I have read a bit about multidimensional arrays would it make sense to solve this problem using such data structures in Java, or how should I proceed?
Problem
I have a text file containing records which contain multiple lines. One record is anything between <SUBBEGIN and <SUBEND.
The lines in the record follow no predefined order and may be absent from a record. In the input file (see below) I am only interested in lines MSISDN, CB,CF and ODBIC fields.
For each of these fields I would like to apply regular expressions to extract the value to the right of the equals.
Output file would be a comma separated file containing these values, example
MSISDN=431234567893 the value 431234567893 is written to the output file
error checking
NoMSISDNnofound when no MSISDN is found in a record
noCFUALLPROVNONE when no CFU-ALL-PROV-NONE is found in a recored
Search and replace operations
CFU-ALL-PROV-NONE should be replaced by CFU-ALL-PROV-1/1/1
CFU-TS10-ACT-914369223311 should be replaced by CFU-TS10-ACT-1/1/0/4369223311
Output for first record
431234567893,BAOC-ALL-PROV,BOIC-ALL-PROV,BOICEXHC-ALL-PROV,BICROAM-ALL-PROV,CFU-ALL-PROV-1/1/1,CFB-ALL-PROV-1/1/1,CFNRY-ALL-PROV-1/1/1,CFNRY-ALL-PROV-1/1/1,CFU-TS10-ACT-1/1/1/4369223311,BAIC,BAOC
Input file
<BEGINFILE>
<SUBBEGIN
IMSI=11111111111111;
MSISDN=431234567893;
CB=BAOC-ALL-PROV;
CB=BOIC-ALL-PROV;
CB=BOICEXHC-ALL-PROV;
CB=BICROAM-ALL-PROV;
IMEISV=4565676567576576;
CW=CW-ALL-PROV;
CF=CFU-ALL-PROV-NONE-YES-NO-NONE-YES-65535-NO-NO-NO-NO-NO-NO-NO-NO-NO-NO;
CF=CFB-ALL-PROV-NONE-YES-YES-NONE-YES-65535-NO-NO-NO-NO-NO-NO-NO-NO-NO-NO;
CF=CFNRY-ALL-PROV-NONE-YES-YES-NONE-YES-65535-NO-NO-NO-NO-NO-NO-NO-NO-NO-NO;
CF=CFNRC-ALL-PROV-NONE-YES-NO-NONE-YES-65535-NO-NO-NO-NO-NO-NO-NO-NO-NO-NO;
CF=CFU-TS10-ACT-914369223311-YES-YES-25-YES-65535-YES-YES-NO-NO-NO-YES-YES-
YES-YES-NO;
ODBIC=BAIC;
ODBOC=BAOC;
ODBROAM=ODBOHC;
ODBPRC=ENTER;
ODBPRC=INFO;
ODBPLMN=NONE;
ODBPOS=NOBPOS-BOTH;
ODBECT=OdbAllECT;
ODBDECT=YES;
ODBMECT=YES;
ODBPREMSMS=YES;
ODBADULTSMS=YES;
<SUBEND
<SUBBEGIN
IMSI=11111111111133;
MSISDN=431234567899;
CB=BAOC-ALL-PROV;
CB=BOIC-ALL-PROV;
CB=BOICEXHC-ALL-PROV;
CB=BICROAM-ALL-PROV;
CW=CW-ALL-PROV;
CF=CFU-ALL-PROV-NONE-YES-NO-NONE-YES-65535-NO-NO-NO-NO-NO-NO-NO-NO+-NO-NO;
CF=CFB-ALL-PROV-NONE-YES-YES-NONE-YES-65535-NO-NO-NO-NO-NO-NO-NO-NO-NO-NO;
CF=CFNRY-ALL-PROV-NONE-YES-YES-NONE-YES-65535-NO-NO-NO-NO-NO-NO-NO-NO-NO-NO;
CF=CFNRC-ALL-PROV-NONE-YES-NO-NONE-YES-65535-NO-NO-NO-NO-NO-NO-NO-NO-NO-NO;
CF=CFU-TS10-ACT-914369223311-YES-NO-NONE-YES-65535-YES-YES-NO-NO-NO-NO-NO-NO-NO-NO;
CF=CFD-TS10-REG-91430000000-YES-YES-25-YES-65535-YES-YES-NO-NO-NO-YES-YES-YES-YES-NO;
ODBIC=BICCROSSDOMESTIC;
ODBOC=BAOC;
ODBROAM=ODBOH;
ODBPRC=INFO;
ODBPLMN=PLMN1
ODBPLMN=PLMN3;
ODBPOS=NOBPOS-BOTH;
ODBECT=OdbAllECT;
ODBDECT=YES;
ODBMECT=YES;
ODBPREMSMS=NO;
ODBADULTSMS=YES;
<SUBEND
From what I understand, you are simply reading a text file and processing it and maybe replacing some words. You do not therefore need a data structure to store the words in. Instead you can simply read the file line by line and pass it through a bunch of if statements (maybe a couple booleans to check if the specific parameters you are searching for have been found?) and then rewrite the line you want to a new file.
