Related
it run corectly but it should have around 500 matches but it only has around 50 and I dont know why!
This is a probelm for my comsci class that I am having isues with
we had to make a function that checks a list for duplication I got that part but then we had to apply it to the birthday paradox( more info here http://en.wikipedia.org/wiki/Birthday_problem) thats where I am runing into problem because my teacher said that the total number of times should be around 500 or 50% but for me its only going around 50-70 times or 5%
duplicateNumber=0
import random
def has_duplicates(listToCheck):
for i in listToCheck:
x=listToCheck.index(i)
del listToCheck[x]
if i in listToCheck:
return True
else:
return False
listA=[1,2,3,4]
listB=[1,2,3,1]
#print has_duplicates(listA)
#print has_duplicates(listB)
for i in range(0,1000):
birthdayList=[]
for i in range(0,23):
birthday=random.randint(1,365)
birthdayList.append(birthday)
x= has_duplicates(birthdayList)
if x==True:
duplicateNumber+=1
else:
pass
print "after 1000 simulations with 23 students there were", duplicateNumber,"simulations with atleast one match. The approximate probibilatiy is", round(((duplicateNumber/1000)*100),3),"%"
This code gave me a result in line with what you were expecting:
import random
duplicateNumber=0
def has_duplicates(listToCheck):
number_set = set(listToCheck)
if len(number_set) is not len(listToCheck):
return True
else:
return False
for i in range(0,1000):
birthdayList=[]
for j in range(0,23):
birthday=random.randint(1,365)
birthdayList.append(birthday)
x = has_duplicates(birthdayList)
if x==True:
duplicateNumber+=1
print "after 1000 simulations with 23 students there were", duplicateNumber,"simulations with atleast one match. The approximate probibilatiy is", round(((duplicateNumber/1000.0)*100),3),"%"
The first change I made was tidying up the indices you were using in those nested for loops. You'll see I changed the second one to j, as they were previously bot i.
The big one, though, was to the has_duplicates function. The basic principle here is that creating a set out of the incoming list gets the unique values in the list. By comparing the number of items in the number_set to the number in listToCheck we can judge whether there are any duplicates or not.
Here is what you are looking for. As this is not standard practice (to just throw code at a new user), I apologize if this offends any other users. However, I believe showing the OP a correct way to write a program should be could all do us a favor if said user keeps the lack of documentation further on in his career.
Thus, please take a careful look at the code, and fill in the blanks. Look up the python doumentation (as dry as it is), and try to understand the things that you don't get right away. Even if you understand something just by the name, it would still be wise to see what is actually happening when some built-in method is being used.
Last, but not least, take a look at this code, and take a look at your code. Note the differences, and keep trying to write your code from scratch (without looking at mine), and if it messes up, see where you went wrong, and start over. This sort of practice is key if you wish to succeed later on in programming!
def same_birthdays():
import random
'''
This is a program that does ________. It is really important
that we tell readers of this code what it does, so that the
reader doesn't have to piece all of the puzzles together,
while the key is right there, in the mind of the programmer.
'''
count = 0
#Count is going to store the number of times that we have the same birthdays
timesToRun = 1000 #timesToRun should probably be in a parameter
#timesToRun is clearly defined in its name as well. Further elaboration
#on its purpose is not necessary.
for i in range(0,timesToRun):
birthdayList = []
for j in range(0,23):
random_birthday = random.randint(1,365)
birthdayList.append(random_birthday)
birthdayList = sorted(birthdayList) #sorting for easier matching
#If we really want to, we could provide a check in the above nester
#for loop to check right away if there is a duplicate.
