists are large.
See chain
and
chain.from_iterable
from the docs. You will need to from itertools import chain first.
Concatenation is linear in memory, so this is the best in terms of
performance and version compatibility. chain.from_iterable was introduced in 2.6.
This method uses Additional Unpacking Generalizations (PEP 448), but cannot
generalize to N lists unless you manually unpack each one yourself.
a += b and a.extend(b) are more or less equivalent for all practical purposes. += when called on a list will internally call
list.__iadd__, which extends the first list by the second.
Performance
2-List Concatenation1
There's not much difference between these methods but that makes sense given they all have the same order of complexity (linear). There's no particular reason to prefer one over the other except as a matter of style.
N-List Concatenation
Plots have been generated using the perfplot module. Code, for your reference.
1. The iadd (+=) and extend methods operate in-place, so a copy has to be generated each time before testing. To keep things fair, all methods have a pre-copy step for the left-hand list which can be ignored.
Comments on Other Solutions
DO NOT USE THE DUNDER METHOD list.__add__ directly in any way, shape or form. In fact, stay clear of dunder methods, and use the operators and operator functions like they were designed for. Python has careful semantics baked into these which are more complicated than just calling the dunder directly. Here is an example. So, to summarise, a.__add__(b) => BAD; a + b => GOOD.
Some answers here offer reduce(operator.add, [a, b]) for pairwise concatenation -- this is the same as sum([a, b], []) only more wordy.
Any method that uses set will drop duplicates and lose ordering. Use with caution.
for i in b: a.append(i) is more wordy, and slower than a.extend(b), which is single function call and more idiomatic. append is slower because of the semantics with which memory is allocated and grown for lists. See here for a similar discussion.
heapq.merge will work, but its use case is for merging sorted lists in linear time. Using it in any other situation is an anti-pattern.
yielding list elements from a function is an acceptable method, but chain does this faster and better (it has a code path in C, so it is fast).
operator.add(a, b) is an acceptable functional equivalent to a + b. It's use cases are mainly for dynamic method dispatch. Otherwise, prefer a + b which is shorter and more readable, in my opinion. YMMV.
You can use sets to obtain merged list of unique values
mergedlist = list(set(listone + listtwo))
This is quite simple, and I think it was even shown in the tutorial:
>>> listone = [1,2,3]
>>> listtwo = [4,5,6]
>>>
>>> listone + listtwo
[1, 2, 3, 4, 5, 6]
This question directly asks about joining two lists. However it's pretty high in search even when you are looking for a way of joining many lists (including the case when you joining zero lists).
I think the best option is to use list comprehensions:
>>> a = [[1,2,3], [4,5,6], [7,8,9]]
>>> [x for xs in a for x in xs]
[1, 2, 3, 4, 5, 6, 7, 8, 9]
You can create generators as well:
>>> map(str, (x for xs in a for x in xs))
['1', '2', '3', '4', '5', '6', '7', '8', '9']
Old Answer
Consider this more generic approach:
a = [[1,2,3], [4,5,6], [7,8,9]]
reduce(lambda c, x: c + x, a, [])
Will output:
[1, 2, 3, 4, 5, 6, 7, 8, 9]
Note, this also works correctly when a is [] or [[1,2,3]].
However, this can be done more efficiently with itertools:
a = [[1,2,3], [4,5,6], [7,8,9]]
list(itertools.chain(*a))
If you don't need a list, but just an iterable, omit list().
