To convert [][]byte to []string, I do this
data, err := ioutil.ReadFile("test.txt")
if err != nil {
return nil, err
}
db := bytes.Split(data, []uint8("\n"))
// Convert [][]byte to []string
s := make([]string, len(db))
for i, val := range db {
s[i] = string(val)
}
fmt.Printf("%v", s)
I am new to golang, I'm not sure is most efficient way to do this.
The most effective way would be to remove this step: db := bytes.Split(data, []uint8("\n")) and instead iterate over data like that:
func main() {
data, _ := ioutil.ReadFile("test.txt")
s := make([]string, 0)
start := 0
for i := range data {
if data[i] == '\n' {
elem := string(data[start : i-1])
s = append(s, elem)
start = i
}
}
fmt.Printf("%v", s)
}
Or if you want to convert [][]byte to []string:
func convert(data [][]byte) []string {
s := make([]string, len(data))
for row := range data {
s[row] = string(data[row])
}
return s
}
[][]byte to []string.
[] [] byte to [] string, I note no difference between your proposal and my code.If you actually want to convert a file content to a []string, you can use bufio.Scanner which is cleaner (IMO) and more efficient than the code you posted:
func readFile(filename string) ([]string, error) {
file, err := os.Open(filename)
if err != nil {
return nil, err
}
defer file.Close()
scanner := bufio.NewScanner(file)
var data []string
for scanner.Scan() {
line := scanner.Text()
data = append(data, line)
}
if err = scanner.Err(); err != nil {
return nil, err
}
return data, nil
}
Here's a benchmark* comparing the original function (readFile1) and my function (readFile2):
BenchmarkReadFile1-8 300 4632189 ns/op 3035552 B/op 10570 allocs/op
BenchmarkReadFile2-8 1000 1695820 ns/op 2169655 B/op 10587 allocs/op
*the benchmark read a sample file of 1.2 MiB and ~10K lines
The new code runs in 36% of the time and 71% of the memory used by the original function.