#But again, we are here
for j in range(0, len(birthdayList)-1):
if (birthdayList[j] == birthdayList[j+1]):
count+=1
break #leaving this nested for-loop
return count
If you wish to find the percent, then get rid of the above return statement and add:
return (count/timesToRun)
Here's a solution that doesn't use set(). It also takes a different approach with the array so that each index represents a day of the year. I also removed the hasDuplicate() function.
import random
sim_total=0
birthdayList=[]
#initialize an array of 0's representing each calendar day
for i in range(365):
birthdayList.append(0)
for i in range(0,1000):
first_dup=True
for n in range(365):
birthdayList[n]=0
for b in range(0, 23):
r = random.randint(0,364)
birthdayList[r]+=1
if (birthdayList[r] > 1) and (first_dup==True):
sim_total+=1
first_dup=False
avg = float(sim_total) / 1000 * 100
print "after 1000 simulations with 23 students there were", sim_total,"simulations with atleast one duplicate. The approximate problibility is", round(avg,3),"%"
I can't seem to find a definitive answer on this and I want to make sure I understand this to the "n'th level" :-)
a = { "a" => "Hello", "b" => "World" }
a.count # 2
a.size # 2
a.length # 2
a = [ 10, 20 ]
a.count # 2
a.size # 2
a.length # 2
So which to use? If I want to know if a has more than one element then it doesn't seem to matter but I want to make sure I understand the real difference. This applies to arrays too. I get the same results.
Also, I realize that count/size/length have different meanings with ActiveRecord. I'm mostly interested in pure Ruby (1.92) right now but if anyone wants to chime in on the difference AR makes that would be appreciated as well.
Thanks!
For arrays and hashes size is an alias for length. They are synonyms and do exactly the same thing.
count is more versatile - it can take an element or predicate and count only those items that match.
> [1,2,3].count{|x| x > 2 }
=> 1
In the case where you don't provide a parameter to count it has basically the same effect as calling length. There can be a performance difference though.
We can see from the source code for Array that they do almost exactly the same thing. Here is the C code for the implementation of array.length:
static VALUE
rb_ary_length(VALUE ary)
{
long len = RARRAY_LEN(ary);
return LONG2NUM(len);
}
And here is the relevant part from the implementation of array.count:
static VALUE
rb_ary_count(int argc, VALUE *argv, VALUE ary)
{
long n = 0;
if (argc == 0) {
VALUE *p, *pend;
if (!rb_block_given_p())
return LONG2NUM(RARRAY_LEN(ary));
// etc..
}
}
The code for array.count does a few extra checks but in the end calls the exact same code: LONG2NUM(RARRAY_LEN(ary)).
Hashes (source code) on the other hand don't seem to implement their own optimized version of count so the implementation from Enumerable (source code) is used, which iterates over all the elements and counts them one-by-one.
In general I'd advise using length (or its alias size) rather than count if you want to know how many elements there are altogether.
Regarding ActiveRecord, on the other hand, there are important differences. check out this post:
Counting ActiveRecord associations: count, size or length?
There is a crucial difference for applications which make use of database connections.
When you are using many ORMs (ActiveRecord, DataMapper, etc.) the general understanding is that .size will generate a query that requests all of the items from the database ('select * from mytable') and then give you the number of items resulting, whereas .count will generate a single query ('select count(*) from mytable') which is considerably faster.
Because these ORMs are so prevalent I following the principle of least astonishment. In general if I have something in memory already, then I use .size, and if my code will generate a request to a database (or external service via an API) I use .count.
In most cases (e.g. Array or String) size is an alias for length.
count normally comes from Enumerable and can take an optional predicate block. Thus enumerable.count {cond} is [roughly] (enumerable.select {cond}).length -- it can of course bypass the intermediate structure as it just needs the count of matching predicates.
Note: I am not sure if count forces an evaluation of the enumeration if the block is not specified or if it short-circuits to the length if possible.
Edit (and thanks to Mark's answer!): count without a block (at least for Arrays) does not force an evaluation. I suppose without formal behavior it's "open" for other implementations, if forcing an evaluation without a predicate ever even really makes sense anyway.