Dealing with big files to implement data in machine learning algorithms, I did it by passing all of the file contents in a variable, and then using the String.split("delimeter") method (Supported from Java 8 and later), I broke the contents in a one-dimensional array, where each cell had the info before the delimeter.
Firstly read the file via a scanner or your way of doing it (let content be the variable with your info), and then break it with
content.split("<SUBEND");
I have a hashMap(guava bimap) in which keys and values both are strings, I wanted to write a program which parses the given file and replaces all the strings in the file which are also in BiMap with corresponding values from Bimap.
for example: i have a file called test.txt has following text
Java is a set of several computer software and specifications developed by Sun Microsystems.
and my BiMap has
"java i" => "value1"
"everal computer" => "value2" etc..
So now i want my program to take test.txt and Bimap as input and give an output which looks something like this
value1s a set of svalue2 software and specifications developed by Sun Microsystems.
please point me towards any algorithm which can do this, the program takes large files as input so brute force may not be a good idea.
Edit: I'm using fixed length strings for keys and values.
That example was just intended to show the operation.
Thanks.
For a batch operation like this, I would avoid putting a lot of data into the memory. Therefore I'd recommend you to write the new content into a new file. If the file in the end must be the exact same file, you can still replace one file by the other, at the end of the process. read, write and flush each new line separately, and you won't have any memory issues.
I have a small program that reads input files that include C-like macros. The processing happens in two passes: the first one searches for the macro definitions and stores them, and the second pass searches for the macro calls and expands / replaces them.
This all works nicely, but it's time consuming. Currently, this is how I do it:
foreach token in file:
foreach macro in macroDefinitions:
if token equals macro.name:
expand()
endif
end foreach
endforeach
In this pseudo-example, 'token' is a word from the source file, and 'macro' is the macro definition from the first pass. There's about 20 000 macro definitions, and 1800 input files, totalling in about 600 000 lines to process (and each line is splitted into n tokens). This means that the total comparison count is (count of tokens) * (count of macro definitions). How could I speed this up? Am I missing something, or do I really have to do all these comparisons?
For additional info, the tokens are Strings in String[] arrays, and the macros are Macro objects in ArrayList type of list. Could I speed the process up with other kinds of data structures?
You need to use a Map that maps from a macro name to its definition.
In pseudo-code:
for each token in file:
if this is a macro defininition:
name, definition <- parse definition
map.put(name, definition)
for each token in file:
if map.contains(token):
definition <- map.get(token):
expand definition
(Update - You can get rid of the contains call and just call get and then test for null. It is worth reading the javadocs to get a better understanding of how the Map, TreeMap and HashMap APIs work.)
Typical implementations of Map use either a balanced binary tree or a hash table, and have lookup and insert operations that have complexity O(logN) or O(1) (under normal circumstances).
I would recommend to create a script e.g. in Perl that actually does the file processing and call that script from your Java code using ProcessBuilder.
Use the best tool for each problem.
Putting the macro definitions in a Map will reduce the time needed to look up the macros considerably.
edit: Klas Lindbäck solution would be better if you can add keys. If you cannot then searching algorithms as i proposed would be one way to improve search speeds.
You could add some search algorithms like Binary search which would drastically improve search results
You can use a HashSet containing the name of the macro definitions and for each token, check that it is contained in the set:
for(String token : token) {
if(macroNamesSet.contains(token)) {
expand();
}
}
The contains method thake O(1) time. So overall and once the set of macro names has been created, it takes (count of tokens) time.
I have a general question on your opinion about my "technique".
There are 2 textfiles (file_1 and file_2) that need to be compared to each other. Both are very huge (3-4 gigabytes, from 30,000,000 to 45,000,000 lines each).
My idea is to read several lines (as many as possible) of file_1 to the memory, then compare those to all lines of file_2. If there's a match, the lines from both files that match shall be written to a new file. Then go on with the next 1000 lines of file_1 and also compare those to all lines of file_2 until I went through file_1 completely.
But this sounds actually really, really time consuming and complicated to me.
Can you think of any other method to compare those two files?