Update
Alternative suggested by Patrick Collins in the comments could also work for you:
sum(a, [])
You could simply use the + or += operator as follows:
a = [1, 2, 3]
b = [4, 5, 6]
c = a + b
Or:
c = []
a = [1, 2, 3]
b = [4, 5, 6]
c += (a + b)
Also, if you want the values in the merged list to be unique you can do:
c = list(set(a + b))
It's worth noting that the itertools.chain function accepts variable number of arguments:
>>> l1 = ['a']; l2 = ['b', 'c']; l3 = ['d', 'e', 'f']
>>> [i for i in itertools.chain(l1, l2)]
['a', 'b', 'c']
>>> [i for i in itertools.chain(l1, l2, l3)]
['a', 'b', 'c', 'd', 'e', 'f']
If an iterable (tuple, list, generator, etc.) is the input, the from_iterable class method may be used:
>>> il = [['a'], ['b', 'c'], ['d', 'e', 'f']]
>>> [i for i in itertools.chain.from_iterable(il)]
['a', 'b', 'c', 'd', 'e', 'f']
For cases with a low number of lists you can simply add the lists together or use in-place unpacking (available in Python-3.5+):
In [1]: listone = [1, 2, 3]
...: listtwo = [4, 5, 6]
In [2]: listone + listtwo
Out[2]: [1, 2, 3, 4, 5, 6]
In [3]: [*listone, *listtwo]
Out[3]: [1, 2, 3, 4, 5, 6]
As a more general way for cases with more number of lists you can use chain.from_iterable()1 function from itertools module. Also, based on this answer this function is the best; or at least a very good way for flatting a nested list as well.
>>> l=[[1, 2, 3], [4, 5, 6], [7, 8, 9]]
>>> import itertools
>>> list(itertools.chain.from_iterable(l))
[1, 2, 3, 4, 5, 6, 7, 8, 9]
1. Note that `chain.from_iterable()` is available in Python 2.6 and later. In other versions, use `chain(*l)`.
With Python 3.3+ you can use yield from:
listone = [1,2,3]
listtwo = [4,5,6]
def merge(l1, l2):
yield from l1
yield from l2
>>> list(merge(listone, listtwo))
[1, 2, 3, 4, 5, 6]
Or, if you want to support an arbitrary number of iterators:
def merge(*iters):
for it in iters:
yield from it
>>> list(merge(listone, listtwo, 'abcd', [20, 21, 22]))
[1, 2, 3, 4, 5, 6, 'a', 'b', 'c', 'd', 20, 21, 22]
If you want to merge the two lists in sorted form, you can use the merge function from the heapq library.
from heapq import merge
a = [1, 2, 4]
b = [2, 4, 6, 7]
print list(merge(a, b))
If you can't use the plus operator (+), you can use the operator import:
import operator
listone = [1,2,3]
listtwo = [4,5,6]
result = operator.add(listone, listtwo)
print(result)
>>> [1, 2, 3, 4, 5, 6]
Alternatively, you could also use the __add__ dunder function:
listone = [1,2,3]
listtwo = [4,5,6]
result = list.__add__(listone, listtwo)
print(result)
>>> [1, 2, 3, 4, 5, 6]
If you need to merge two ordered lists with complicated sorting rules, you might have to roll it yourself like in the following code (using a simple sorting rule for readability :-) ).
list1 = [1,2,5]
list2 = [2,3,4]
newlist = []
while list1 and list2:
if list1[0] == list2[0]:
newlist.append(list1.pop(0))
list2.pop(0)
elif list1[0] < list2[0]:
newlist.append(list1.pop(0))
else:
newlist.append(list2.pop(0))
if list1:
newlist.extend(list1)
if list2:
newlist.extend(list2)
assert(newlist == [1, 2, 3, 4, 5])
If you are using NumPy, you can concatenate two arrays of compatible dimensions with this command:
numpy.concatenate([a,b])
Use a simple list comprehension:
joined_list = [item for list_ in [list_one, list_two] for item in list_]
It has all the advantages of the newest approach of using Additional Unpacking Generalizations - i.e. you can concatenate an arbitrary number of different iterables (for example, lists, tuples, ranges, and generators) that way - and it's not limited to Python 3.5 or later.
Another way:
>>> listone = [1, 2, 3]
>>> listtwo = [4, 5, 6]
>>> joinedlist = [*listone, *listtwo]
>>> joinedlist
[1, 2, 3, 4, 5, 6]
>>>
list(set(listone) | set(listtwo))
The above code does not preserve order and removes duplicates from each list (but not from the concatenated list).
