How can I get the count of line in a file in an efficient way

Question

I have a big file  It includes approximately 3 000-20 000 lines  How can I get the total count of lines in the file using Java

User · Answer

Try the unix  wc  command  I don t mean use it  I mean download the source and see how they do it  It s probably in c  but you can easily port the behavior to java  The problem with making your own is to account for the ending cr lf problem

User · Answer

If the already posted answers aren t fast enough you ll probably have to look for a solution specific to your particular problem   For example if these text files are logs that are only appended to and you regularly need to know the number of lines in them you could create an index   This index would contain the number of lines in the file  when the file was last modified and how large the file was then   This would allow you to recalculate the number of lines in the file by skipping over all the lines you had already seen and just reading the new lines

User · Answer

All previous answers suggest to read though the whole file and count the amount of newlines you find while doing this  You commented some as  not effective  but thats the only way you can do that  A  line  is nothing else as a simple character inside the file  And to count that character you must have a look at every single character within the file   I m sorry  but you have no choice   -

User · Answer

The buffered reader is overkill  Reader r   new FileReader  f txt     int count   0  int nextchar   0  while  nextchar    -1           nextchar   r read            if  nextchar    Character getNumericValue   n                  count                      My search for a simple example has createde one thats actually quite poor  calling read   repeadedly for a single character is less than optimal  see here for examples and measurements

User · Answer

This is about as efficient as it can get  buffered binary read  no string conversion   FileInputStream stream   new FileInputStream   tmp test txt    byte   buffer   new byte 8192   int count   0  int n  while   n   stream read buffer    gt  0        for  int i   0  i  lt  n  i              if  buffer i       n   count            stream close    System out println  Number of lines      count

User · Answer

Read the file line by line and increment a counter for each line until you have read the entire file

User · Answer

use LineNumberReader  something like  public static int countLines File aFile  throws IOException       LineNumberReader reader   null      try           reader   new LineNumberReader new FileReader aFile            while   reader readLine       null           return reader getLineNumber          catch  Exception ex            return -1        finally            if reader    null               reader close

User · Answer

Old post  but I have a solution that could be usefull for next people  Why not just use file length to know what is the progression  Of course  lines has to be almost the same size  but it works very well for big files   public static void main String   args  throws IOException       File file   new File  yourfilehere        double fileSize   file length        System out println          gt  File size       fileSize       InputStream inputStream   new FileInputStream file       InputStreamReader inputStreamReader   new InputStreamReader inputStream   iso-8859-1        BufferedReader bufferedReader   new BufferedReader inputStreamReader       int totalRead   0      try           while  bufferedReader ready                  String line   bufferedReader readLine                   LINE PROCESSING HERE             totalRead    line length     1     we add  1 byte for the newline char              System out println  Progress     gt        totalRead   fileSize    100                           finally           bufferedReader close              It allows to see the progression without doing any full read on the file  I know it depends on lot of elements  but I hope it will be usefull       Edition  Here is a version with estimated time  I put some SYSO to show progress and estimation  I see that you have a good time estimation errors after you have treated enough line  I try with 10M lines  and after 1  of the treatment  the time estimation was exact at 95    I know  some values has to be set in variable  This code is quickly written but has be usefull for me  Hope it will be for you too      long startProcessLine   System currentTimeMillis        int totalRead   0      long progressTime   0      double percent   0      int i   0      int j   0      int fullEstimation   0      try           while  bufferedReader ready                  String line   bufferedReader readLine                totalRead    line length     1              progressTime   System currentTimeMillis   - startProcessLine              percent    double  totalRead   fileSize   100              if   percent  gt  1   amp  amp  i   10000    0                    int estimation    int    progressTime   percent     100 - percent                    fullEstimation    progressTime   estimation                  j                    System out print  Progress     gt      percent                          System out print   - current progress        progressTime      milliseconds                    System out print   - Will be finished in     gt      estimation     milliseconds                    System out println   - estimated full time   gt       progressTime   estimation                              i                    finally           bufferedReader close              System out println  Ended in      progressTime      seconds        System out println  Estimative average     gt       fullEstimation   j        System out println  Difference          double  100    double  progressTime      progressTime -  fullEstimation   j              Feel free to improve this code if you think it s a good solution

User · Answer

Files lines Java 8  has a nice and short way using NIO using Files lines  Note that you have to close the stream using try-with-resources  long lineCount  try  Stream lt String gt  stream   Files lines path  StandardCharsets UTF 8       lineCount   stream count       If you don t specify the character encoding  the default one used is UTF-8  You may specify an alternate encoding to match your particular data file as shown in the example above

User · Answer

Read the file through and count the number of newline characters  An easy way to read a file in Java  one line at a time  is the java util Scanner class

User · Answer

I found some solution for this  it might useful for you  Below is the code snippet for  count the no of lines from the file     File file   new File   mnt sdcard abc txt      LineNumberReader lineNumberReader   new LineNumberReader new FileReader file      lineNumberReader skip Long MAX VALUE     int lines   lineNumberReader getLineNumber      lineNumberReader close

User · Answer

This solution is about 3 6   faster than the top rated answer when tested on a file with 13 8 million lines  It simply reads the bytes into a buffer and counts the  n characters  You could play with the buffer size  but on my machine  anything above 8KB didn t make the code faster   private int countLines File file  throws IOException       int lines   0       FileInputStream fis   new FileInputStream file       byte   buffer   new byte BUFFER SIZE      BUFFER SIZE   8   1024     int read       while   read   fis read buffer      -1            for  int i   0  i  lt  read  i                  if  buffer i       n   lines                         fis close         return lines

User · Answer

Probably the fastest solution in pure Java would be to read the file as bytes using a NIO Channel into large ByteBuffer   Then using your knowledge of the file encoding scheme s  count the encoded CR and or NL bytes  per the relevant line separator convention     The keys to maximising throughput will be    make sure that you read the file in large chunks  avoid copying the bytes from one buffer to another   avoid copying   converting bytes into characters  and  avoid allocating objects to represent the file lines    The actual code is too complicated for me to write on the fly   Besides  the OP is not asking for the fastest solution

User · Answer

BufferedReader reader   new BufferedReader new FileReader  file txt     int lines   0  while  reader readLine      null  lines    reader close      Update  To answer the performance-question raised here  I made a measurement  First thing  20 000 lines are too few  to get the program running for a noticeable time  I created a text-file with 5 million lines  This solution  started with java without parameters like -server or -XX-options  needed around 11 seconds on my box  The same with wc -l  UNIX command-line-tool to count lines   11 seconds  The solution reading every single character and looking for   n  needed 104 seconds  9-10 times as much

User · Answer

Quick and dirty  but it does the job   import java io     public class Counter        public final static void main String   args  throws IOException           if  args length  gt  0                File file   new File args 0                System out println countLines file                         public final static int countLines File file  throws IOException           ProcessBuilder builder   new ProcessBuilder  wc    -l   file getAbsolutePath             Process process   builder start            InputStream in   process getInputStream            LineNumberReader reader   new LineNumberReader new InputStreamReader in            String line   reader readLine            if  line    null                return Integer parseInt line trim   split      0              else               return -1

User · Answer

Do You need exact number of lines or only its approximation  I happen to process large files in parallel and often I don t need to know exact count of lines - I then revert to sampling  Split the file into ten 1MB chunks and count lines in each chunk  then multiply it by 10 and You ll receive pretty good approximation of line count

[java] How can I get the count of line in a file in an efficient way?

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