I found a good answare at http://blog.hasmanythrough.com/2008/2/27/count-length-size
In ActiveRecord, there are several ways to find out how many records
are in an association, and there are some subtle differences in how
they work.
post.comments.count - Determine the number of elements with an SQL
COUNT query. You can also specify conditions to count only a subset of
the associated elements (e.g. :conditions => {:author_name =>
"josh"}). If you set up a counter cache on the association, #count
will return that cached value instead of executing a new query.
post.comments.length - This always loads the contents of the
association into memory, then returns the number of elements loaded.
Note that this won't force an update if the association had been
previously loaded and then new comments were created through another
way (e.g. Comment.create(...) instead of post.comments.create(...)).
post.comments.size - This works as a combination of the two previous
options. If the collection has already been loaded, it will return its
length just like calling #length. If it hasn't been loaded yet, it's
like calling #count.
Also I have a personal experience:
<%= h(params.size.to_s) %> # works_like_that !
<%= h(params.count.to_s) %> # does_not_work_like_that !
We have a several ways to find out how many elements in an array like .length, .count and .size. However, It's better to use array.size rather than array.count. Because .size is better in performance.
Adding more to Mark Byers answer. In Ruby the method array.size is an alias to Array#length method. There is no technical difference in using any of these two methods. Possibly you won't see any difference in performance as well. However, the array.count also does the same job but with some extra functionalities Array#count
It can be used to get total no of elements based on some condition. Count can be called in three ways:
Array#count # Returns number of elements in Array
Array#count n # Returns number of elements having value n in Array
Array#count{|i| i.even?} Returns count based on condition invoked on each element array
array = [1,2,3,4,5,6,7,4,3,2,4,5,6,7,1,2,4]
array.size # => 17
array.length # => 17
array.count # => 17
Here all three methods do the same job. However here is where the count gets interesting.
Let us say, I want to find how many array elements does the array contains with value 2
array.count 2 # => 3
The array has a total of three elements with value as 2.
Now, I want to find all the array elements greater than 4
array.count{|i| i > 4} # =>6
The array has total 6 elements which are > than 4.
I hope it gives some info about count method.
What is the right way to split a string into words ?
(string doesn't contain any spaces or punctuation marks)
For example: "stringintowords" -> "String Into Words"
Could you please advise what algorithm should be used here ?
! Update: For those who think this question is just for curiosity. This algorithm could be used to camеlcase domain names ("sportandfishing .com" -> "SportAndFishing .com") and this algo is currently used by aboutus dot org to do this conversion dynamically.
Let's assume that you have a function isWord(w), which checks if w is a word using a dictionary. Let's for simplicity also assume for now that you only want to know whether for some word w such a splitting is possible. This can be easily done with dynamic programming.
Let S[1..length(w)] be a table with Boolean entries. S[i] is true if the word w[1..i] can be split. Then set S[1] = isWord(w[1]) and for i=2 to length(w) calculate
S[i] = (isWord[w[1..i] or for any j in {2..i}: S[j-1] and isWord[j..i]).
This takes O(length(w)^2) time, if dictionary queries are constant time. To actually find the splitting, just store the winning split in each S[i] that is set to true. This can also be adapted to enumerate all solution by storing all such splits.
As mentioned by many people here, this is a standard, easy dynamic programming problem: the best solution is given by Falk Hüffner. Additional info though:
(a) you should consider implementing isWord with a trie, which will save you a lot of time if you use properly (that is by incrementally testing for words).
(b) typing "segmentation dynamic programming" yields a score of more detail answers, from university level lectures with pseudo-code algorithm, such as this lecture at Duke's (which even goes so far as to provide a simple probabilistic approach to deal with what to do when you have words that won't be contained in any dictionary).
There should be a fair bit in the academic literature on this. The key words you want to search for are word segmentation. This paper looks promising, for example.
In general, you'll probably want to learn about markov models and the viterbi algorithm. The latter is a dynamic programming algorithm that may allow you to find plausible segmentations for a string without exhaustively testing every possible segmentation. The essential insight here is that if you have n possible segmentations for the first m characters, and you only want to find the most likely segmentation, you don't need to evaluate every one of these against subsequent characters - you only need to continue evaluating the most likely one.