As already pointed out by many, itertools.chain() is the way to go if one needs to apply exactly the same treatment to both lists. In my case, I had a label and a flag which were different from one list to the other, so I needed something slightly more complex. As it turns out, behind the scenes itertools.chain() simply does the following:
for it in iterables:
for element in it:
yield element
(see https://docs.python.org/2/library/itertools.html), so I took inspiration from here and wrote something along these lines:
for iterable, header, flag in ( (newList, 'New', ''), (modList, 'Modified', '-f')):
print header + ':'
for path in iterable:
[...]
command = 'cp -r' if os.path.isdir(srcPath) else 'cp'
print >> SCRIPT , command, flag, srcPath, mergedDirPath
[...]
The main points to understand here are that lists are just a special case of iterable, which are objects like any other; and that for ... in loops in python can work with tuple variables, so it is simple to loop on multiple variables at the same time.
You could use the append() method defined on list objects:
mergedlist =[]
for elem in listone:
mergedlist.append(elem)
for elem in listtwo:
mergedlist.append(elem)
a = [1, 2, 3]
b = [4, 5, 6]
c = a + b
print(c)
Output
>>> [1, 2, 3, 4, 5, 6]
In the above code, the "+" operator is used to concatenate the two lists into a single list.
Another solution
a = [1, 2, 3]
b = [4, 5, 6]
c = [] # Empty list in which we are going to append the values of list (a) and (b)
for i in a:
c.append(i)
for j in b:
c.append(j)
print(c)
Output
>>> [1, 2, 3, 4, 5, 6]
All the possible ways to join lists that I could find
import itertools
A = [1,3,5,7,9] + [2,4,6,8,10]
B = [1,3,5,7,9]
B.append([2,4,6,8,10])
C = [1,3,5,7,9]
C.extend([2,4,6,8,10])
D = list(zip([1,3,5,7,9],[2,4,6,8,10]))
E = [1,3,5,7,9]+[2,4,6,8,10]
F = list(set([1,3,5,7,9] + [2,4,6,8,10]))
G = []
for a in itertools.chain([1,3,5,7,9], [2,4,6,8,10]):
G.append(a)
print("A: " + str(A))
print("B: " + str(B))
print("C: " + str(C))
print("D: " + str(D))
print("E: " + str(E))
print("F: " + str(F))
print("G: " + str(G))
Output
A: [1, 3, 5, 7, 9, 2, 4, 6, 8, 10]
B: [1, 3, 5, 7, 9, [2, 4, 6, 8, 10]]
C: [1, 3, 5, 7, 9, 2, 4, 6, 8, 10]
D: [(1, 2), (3, 4), (5, 6), (7, 8), (9, 10)]
E: [1, 3, 5, 7, 9, 2, 4, 6, 8, 10]
F: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
G: [1, 3, 5, 7, 9, 2, 4, 6, 8, 10]
I recommend three methods to concatenate the list, but the first method is most recommended,
# Easiest and least complexity method <= recommended
listone = [1, 2, 3]
listtwo = [4, 5, 6]
newlist = listone + listtwo
print(newlist)
# Second-easiest method
newlist = listone.copy()
newlist.extend(listtwo)
print(newlist)
In the second method, I assign newlist to a copy of the listone, because I don't want to change listone.
# Third method
newlist = listone.copy()
for j in listtwo:
newlist.append(j)
print(newlist)
This is not a good way to concatenate lists because we are using a for loop to concatenate the lists. So time complexity is much higher than with the other two methods.
The most common method used to concatenate lists are the plus operator and the built-in method append, for example:
list = [1,2]
list = list + [3]
# list = [1,2,3]
list.append(3)
# list = [1,2,3]
list.append([3,4])
# list = [1,2,[3,4]]
For most of the cases, this will work, but the append function will not extend a list if one was added. Because that is not expected, you can use another method called extend. It should work with structures:
list = [1,2]
list.extend([3,4])
# list = [1,2,3,4]
A really concise way to combine a list of lists is
list_of_lists = [[1,2,3], [4,5,6], [7,8,9]]
reduce(list.__add__, list_of_lists)
which gives us
[1, 2, 3, 4, 5, 6, 7, 8, 9]
So there are two easy ways.
Using +: It creates a new list from provided lists
Example:
In [1]: a = [1, 2, 3]
In [2]: b = [4, 5, 6]
In [3]: a + b
Out[3]: [1, 2, 3, 4, 5, 6]
In [4]: %timeit a + b
10000000 loops, best of 3: 126 ns per loop
Using extend: It appends new list to existing list. That means it does not create a separate list.
