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Golang: How do I determine the number of lines in a file efficiently?

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go

In Golang, I am looking for an efficient way to determine the number of lines a file has.

Of course, I can always loop through the entire file, but does not seem very efficient.

file, _ := os.Open("/path/to/filename") fileScanner := bufio.NewScanner(file) lineCount := 0 for fileScanner.Scan() {     lineCount++ } fmt.Println("number of lines:", lineCount) 

Is there a better (quicker, less expensive) way to find out how many lines a file has?

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SunSparc Avatar asked Jul 03 '14 20:07

SunSparc


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1 Answers

Here's a faster line counter using bytes.Count to find the newline characters.

It's faster because it takes away all the extra logic and buffering required to return whole lines, and takes advantage of some assembly optimized functions offered by the bytes package to search characters in a byte slice.

Larger buffers also help here, especially with larger files. On my system, with the file I used for testing, a 32k buffer was fastest.

func lineCounter(r io.Reader) (int, error) {     buf := make([]byte, 32*1024)     count := 0     lineSep := []byte{'\n'}      for {         c, err := r.Read(buf)         count += bytes.Count(buf[:c], lineSep)          switch {         case err == io.EOF:             return count, nil          case err != nil:             return count, err         }     } } 

and the benchmark output:

BenchmarkBuffioScan   500      6408963 ns/op     4208 B/op    2 allocs/op BenchmarkBytesCount   500      4323397 ns/op     8200 B/op    1 allocs/op BenchmarkBytes32k     500      3650818 ns/op     65545 B/op   1 allocs/op 
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JimB Avatar answered Sep 21 '22 22:09

JimB