I am currently trying to process a large txt file (a bit less than 2GB) containing lines of strings.
I am loading all its content from an InputStream to a List<String>. I do that via the following snippet :
try(BufferedReader reader = new BufferedReader(new InputStreamReader(zipInputStream))) {
List<String> data = reader.lines()
.collect(Collectors.toList());
}
The problem is, the file itsef is less than 2GB, but when I look at the memory, the JVM is allocating twice the size of the file :
Also, here are the heaviest objects in memory :
So what I Understand is that Java is allocating twice the memory needed for the operation, one to put the content of the file in a byte array and another one to instanciate the string list.
My question is : can we optimize that ? avoid having twice the memory size needed ?
tl;dr String objects can take 2 bytes per character.
The long answer: conceptually a String is a sequence of char. Each char will represent one Codepoint (or half of one, but we can ignore that detail for now).
Each codepoint tends to represent a character (sometimes multiple codepoints make up one "character", but that's another detail we can ignore for this answer).
That means that if you read a 2 GB text file that was stored with a single-byte encoding (usually a member of the ISO-8859-* family) or variable-byte encoding (mostly UTF-8), then the size in memory in Java can easily be twice the size on disk.
Now there's a good amount on caveats on this, primarily that Java can (as an internal, invisible operation) use a single byte for each character in a String if and only if the characters used allow it (effectively if they fit into the fixed internal encoding that the JVM picked for this). But that didn't seem to happen for you.
What can you do to avoid that? That depends on what your use-case is:
Don't use String to store the data in the first place. Odds are that this data is actually representing some structure, and if you parse it into a dedicated format, you might get away with way less memory usage.
Don't keep the whole thing in memory: more often then not, you don't actually need everything in memory at once. Instead process and write away the data as you read it, thus never having more than a hand-full of records in memory at once.
Build your own string-like data type for your specific use-case. While building a full string replacement is a massive undertaking, if you know what subset of features you need it might actually be a quite surmountable challenge.
try to make sure that the data is stored as compact strings, if possible, by figuring out why that's not already happening (this requires digging deep in to the details of your JVM).
I have two large txt files around 150 mb. I want to read some data from each line of file1 and scan through all the lines of file2 till I find the matching data. If the matching data is not found, I want to output that line to another file.
I want the program to use as less memory as possible. Time is not a constraint.
Edit1
I have tried couple of options
Option1 : I have read the file2 using BufferedReader, Scanner and apache commons FileUtils.lineIterator. Loaded data of file2 into HashMap by reading each line. Read the data from file1 one line at a time and compared with data in HashMap. If it didn't match, wrote the line in a file3.
Option 2 : Read the file2 n times for every records in File 1 using the above mentioned three Readers.After every read I had to close the file and read again. I am wondering what's the best way. Is there any other option I can look into
I have to make some assumptions about the file.
I am going to assume the lines are long, and you want the lines that are not the same in the 2 files.
I would read the files 4 times (2 times per file).
Of course, it's not as efficient as reading it 2 times (1 time per file), but reading it 2 times means lots of memory is used.
Pseudo code for 1st read of each file:
Map<MyComparableByteArray, Long> digestMap = new HashMap<>();
try (BufferedReader br = ...)
{
long lineNr = 0;
String line;
while ((line = br.readLine()) != null)
{
digestMap.put(CreateDigest(line), lineNr);
}
}
If the digests are different/unique, I know that the line does not occur in the other file.
If the digests are the same, we will need to check the lines and actually compare them to make sure that they are really the same - this can occur during the second read.
Now what is also important is that we need to be careful of the digest we choose.
If we choose a short digest (i.e. md-5), we might run into lots of collisions, but this is appropriate for files with short lines, and we will need to handle the collisions separately (i.e. convert the map to a map<digest, list> structure.
If we choose a long digest (i.e. sha2-512), we won't run into lots of collisions (still safer to handle it like I mentioned above), BUT we will have the problem of not saving as much memory unless the file lines are very long.
So the general technique is:
Read each file and generate hashes.
Compare the hashes to mark the lines that need to be compared.
Read each file again and generate the output. Recheck all collisions found by the hashes in this step.
By the way, MyComparableByteArray is a custom wrapper around a byte[], to enable it to be a HashMap key (i.e. by implementing equals() and hashCode() methods). The byte[] cannot be used as a key, as it doesn't work with equals() and hashCode(). There are 2 ways to handle this:
custom wrapper as I've mentioned - this will be more efficient than the alternative.
convert it to a string using base64. This will make the memory usage around 2.5x worse than option 1, but does not need the custom code.
This question already has answers here:
Sort a file with huge volume of data given memory constraint
(14 answers)
Closed 2 years ago.
I have a text file that has text on each line, for example:
1245 Dog Husky
2356 Cat Tabby
3476 Dog Pug
with a huge amount of kind of arbitrary data repeated per line, about 10,000 lines, for the sake of argument, lets say it tends to infinity.