If you want to ensure that you get this right, you'll have to use a dictionary based approach and it'll be horrendously inefficient. You'll also have to expect to receive multiple results from your algorithm.
For example: windowsteamblog (of http://windowsteamblog.com/ fame)
windows team blog
window steam blog
Consider the sheer number of possible splittings for a given string. If you have n characters in the string, there are n-1 possible places to split. For example, for the string cat, you can split before the a and you can split before the t. This results in 4 possible splittings.
You could look at this problem as choosing where you need to split the string. You also need to choose how many splits there will be. So there are Sum(i = 0 to n - 1, n - 1 choose i) possible splittings. By the Binomial Coefficient Theorem, with x and y both being 1, this is equal to pow(2, n-1).
Granted, a lot of this computation rests on common subproblems, so Dynamic Programming might speed up your algorithm. Off the top of my head, computing a boolean matrix M such M[i,j] is true if and only if the substring of your given string from i to j is a word would help out quite a bit. You still have an exponential number of possible segmentations, but you would quickly be able to eliminate a segmentation if an early split did not form a word. A solution would then be a sequence of integers (i0, j0, i1, j1, ...) with the condition that j sub k = i sub (k + 1).
If your goal is correctly camel case URL's, I would sidestep the problem and go for something a little more direct: Get the homepage for the URL, remove any spaces and capitalization from the source HTML, and search for your string. If there is a match, find that section in the original HTML and return it. You'd need an array of NumSpaces that declares how much whitespace occurs in the original string like so:
Needle: isashort
Haystack: This is a short phrase
Preprocessed: thisisashortphrase
NumSpaces : 000011233333444444
And your answer would come from:
location = prepocessed.Search(Needle)
locationInOriginal = location + NumSpaces[location]
originalLength = Needle.length() + NumSpaces[location + needle.length()] - NumSpaces[location]
Haystack.substring(locationInOriginal, originalLength)
Of course, this would break if madduckets.com did not have "Mad Duckets" somewhere on the home page. Alas, that is the price you pay for avoiding an exponential problem.
This can be actually done (to a certain degree) without dictionary. Essentially, this is an unsupervised word segmentation problem. You need to collect a large list of domain names, apply an unsupervised segmentation learning algorithm (e.g. Morfessor) and apply the learned model for new domain names. I'm not sure how well it would work, though (but it would be interesting).
This is basically a variation of a knapsack problem, so what you need is a comprehensive list of words and any of the solutions covered in Wiki.
With fairly-sized dictionary this is going to be insanely resource-intensive and lengthy operation, and you cannot even be sure that this problem will be solved.
Create a list of possible words, sort it from long words to short words.
Check if each entry in the list against the first part of the string. If it equals, remove this and append it at your sentence with a space. Repeat this.
A simple Java solution which has O(n^2) running time.
public class Solution {
// should contain the list of all words, or you can use any other data structure (e.g. a Trie)
private HashSet<String> dictionary;
public String parse(String s) {
return parse(s, new HashMap<String, String>());
}
public String parse(String s, HashMap<String, String> map) {
if (map.containsKey(s)) {
return map.get(s);
}
if (dictionary.contains(s)) {
return s;
}
for (int left = 1; left < s.length(); left++) {
String leftSub = s.substring(0, left);
if (!dictionary.contains(leftSub)) {
continue;
}
String rightSub = s.substring(left);
String rightParsed = parse(rightSub, map);
if (rightParsed != null) {
String parsed = leftSub + " " + rightParsed;
map.put(s, parsed);
return parsed;
}
}
map.put(s, null);
return null;
}
}
I was looking at the problem and thought maybe I could share how I did it.