Example:
In [1]: a = [1, 2, 3]
In [2]: b = [4, 5, 6]
In [3]: %timeit a.extend(b)
10000000 loops, best of 3: 91.1 ns per loop
Thus we see that out of two of most popular methods, extend is efficient.
You could also just use sum.
>>> a = [1, 2, 3]
>>> b = [4, 5, 6]
>>> sum([a, b], [])
[1, 2, 3, 4, 5, 6]
>>>
This works for any length and any element type of list:
>>> a = ['a', 'b', 'c', 'd']
>>> b = [1, 2, 3, 4]
>>> c = [1, 2]
>>> sum([a, b, c], [])
['a', 'b', 'c', 'd', 1, 2, 3, 4, 1, 2]
>>>
The reason I add [], is because the start argument is set to 0 by default, so it loops through the list and adds to start, but 0 + [1, 2, 3] would give an error, so if we set the start to []. It would add to [], and [] + [1, 2, 3] would work as expected.
I assume you want one of the two methods:
Keep duplicate elements
It is very easy. Just concatenate like a string:
def concat_list(l1,l2):
l3 = l1+l2
return l3
Next, if you want to eliminate duplicate elements
def concat_list(l1,l2):
l3 = []
for i in [l1,l2]:
for j in i:
if j not in l3:
# Check if element exists in final list, if no then add element to list
l3.append(j)
return l3
The solutions provided are for a single list. In case there are lists within a list and the merging of corresponding lists is required, the "+" operation through a for loop does the work.
a = [[1,2,3], [4,5,6]]
b = [[0,1,2], [7,8,9]]
for i in range(len(a)):
cc.append(a[i] + b[i])
Output: [[1, 2, 3, 0, 1, 2], [4, 5, 6, 7, 8, 9]]

two-dimensional array inside two-dimensional arrays in php

Is it possible to achieve something like this? A two-dimensional array containing two-dimensional jagged array?
$jobOrder = array(array(1, "Web Developer", 100,
array(array(1, "PHP", 1),
array(2, "HTML", 1), array(3, "JAVA", 1)),
array(array(1, "pleasing personality", 1),
array(2, "english skills", 1)), 0),
array(2, "Senior Programmer", 50,
array(array(3, "Phython", 1),
array(5, "RUBY", 1),
array(10, "c#", 1)),
array(array(5, "good social skills", 1),
array(11, "management skills", 1))));
I want to store Job Order details into an array that should contain an orderID, job title, number of openings, skills(may have multiple skills so stored in an array; 2-d because I also wanted to store skillID, skill name and flag: if its been removed or not), qualifications(may have multiple qualifications same with skills), requirements and benefits (also same with skills). I would like to know how to access it.
You can access element like this:
$jobOrder = array(array(1, "Web Developer", 100,
array(array(1, "PHP", 1),
array(2, "HTML", 1), array(3, "JAVA", 1)),
array(array(1, "pleasing personality", 1),
array(2, "english skills", 1)), 0),
array(2, "Senior Programmer", 50,
array(array(3, "Phython", 1),
array(5, "RUBY", 1),
array(10, "c#", 1)),
array(array(5, "good social skills", 1),
array(11, "management skills", 1))));
print_r($jobOrder[0][3]);
I will print the array and also you can access the elements in the array also by adding further indexes.
Yes It is possible as you can store number of array rows in the array: means you can add number of rows as well number of elements inside the array even it may multidimensional array as you defined.

How many times can x unique values be pulled from an array

I have an array of values (non unique) and I need to work out how many times I can pull out x (e.g 3) unique items from that array.
e.g.
[5, 5, 4, 1, 2, 3, 4, 2, 1, 3, 5, 5]
What is the largest quantity of unique items of length 3 (e.g. [5, 4, 1]) I can retrieve?
For context, this is for an offer system in a shopping cart. The array of values is the product ids, and I need to know how many times I can apply a specific offer that requires 3 different items from the array of ids in order to be valid.
Thanks for any help - just ask if anything is unclear and I'll try to explain. If I've missed an existing question that answers this please let me know and I'll close this question.