How long do you think the comparison could take?
For my program, time does not matter that much. I have no experience in working with such huge files, therefore I have no idea how long this might take. It shouldn't take more than a day though. ;-) But I am afraid my technique could take forever...
Antoher question that just came to my mind: how many lines would you read into the memory? As many as possible? Is there a way to determine the number of possible lines before actually trying it?
I want to read as many as possible (because I think that's faster) but I've ran out of memory quite often.
Thanks in advance.
EDIT
I think I have to explain my problem a bit more.
The purpose is not to see if the two files in general are identical (they are not).
There are some lines in each file that share the same "characteristic".
Here's an example:
file_1 looks somewhat like this:
mat1 1000 2000 TEXT //this means the range is from 1000 - 2000
mat1 2040 2050 TEXT
mat3 10000 10010 TEXT
mat2 20 500 TEXT
file_2looks like this:
mat3 10009 TEXT
mat3 200 TEXT
mat1 999 TEXT
TEXT refers to characters and digits that are of no interest for me, mat can go from mat1 - mat50 and are in no order; also there can be 1000x mat2 (but the numbers in the next column are different). I need to find the fitting lines in a way that: matX is the same in both compared lines an the number mentioned in file_2 fits into the range mentioned in file_1.
So in my example I would find one match: line 3 of file_1and line 1 of file_2 (because both are mat3 and 10009 is between 10000 and 10010).
I hope this makes it clear to you!
So my question is: how would you search for the matching lines?
Yes, I use Java as my programming language.
EDIT
I now divided the huge files first so that I have no problems with being out of memory. I also think it is faster to compare (many) smaller files to each other than those two huge files. After that I can compare them the way I mentioned above. It may not be the perfect way, but I am still learning ;-)
Nonentheless all your approaches were very helpful to me, thank you for your replies!
I think, your way is rather reasonable.
I can imagine different strategies -- for example, you can sort both files before compare (where is efficient implementation of filesort, and unix sort utility can sort several Gbs files in minutes), and, while sorted, you can compare files sequentally, reading line by line.
But this is rather complex way to go -- you need to run external program (sort), or write comparable efficient implementation of filesort in java by yourself -- which is by itself not an easy task. So, for the sake of simplicity, I think you way of chunked read is very promising;
As for how to find reasonable block -- first of all, it may not be correct what "the more -- the better" -- I think, time of all work will grow asymptotically, to some constant line. So, may be you'll be close to that line faster then you think -- you need benchmark for this.
Next -- you may read lines to buffer like this:
final List<String> lines = new ArrayList<>();
try{
final List<String> block = new ArrayList<>(BLOCK_SIZE);
for(int i=0;i<BLOCK_SIZE;i++){
final String line = ...;//read line from file
block.add(line);
}
lines.addAll(block);
}catch(OutOfMemory ooe){
//break
}
So you read as many lines, as you can -- leaving last BLOCK_SIZE of free memory. BLOCK_SIZE should be big enouth to the rest of you program to run without OOM
In an ideal world, you would be able to read in every line of file_2 into memory (probably using a fast lookup object like a HashSet, depending on your needs), then read in each line from file_1 one at a time and compare it to your data structure holding the lines from file_2.
As you have said you run out of memory however, I think a divide-and-conquer type strategy would be best. You could use the same method as I mentioned above, but read in a half (or a third, a quarter... depending on how much memory you can use) of the lines from file_2 and store them, then compare all of the lines in file_1. Then read in the next half/third/quarter/whatever into memory (replacing the old lines) and go through file_1 again. It means you have to go through file_1 more, but you have to work with your memory constraints.
EDIT: In response to the added detail in your question, I would change my answer in part. Instead of reading in all of file_2 (or in chunks) and reading in file_1 a line at a time, reverse that, as file_1 holds the data to check against.
Also, with regards searching the matching lines. I think the best way would be to do some processing on file_1. Create a HashMap<List<Range>> that maps a String ("mat1" - "mat50") to a list of Ranges (just a wrapper for a startOfRange int and an endOfRange int) and populate it with the data from file_1. Then write a function like (ignoring error checking)
boolean isInRange(String material, int value)
{
List<Range> ranges = hashMapName.get(material);
for (Range range : ranges)
{
if (value >= range.getStart() && value <= range.getEnd())
{
return true;
}
}
return false;
}
and call it for each (parsed) line of file_2.
Now that you've given us more specifics, the approach I would take relies upon pre-partitioning, and optionally, sorting before searching for matches.
