How can I parse RGB color in web color format (3 or 6 hex digits) to Color from image/color? Does go have any built-in parser for that? I want to be able to parse both #XXXXXX and #XXX colors formats. color docs says nothing about it: https://golang.org/pkg/image/color/ but this task is very common, so I beleive that go has some functions for that (which I just didn't find).
Here's another solution using fmt.Sscanf(). It certainly not the fastest solution, but it is elegant. It scans right into the fields of a color.RGBA struct:
func ParseHexColor(s string) (c color.RGBA, err error) {
c.A = 0xff
switch len(s) {
case 7:
_, err = fmt.Sscanf(s, "#%02x%02x%02x", &c.R, &c.G, &c.B)
case 4:
_, err = fmt.Sscanf(s, "#%1x%1x%1x", &c.R, &c.G, &c.B)
// Double the hex digits:
c.R *= 17
c.G *= 17
c.B *= 17
default:
err = fmt.Errorf("invalid length, must be 7 or 4")
}
return
}
Testing it:
hexCols := []string{
"#112233",
"#123",
"#000233",
"#023",
"invalid",
"#abcd",
"#-12",
}
for _, hc := range hexCols {
c, err := ParseHexColor(hc)
fmt.Printf("%-7s = %3v, %v
", hc, c, err)
}
Output (try it on the Go Playground):
#112233 = { 17 34 51 255}, <nil>
#123 = { 17 34 51 255}, <nil>
#000233 = { 0 2 51 255}, <nil>
#023 = { 0 34 51 255}, <nil>
invalid = { 0 0 0 255}, input does not match format
#abcd = { 0 0 0 255}, invalid length, must be 7 or 4
#-12 = { 0 0 0 255}, expected integer
If performance does matter, fmt.Sscanf() is a really bad choice. It requires a format string which the implementation has to parse, and according to it parse the input, and use reflection to store the result to the pointed values.
Since the task is basically just "parsing" a hexadecimal value, we can do better than this. We don't even have to call into a general hex parsing library (such as encoding/hex), we can do that on our own. We don't even have to treat the input as a string, or even as a series of runes, we may lower to the level of treating it as a series of bytes. Yes, Go stores string values as UTF-8 byte sequences in memory, but if the input is a valid color string, all its bytes must be in the range of 0..127 which map to bytes 1-to-1. If that would not be the case, the input would already be invalid, which we will detect, but what color we return in that case should not matter (does not matter).
Now let's see a fast implementation:
var errInvalidFormat = errors.New("invalid format")
func ParseHexColorFast(s string) (c color.RGBA, err error) {
c.A = 0xff
if s[0] != '#' {
return c, errInvalidFormat
}
hexToByte := func(b byte) byte {
switch {
case b >= '0' && b <= '9':
return b - '0'
case b >= 'a' && b <= 'f':
return b - 'a' + 10
case b >= 'A' && b <= 'F':
return b - 'A' + 10
}
err = errInvalidFormat
return 0
}
switch len(s) {
case 7:
c.R = hexToByte(s[1])<<4 + hexToByte(s[2])
c.G = hexToByte(s[3])<<4 + hexToByte(s[4])
c.B = hexToByte(s[5])<<4 + hexToByte(s[6])
case 4:
c.R = hexToByte(s[1]) * 17
c.G = hexToByte(s[2]) * 17
c.B = hexToByte(s[3]) * 17
default:
err = errInvalidFormat
}
return
}
Testing it with the same inputs as in the first example, the output is (try it on the Go Playground):
#112233 = { 17 34 51 255}, <nil>
#123 = { 17 34 51 255}, <nil>
#000233 = { 0 2 51 255}, <nil>
#023 = { 0 34 51 255}, <nil>
invalid = { 0 0 0 255}, invalid format
#abcd = { 0 0 0 255}, invalid format
#-12 = { 0 17 34 255}, invalid format
Let's benchmark these 2 solutions. The benchmarking code will include calling them with long and short formats. Error case is excluded.
func BenchmarkParseHexColor(b *testing.B) {
for i := 0; i < b.N; i++ {
ParseHexColor("#112233")
ParseHexColor("#123")
}
}
func BenchmarkParseHexColorFast(b *testing.B) {
for i := 0; i < b.N; i++ {
ParseHexColorFast("#112233")
ParseHexColorFast("#123")
}
}
And here are the benchmark results:
go test -bench . -benchmem
BenchmarkParseHexColor-4 500000 2557 ns/op 144 B/op 9 allocs/op
BenchmarkParseHexColorFast-4 100000000 10.3 ns/op 0 B/op 0 allocs/op
As we can see, the "fast" solution is roughly 250 times faster and uses no allocation (unlike the "elegant" solution).
You can convert any 2 hex digits into an integer using strconv.ParseUint
strconv.ParseUint(str, 16, 8)
The 16 indicates base 16 (hex) and the 8 indicates the bit count, in this case, one byte.
You can use this to parse each 2 characters into their components
https://play.golang.org/p/B56B8_NvnVR
func ParseHexColor(v string) (out color.RGBA, err error) {
if len(v) != 7 {
return out, errors.New("hex color must be 7 characters")
}
if v[0] != '#' {
return out, errors.New("hex color must start with '#'")
}
var red, redError = strconv.ParseUint(v[1:3], 16, 8)
if redError != nil {
return out, errors.New("red component invalid")
}
out.R = uint8(red)
var green, greenError = strconv.ParseUint(v[3:5], 16, 8)
if greenError != nil {
return out, errors.New("green component invalid")
}
out.G = uint8(green)
var blue, blueError = strconv.ParseUint(v[5:7], 16, 8)
if blueError != nil {
return out, errors.New("blue component invalid")
}
out.B = uint8(blue)
return
}
Edit: Thanks to Peter for the correction
An RGBA color is just 4 bytes, one each for the red, green, blue, and alpha channels. For three or six hex digits the alpha byte is usually implied to be 0xFF (AABBCC is considered the same as AABBCCFF, as is ABC).
So parsing the color string is as simple as normalizing it, such that it is of the form "RRGGBBAA" (4 hex encoded bytes), and then decoding it:
package main
import (
"encoding/hex"
"fmt"
"image/color"
"log"
)
func main() {
colorStr := "102030FF"
colorStr, err := normalize(colorStr)
if err != nil {
log.Fatal(err)
}
b, err := hex.DecodeString(colorStr)
if err != nil {
log.Fatal(err)
}
color := color.RGBA{b[0], b[1], b[2], b[3]}
fmt.Println(color) // Output: {16 32 48 255}
}
func normalize(colorStr string) (string, error) {
// left as an exercise for the reader
return colorStr, nil
}
Try it on the playground: https://play.golang.org/p/aCX-vyfMG4G