It's a little too hard to explain my algorithm in words so maybe I could share my optimized solution in pseudocode:
string mainword = "stringintowords";
array substrings = get_all_substrings(mainword);
/** this way, one does not check the dictionary to check for word validity
* on every substring; It would only be queried once and for all,
* eliminating multiple travels to the data storage
*/
string query = "select word from dictionary where word in " + substrings;
array validwords = execute(query).getArray();
validwords = validwords.sort(length, desc);
array segments = [];
while(mainword != ""){
for(x = 0; x < validwords.length; x++){
if(mainword.startswith(validwords[x])) {
segments.push(validwords[x]);
mainword = mainword.remove(v);
x = 0;
}
}
/**
* remove the first character if any of valid words do not match, then start again
* you may need to add the first character to the result if you want to
*/
mainword = mainword.substring(1);
}
string result = segments.join(" ");
Given a set of strings, for example:
EFgreen
EFgrey
EntireS1
EntireS2
J27RedP1
J27GreenP1
J27RedP2
J27GreenP2
JournalP1Black
JournalP1Blue
JournalP1Green
JournalP1Red
JournalP2Black
JournalP2Blue
JournalP2Green
I want to be able to detect that these are three sets of files:
EntireS[1,2]
J27[Red,Green]P[1,2]
JournalP[1,2][Red,Green,Blue]
Are there any known ways of approaching this problem - any published papers I can read on this?
The approach I am considering is for each string look at all other strings and find the common characters and where differing characters are, trying to find sets of strings that have the most in common, but I fear that this is not very efficient and may give false positives.
Note that this is not the same as 'How do I detect groups of common strings in filenames' because that assumes that a string will always have a series of digits following it.
I would start here: http://en.wikipedia.org/wiki/Longest_common_substring_problem
There are links to supplemental information in the external links, including Perl implementations of the two algorithms explained in the article.
Edited to add:
Based on the discussion, I still think Longest Common Substring could be at the heart of this problem. Even in the Journal example you reference in your comment, the defining characteristic of that set is the substring 'Journal'.
I would first consider what defines a set as separate from the other sets. That gives you your partition to divide up the data, and then the problem is in measuring how much commonality exists within a set. If the defining characteristic is a common substring, then Longest Common Substring would be a logical starting point.
To automate the process of set detection, in general, you will need a pairwise measure of commonality which you can use to measure the 'difference' between all possible pairs. Then you need an algorithm to compute the partition that results in the overall lowest total difference. If the difference measure is not Longest Common Substring, that's fine, but then you need to determine what it will be. Obviously it needs to be something concrete that you can measure.
Bear in mind also that the properties of your difference measurement will bear on the algorithms that can be used to make the partition. For example, assume diff(X,Y) gives the measure of difference between X and Y. Then it would probably be useful if your measure of distance was such that diff(A,C) <= diff(A,B) + diff(B,C). And obviously diff(A,C) should be the same as diff(C,A).
In thinking about this, I also begin to wonder whether we could conceive of the 'difference' as a distance between any two strings, and, with a rigorous definition of the distance, could we then attempt some kind of cluster analysis on the input strings. Just a thought.
Great question! The steps to a solution are:
tokenizing input
using tokens to build an appropriate data structure. a DAWG is ideal, but a Trie is simpler and a decent starting point.
optional post-processing of the data structure for simplification or clustering of subtrees into separate outputs
serialization of the data structure to a regular expression or similar syntax
I've implemented this approach in regroup.py. Here's an example:
$ cat | ./regroup.py --cluster-prefix-len=2
EFgreen
EFgrey
EntireS1
EntireS2
J27RedP1
J27GreenP1
J27RedP2
J27GreenP2
JournalP1Black
JournalP1Blue
JournalP1Green
JournalP1Red
JournalP2Black
JournalP2Blue
JournalP2Green
^D
EFgre(en|y)
EntireS[12]
J27(Green|Red)P[12]
JournalP[12](Bl(ack|ue)|(Green|Red))
Something like that might work.