I have code that reads this data and stores it in an object, pseudo code follows;
Pet P;
lineInput = reader.readLine(); //where reader is reading the above mentionedfile
P.id = lineInput.split('\t')[0]
P.type = lineInput.split('\t')[1] //Assigning the parts of the line to it's relevant data members
P.breed = lineInput.split('\t')[2]
Now here's the problem, considering I need to be able to sort, search and display these values as fast as possible, I don't know what my best option is, I came up with two methods which can be seen below
Method 1: Store all the Objects in an array list based on their starting id number
ArrayList<Pet> idStartsWith1;
if(P.id starts with 1)
idStartsWith1.add(P); // "1245 Dog Husky" will be added here
ArrayList<Pet> idStartsWith2;
if(P.id starts with 2)
idStartsWith2.add(P); // "2356 Cat Tabby" will be added here
ArrayList<Pet> idStartsWith3;
if(P.id starts with 3)
idStartsWith3.add(P); // "3476 Dog Pug" will be added here
I think this would be the faster method, as these arraylists are already in the process memory, but I fear that it would overload the memory, and cause issues. (Remember, the number of lines in the text file tends to infinite)
Method 2:Write all the Objects to a .dat file based on their starting id number
Writer writer1 = new Writer("idStartsWith1.dat"); //writer1 will write to file "idStartsWith1.dat"
if(P.id starts with 1)
writer1.write(P); // "1245 Dog Husky" will be writen to this file
Writer writer2 = new Writer("idStartsWith2.dat"); //writer2 will write to file "idStartsWith2.dat"
if(P.id starts with 2)
writer2.write(P);
Writer writer3 = new Writer("idStartsWith3.dat"); //writer3 will write to file "idStartsWith3.dat"
if(P.id starts with 3)
writer3.write(P);
This will prevent the process memory from being overloaded but I fear that having to open, then read, then close the file each time I need to search and display a Pet, will add significant delays to the runtime.
Which of these two methods would work better? or is there another more efficient method that would not occur to a java novice like me?
Data of many applications are small enough to fit into main memory of a desktop computed. When your file has 1 GB, then you need some 3 GB of main memory and that's no problem for most desktops. On mobile, it's different.
Nothing can be as fast as working with main memory, when done right. ArrayList is not usable for searching, but a Map is.
You can use a database instead and you probably should. It's much slower than having all data in main memory, but still very fast, assuming you do it right (learn about indexes etc.). Most database can import a CSV file directly and are able to answer all your queries - filtering, sorting and joining other tables are what databases exists for.
I was supposed to write a method that reads a DNA sequence in order to test some string matching algorithms on it.
I took some existing code I use to read text files (don't really know any others):
try {
FileReader fr = new FileReader(file);
BufferedReader br = new BufferedReader(fr);
while((line = br.readLine()) != null) {
seq += line;
}
br.close();
}
catch(FileNotFoundException e) { e.printStackTrace(); }
catch(IOException e) { e.printStackTrace(); }
This seems to work just fine for small text files with ~3000 characters, but it takes forever (I just cancelled it after 10 minutes) to read files containing more than 45 million characters.
Is there a more efficient way of doing this?
One thing I notice is that you are doing seq+=line. seq is probably a String? If so, then you have to remember that strings are immutable. So in fact what you are doing is creating a new String each time you are trying to append a line to it. Please use StringBuilder instead. Also, if possible you don't want to do create a string and then process. That way you have to do it twice. Ideally you want to process as you read, but I don't know your situation.
The main element slowing your progress is the "concatenation" of the String seq and line when you call seq+=line. I use quotes for concatenation because in Java, Strings cannot be modified once they are created (e.g. immutable as user1598503 mentioned). Initially, this is not an issue, as the Strings are small, however once the Strings become very long, e.e. hundreds of thousands of characters, memory must be reallocated for the new String, which takes quite a bit of time. StringBuilder will allow you to do these concatenations in place, meaning you will not be creating a new Object every single time.
Your problem is not that the reading takes too much time, but the concatenating takes too much time. Just to verify this I ran your code (didn't finish) and then simply comented line 8 (seq += line) and it ran in under a second. You could try using seq = seq.concat(line) since it has been reported to be quite a bit faster most of the times, but I tried that too and didn't ran under 1-2 minutes (for a 9.6mb input file). My solution would be to store your lines in an ArrayList (or a container of your choice). The ArrayList example worked in about 2-3 seconds with the same input file. (so the content of your while loop would be list.add(line);). If you really, really want to store your entire file in a string you could do something like this (using the Scanner class):
String content = new Scanner(new File("input")).useDelimiter("\\Z").next();
^^This works in a matter of seconds as well. I should mention that "\Z" is the end of file delimiter so that's why it reads the whole thing in one swoop.
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.