How to compare large text files?

Question

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!

Answer 1

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.

Answer 2

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

Answer 3

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.

Answer 4

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

Answer 5

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.

Answer 6

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.

Answer 7

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...

Answer 8

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.

Answer 9

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.

Answer 10

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:

1. Create a hash map which maps Long to a List of Longs (HashMap<Long, List<Long>>)
2. Get the hash of each line in the first file (Object.hashCode should be sufficient)
3. Get the offset in the file of the line so you can find it again later
4. Add the offset to the list of lines with matching hashCodes in the hash map
5. Compare each line of the second file to the set of line offsets in the index
6. 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.

Answer 11

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.

Answer 12

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.

Answer 13

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.

Answer 14

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.