Here's one way you can follow:
$basket = array(5, 5, 4, 1, 2, 3, 4, 2, 1, 3, 5, 5);
$length = 3; //unique set size
function getSetCount($basket, $length) {
$counts = array_count_values($basket); //count of each ID
$totalCount = count($basket); //count of all elements
//maximum amount of sets
$max = floor($totalCount / $length);
//since every ID can be only once in a set, it all above $max occurences won't be able to fit any set.
$reducedCounts = array_map(function ($el) use ($max) {
return min($el, $max);
}, $counts);
//final result is "all elements that should fit in sets / size of set
return floor(array_sum($reducedCounts) / $length);
}
If you would like to print them:
function printSets($basket, $length) {
$counts = array_count_values($basket); //count of each ID
while(!empty($counts)) {
arsort($counts);//reverse sort with keeping the keys
$set = array_slice(array_keys($counts),0,$length); //get the set
echo implode(',', $set) . '<br/>';
foreach($set as $id) { //reduce counts
$counts[$id]--;
}
$counts = array_filter($counts); //clean zeros
}
}
The code above may not handle proparly some edge cases. But this is the idea.
Basically array_count_values() counts values' occurences and returns an array of value => count pairs. Then its easy to manipulate this data.
If i understand you correctly:
function getChunksCount($products)
{
// get unique ids
$uniqueProducts = array_unique($products);
// count unique ids
$count = count($uniqueProducts);
// let's get the number of chucks available
$chunkSize = 3;
// round to the floor
return floor($count / $chunkSize);
}
No complex logic or processing at all. Next time try to write down what exactly needs to be done in what order, and the solution might become pretty obvious :)
You can do it using array_unique and array_slice.
$arr = [5, 5, 4, 1, 2, 3, 4, 2, 1, 3, 5, 5];
$new_arr = array_slice(array_unique($arr), 0, 3);
print_r($new_arr); //Array ( [0] => 5 [1] => 4 [2] => 1 )
You can use array_count_values.
<?php `$array = array(5, 5, 4, 1, 2, 3, 4, 2, 1, 3, 5, 5); $vals = array_count_values($array); echo 'No. of NON Duplicate Items: '.count($vals).'<br><br>'; print_r($vals);`?>
Output is -Array ( [5] => 4 [4] => 2 1 => 2 [2] => 2 [3] => 2 )

Sorting an array using a storage

I have an array like this:
$a = [2, 1, 1, 2, 3, 1, 3, 2];
I need to sort it, by using an external variable. But I need the following output:
$output = [
[0, s], // Move $a[0] to storage
[5, 0], // Move $a[5] to $[0]
[s, 5], // Move storage to $a[5]
[4, s], // Move $a[4] to storage
[7, 4], // Move $a[7] to $a[4]
[s, 7] // Move storage to $[7]
];
I need an algorithm to make an array, a delimitered string, or any kind of output, containing the steps to sort the array.
Mainly in PHP but I can implement it from any lang.
A fun if perhaps not that efficient idea:
Determine the count and sorted offset of each element:
$a = [2, 1, 1, 2, 3, 1, 3, 2];
$count_and_offset = [1 => [3,0], 2 => [3,3], 3 => [2,6]]
Determine the permutation,
$a_permutation = [5, 2, 3, 4, 8, 1, 7, 6];
Enumerate the cycles (http://en.wikipedia.org/wiki/Cyclic_permutation),
(1586)(2)(3)(4)(7)
Make a list (example is not zero-based),
[[6, s]
,[8, 6]
,[5, 8]
,[1, 5]
,[s, 1]]
You can use insertion sort (or do similar thing for other sorting algorithm like bubble sort or quicksort)
Pseudo code (which I took from wiki)
for i ← 1 to length(A)
j ← i
while j > 0 and A[j-1] > A[j]
swap A[j] and A[j-1]
j ← j - 1
So for the each swap step, you just need to print
[j , s]
[j - 1,j]
[s , j - 1]
You are looking for array_map : http://php.net/manual/fr/function.array-map.php (old answer : usort : http://php.net/manual/fr/function.usort.php) : create a custom function which make what you are looking for, and call it with array_map.

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