Sort months ( with strings ) algorithm

Question

I have this months array:

["January", "March", "December" , "October" ]

And I want to have it sorted like this:

["January", "March", "October", "December" ] 

I'm currently thinking in a "if/else" horrible cascade but I wonder if there is some other way to do this.

The bad part is that I need to do this only with "string" ( that is, without using Date object or anything like that )

What would be a good approach?

Answer 1

Have an array with the proper sort, and sort based on it.

Another solution (if your language supports it) is to have an associative array from month names to numbers (1..12) and use a custom comparator running sort on your array.

Solution in Perl :D

my @mon = qw( January February March April May June July August September October November December );
my $mon;
@{$mon}{@mon} = (0..$#mon);

sub by_month {
    $mon->{$a} <=> $mon->{$b};
}

sort by_month @data_to_sort

(although I'm sure a golfer could do that in < 30 characters)

And here's a solution in plain C : http://www.pnambic.com/CPS/SortAnal/html/MonOrder.html

Answer 2

Create a table with name->index, then sort the array based on its value in the table.

A couple of examples may be useful, in C# arr.sort(myCompare), in Java Collections.sort(arr, myCompare), in Python arr.sort(myCompare), in PHP usort($arr, 'myCompare'), in C++ sort(vec.begin(), vec.end(), myCompare).

Answer 3

For something like months, I'd just hard-code the arrays I needed...

It's not like months are gonna change any time soon.

Answer 4

If I had a way to supply a custom sorting order, I'd create a list defining the correct order:

correct = List("January", "February", "March", ...)

And then sort by the position in that list, something like:

toSort.sort(a, b) => compare(correct.index(a), correct.index(b))

Answer 5

Add a prefix for each month:

Jan -> aJan
Feb -> bFeb
...

Sort, then remove the prefix.

Answer 6

Create a mapping:

month_map = {"January":1,
             "February":2,
             "March":3,
             "April":4} # etc..

Use the mapping to compare one month to the other.

OR

Most languages/frameworks have objects for handling dates. Create date objects for all the months and compare them using the native (if available) inequality operators or basic sorting functions:

import datetime
January  = datetime.date(2010,1,1)
February = datetime.date(2010,2,1)
if February < January: print("The world explodes.")

Answer 7

Thanks all for the suggestions, I would like to mark you all as accepted.

Here's the resulting code:

// correct order 
months as String[] = ["jan", "feb", "mar", "apr", "may", "jun",
                      "jul", "aug", "sep", "oct", "nov", "dec"]
// my unsorted months 
myMonths as String[] = ["mar", "dec", "jul", "jan", "sep"]

// poor substitute for Map
mappedIndex as Int[]

// create an array with the corresponding index 
for each m in myMonths do
    i as Int = 0;
    for each month in months do 
       if m == month then 
           mappedIndex[] = i // no break, so I should use "else"
       else
          i = i + 1
       end
    end
end

// At this point mapped index has the same order as my "unsorted" array
// in this case [2,11,5,0,8]

// Fortunately this language has "sort" otherwise I would jump out of the window
mappedIndex.sort()

// create a new array to hold the sorted values
myMonthsSorted as String[]

// and fill it with the correct value
for each i in mappedIndex do 
   myMonthsSorted[] = months[i]
end

Answer 8

Talking from a Java POV, I'm going to pimp (as I often do) google-collections (soon to be replaced with Guava):

Arrays.sort(myMonths, Ordering.explicit("Jan", "Feb", "Mar", ....));

... and you're done.

Don't write it yourself if someone else has done it, probably more efficiently and with a nicer API than you probably would.

Not helpful in the general case, but just in case any Java folk have the same problem...