This should eliminate a substantial amount of comparisons that wouldn't otherwise match anyway in the naive, brute-force approach. For the sake of argument, lets peg both files at 40 million lines each.
Partitioning: Read through file_1 and send all lines starting with mat1 to file_1_mat1, and so on. Do the same for file_2. This is trivial with a little grep, or should you wish to do it programmatically in Java it's a beginner's exercise.
That's one pass through two files for a total of 80million lines read, yielding two sets of 50 files of 800,000 lines each on average.
Sorting: For each partition, sort according to the numeric value in the second column only (the lower bound from file_1 and the actual number from file_2). Even if 800,000 lines can't fit into memory I suppose we can adapt 2-way external merge sort and perform this faster (fewer overall reads) than a sort of the entire unpartitioned space.
Comparison: Now you just have to iterate once through both pairs of file_1_mat1 and file_2_mat1, without need to keep anything in memory, outputting matches to your output file. Repeat for the rest of the partitions in turn. No need for a final 'merge' step (unless you're processing partitions in parallel).
Even without the sorting stage the naive comparison you're already doing should work faster across 50 pairs of files with 800,000 lines each rather than with two files with 40 million lines each.
there is a tradeoff: if you read a big chunk of the file, you save the disc seek time, but you may have read information you will not need, since the change was encountered on the first lines.
You should probably run some experiments [benchmarks], with varying chunk size, to find out what is the optimal chunk to read, in the average case.
No sure how good an answer this would be - but have a look at this page: http://c2.com/cgi/wiki?DiffAlgorithm - it summarises a few diff algorithms. Hunt-McIlroy algorithm is probably the better implementation. From that page there's also a link to a java implementation of the GNU diff. However, I think an implementation in C/C++ and compiled into native code will be much faster. If you're stuck with java, you may want to consider JNI.
Indeed, that could take a while. You have to make 1,200.000,000 line comparisions.
There are several possibilities to speed that up by an order of magnitute:
One would be to sort file2 and do kind of a binary search on file level.
Another approach: compute a checksum of each line, and search that. Depending on average line length, the file in question would be much smaller and you really can do a binary search if you store the checksums in a fixed format (i.e. a long)
The number of lines you read at once from file_1 does not matter, however. This is micro-optimization in the face of great complexity.
If you want a simple approach: you can hash both of the files and compare the hash. But it's probably faster (especially if the files differ) to use your approach. About the memory consumption: just make sure you use enough memory, using no buffer for this kind a thing is a bad idea..
And all those answers about hashes, checksums etc: those are not faster. You have to read the whole file in both cases. With hashes/checksums you even have to compute something...
What you can do is sort each individual file. e.g. the UNIX sort or similar in Java. You can read the sorted files one line at a time to perform a merge sort.
I have never worked with such huge files but this is my idea and should work.
You could look into hash. Using SHA-1 Hashing.
Import the following
import java.io.FileInputStream;
import java.security.MessageDigest;
Once your text file etc has been loaded have it loop through each line and at the end print out the hash. The example links below will go into more depth.
StringBuffer myBuffer = new StringBuffer("");
//For each line loop through
for (int i = 0; i < mdbytes.length; i++) {
myBuffer.append(Integer.toString((mdbytes[i] & 0xff) + 0x100, 16).substring(1));
}
System.out.println("Computed Hash = " + sb.toString());
SHA Code example focusing on Text File
SO Question about computing SHA in JAVA (Possibly helpful)
Another sample of hashing code.
Simple read each file seperatley, if the hash value for each file is the same at the end of the process then the two files are identical. If not then something is wrong.
Then if you get a different value you can do the super time consuming line by line check.
Overall, It seems that reading line by line by line by line etc would take forever. I would do this if you are trying to find each individual difference. But I think hashing would be quicker to see if they are the same.
SHA checksum
If you want to know exactly if the files are different or not then there isn't a better solution than yours -- comparing sequentially.
However you can make some heuristics that can tell you with some kind of probability if the files are identical.
1) Check file size; that's the easiest.
2) Take a random file position and compare block of bytes starting at this position in the two files.
3) Repeat step 2) to achieve the needed probability.
You should compute and test how many reads (and size of block) are useful for your program.
My solution would be to produce an index of one file first, then use that to do the comparison. This is similar to some of the other answers in that it uses hashing.
You mention that the number of lines is up to about 45 million. This means that you could (potentially) store an index which uses 16 bytes per entry (128 bits) and it would use about 45,000,000*16 = ~685MB of RAM, which isn't unreasonable on a modern system. There are overheads in using the solution I describe below, so you might still find you need to use other techniques such as memory mapped files or disk based tables to create the index. See Hypertable or HBase for an example of how to store the index in a fast disk-based hash table.