Build a trie that represents all your strings.
In the example you gave, there would be two edges from the root: "E" and "J". The "J" branch would then split into "Jo" and "J2".
A single strand that forks, e.g. E-n-t-i-r-e-S-(forks to 1, 2) indicates a choice, so that would be EntireS[1,2]
If the strand is "too short" in relation to the fork, e.g. B-A-(forks to N-A-N-A and H-A-M-A-S), we list two words ("banana, bahamas") rather than a choice ("ba[nana,hamas]"). "Too short" might be as simple as "if the part after the fork is longer than the part before", or maybe weighted by the number of words that have a given prefix.
If two subtrees are "sufficiently similar" then they can be merged so that instead of a tree, you now have a general graph. For example if you have ABRed,ABBlue,ABGreen,CDRed,CDBlue,CDGreen, you may find that the subtree rooted at "AB" is the same as the subtree rooted at "CD", so you'd merge them. In your output this will look like this: [left branch, right branch][subtree], so: [AB,CD][Red,Blue,Green]. How to deal with subtrees that are close but not exactly the same? There's probably no absolute answer but someone here may have a good idea.
I'm marking this answer community wiki. Please feel free to extend it so that, together, we may have a reasonable answer to the question.
try "frak" . It creates regex expression from set of strings. Maybe some modification of it will help you.
https://github.com/noprompt/frak
Hope it helps.
There are many many approaches to string similarity. I would suggest taking a look at this open-source library that implements a lot of metrics like Levenshtein distance.
http://sourceforge.net/projects/simmetrics/
You should be able to achieve this with generalized suffix trees: look for long paths in the suffix tree which come from multiple source strings.
There are many solutions proposed that solve the general case of finding common substrings. However, the problem here is more specialized. You're looking for common prefixes, not just substrings. This makes it a little simpler.
A nice explanation for finding longest common prefix can be found at
http://www.geeksforgeeks.org/longest-common-prefix-set-1-word-by-word-matching/
So my proposed "pythonese" pseudo-code is something like this (refer to the link for an implementation of find_lcp:
def count_groups(items):
sorted_list = sorted(items)
prefix = sorted_list[0]
ctr = 0
groups = {}
saved_common_prefix = ""
for i in range(1, sorted_list):
common_prefix = find_lcp(prefix, sorted_list[i])
if len(common_prefix) > 0: #we are still in the same group of items
ctr += 1
saved_common_prefix = common_prefix
prefix = common_prefix
else: # we must have switched to a new group
groups[saved_common_prefix] = ctr
ctr = 0
saved_common_prefix = ""
prefix = sorted_list[i]
return groups
For this particular example of strings to keep it extremely simple consider using simple word/digit -separation.
A non-digit sequence apparently can begin with capital letter (Entire). After breaking all strings into groups of sequences, something like
[Entire][S][1]
[Entire][S][2]
[J][27][Red][P][1]
[J][27][Green][P][1]
[J][27][Red][P][2]
....
[Journal][P][1][Blue]
[Journal][P][1][Green]
Then start grouping by groups, you can fairly soon see that prefix "Entire" is a common for some group and that all subgroups have S as headgroup, so only variable for those is 1,2.
For J27 case you can see that J27 is only leaf, but that it then branches at Red and Green.
So somekind of List<Pair<list, string>> -structure (composite pattern if I recall correctly).