So, in full, the algorithm would be something like:
Create a hash map which maps Long to a List of Longs (HashMap<Long, List<Long>>)
Get the hash of each line in the first file (Object.hashCode should be sufficient)
Get the offset in the file of the line so you can find it again later
Add the offset to the list of lines with matching hashCodes in the hash map
Compare each line of the second file to the set of line offsets in the index
Keep any lines which have matching entries
EDIT:
In response to your edited question, this wouldn't really help in itself. You could just hash the first part of the line, but it would only create 50 different entries. You could then create another level in the data structure though, which would map the start of each range to the offset of the line it came from.
So something like index.get("mat32") would return a TreeMap of ranges. You could look for the range preceding the value you are looking for lowerEntry(). Together this would give you a pretty fast check to see if a given matX/number combination was in one of the ranges you are checking for.
try to avoid memory consuming and make it disc consuming.
i mean divide each file into loadable size parts and compare them, this may take some extra time but will keep you safe dealing with memory limits.
What about using source control like Mercurial? I don't know, maybe it isn't exactly what you want, but this is a tool that is designed to track changes between revisions. You can create a repository, commit the first file, then overwrite it with another one an commit the second one:
hg init some_repo
cd some_repo
cp ~/huge_file1.txt .
hg ci -Am "Committing first huge file."
cp ~/huge_file2.txt huge_file1.txt
hg ci -m "Committing second huge file."
From here you can get a diff, telling you what lines differ. If you could somehow use that diff to determine what lines were the same, you would be all set.
That's just an idea, someone correct me if I'm wrong.
I would try the following: for each file that you are comparing, create temporary files (i refer to it as partial file later) on disk representing each alphabetic letter and an additional file for all other characters. then read the whole file line by line. while doing so, insert the line into the relevant file that corresponds to the letter it starts with. since you have done that for both files, you can now limit the comparison for loading two smaller files at a time. a line starting with A for example can appear only in one partial file and there will not be a need to compare each partial file more than once. If the resulting files are still very large, you can apply the same methodology on the resulting partial files (letter specific files) that are being compared by creating files according to the second letter in them. the trade-of here would be usage of large disk space temporarily until the process is finished. in this process, approaches mentioned in other posts here can help in dealing with the partial files more efficiently.
This question is not about 'best' barcode library recommendation, we use various products on different platforms, and need a simple way to verify if a given barcode is correct (according to its specification).
We have found cases where a barcode is rendered differently by different barcode libraries and free online barcode generators in the Internet. For example, a new release of a Delphi reporting library outputs non-numeric characters in Code128 as '0' or simply skips them in the text area. Before we do the migration, we want to check if these changes are caused by a broken implementation in the new library so we can report this as a bug to the author.
We mainly need Code128 and UCC/EAN-128 with A/B/C subcodes.
Online resources I checked so far are:
IDAutomation.com (displays ABC123 as 0123 with Code128-C)
Morovia.com
BarcodesInc (does not accept comma)
TEC-IT
They show different results too, for example in support for characters like comma or plus signs, at least in the human readable text.
For Code128 there isn't a single correct answer. If you use Code128-A you can get a different result than Code128-C. By result I mean how it looks. Take "803150" as an example. In Code128-A you'll need 6 characters (+ start, checksum, stop) to represent this number. Code128-C only consists of numbers, so you can compress two digits into one character. Hence you'll need only 3 characters (+ start, checksum, stop) to represent the same number. The barcodes will look different (A being longer in this case), but if you scan them both will give the correct number.
Further, Code128 doesn't need to be just A, B or C. You can actually combine the different subsets. This is common for cases like "US123457890", where Code128-A or B is used on "US" and Code128-C is used on the remaining digits. This is sometime referred to as Code-128 Auto, or just Code-128. The result is a "compressed" barcode in terms of width. You could represent the same data with A/B but again that would give you a longer barcode.
Take two online generators:
IDAutomation
BarcodesInc
I recommend the first one, where you can select between Auto/A/B/C. Here is an example image illustrating the differences:
On IDAutomation, Auto is default while A is default on Barcodes-Inc. Both are correct, you just need to be careful what subset you have selected when comparing output. I also recommend a barcode reader for use in development to test the output. Also, see this page for a comparision of the different subsets with ASCII values. I also find grandzebu.net useful, which has a free Code128 font you can use as well.
It sounds like your Delphi library always use Code128-C, since it's only possible to represent numbers in this subset.
Why not just scan them and see what comes back?