import java.util.*;
class StringProblem
{
public List<String> subString(String name)
{
List<String> list=new ArrayList<String>();
for(int i=0;i<=name.length();i++)
{
for(int j=i+1;j<=name.length();j++)
{
String s=name.substring(i,j);
list.add(s);
}
}
return list;
}
public String commonString(List<String> list1,List<String> list2,List<String> list3)
{
list2.retainAll(list1);
list3.retainAll(list2);
Iterator it=list3.iterator();
String s="";
int length=0;
System.out.println(list3); // 1 1 2 3 1 2 1
while(it.hasNext())
{
if((s=it.next().toString()).length()>length)
{
length=s.length();
}
}
return s;
}
public static void main(String args[])
{
Scanner sc=new Scanner(System.in);
System.out.println("Enter the String1:");
String name1=sc.nextLine();
System.out.println("Enter the String2:");
String name2=sc.nextLine();
System.out.println("Enter the String3:");
String name3=sc.nextLine();
// String name1="salman";
// String name2="manmohan";
// String name3="rahman";
StringProblem sp=new StringProblem();
List<String> list1=new ArrayList<String>();
list1=sp.subString(name1);
List<String> list2=new ArrayList<String>();
list2=sp.subString(name2);
List<String> list3=new ArrayList<String>();
list3=sp.subString(name3);
sp.commonString(list1,list2,list3);
System.out.println(" "+sp.commonString(list1,list2,list3));
}
}
Being used to the standard way of sorting strings, I was surprised when I noticed that Windows sorts files by their names in a kind of advanced way. Let me give you an example:
Track1.mp3
Track2.mp3
Track10.mp3
Track20.mp3
I think that those names are compared (during sorting) based on letters and by numbers separately.
On the other hand, the following is the same list sorted in a standard way:
Track1.mp3
Track10.mp3
Track2.mp3
Track20.mp3
I would like to create a comparing alogorithm in Delphi that would let me sort strings in the same way. At first I thought it would be enough to compare consecutive characters of two strings while they are letters. When a digit would be found at some position of both the strings, I would read all digits following them to form a number and then compare the numbers.
To give you an example, I'll compare "Track10" and "Track2" strings this way:
1) read characters while they are equal and while they are letters: "Track", "Track"
2) if a digit is found, read all following digits: "10", "2"
2a) if they are equal, go to 1 or else finish
Ten is greater than two, so "Track10" is greater than "Track2"
It had seemed that everything would be all right until I noticed, during my tests, that Windows considered "Track010" lower than "Track10", while I thought the first one was greater as it was longer (not mentioning that according to my algorithm both the strings would be equal, which is wrong).
Could you provide me with the idea how exactly Windows sorts files by names or maybe you have a ready-to-use algorithm (in any programming language) that I could base on?
Thanks a lot!
Mariusz
Jeff wrote up an article about this on Coding Horror. This is called natural sorting, where you effectively treat a group of digits as a single "character". There are implementations out there in every language under the sun, but strangely it's not usually built-in to most languages' standard libraries.
The mother of all sorts:
ls '*.mp3' | sort --version-sort
The absolute easiest way, I found, was isolate the string you want, so in the OP's case, Path.GetFileNameWithoutExtension(), remove the non-digits, convert to int, and sort. Using LINQ and some extension methods, it's a one-liner. In my case, I was going on directories:
Directory.GetDirectories(#"a:\b\c").OrderBy(x => x.RemoveNonDigits().ToIntOrZero())
Where RemoveNonDigits and ToIntOrZero are extensions methods:
public static string RemoveNonDigits(this string value) {
return Regex.Replace(value, "[^0-9]", string.Empty);
}
public static int ToIntOrZero(this string toConvert) {
try {
if (toConvert == null || toConvert.Trim() == string.Empty) return 0;
return int.Parse(toConvert);
} catch (Exception) {
return 0;
}
}
The extension methods are common tools I use everywhere. YMMV.
Here's a Python approach:
import re
def tryint(s):
"""
Return an int if possible, or `s` unchanged.
"""
try:
return int(s)
except ValueError:
return s
def alphanum_key(s):
"""
Turn a string into a list of string and number chunks.
>>> alphanum_key("z23a")
["z", 23, "a"]
"""
return [ tryint(c) for c in re.split('([0-9]+)', s) ]
def human_sort(l):
"""
Sort a list in the way that humans expect.
"""
l.sort(key=alphanum_key)
And a blog post with more detail: https://nedbatchelder.com/blog/200712/human_sorting.html