Convert byte slice "[]uint8" to float64 in GoLang

I'm trying to convert a []uint8 byte slice into a float64 in GoLang. I can't find a solution for this issue online. I've seen suggestions of converting to a string first and then to a float64 but this doesn't seem to work, it loses it's value and I end up with zeroes.

Example:

metric.Value, _ = strconv.ParseFloat(string(column.Value), 64)

And it doesn't work...

For example,

package main

import (
	"encoding/binary"
	"fmt"
	"math"
)

func Float64frombytes(bytes []byte) float64 {
	bits := binary.LittleEndian.Uint64(bytes)
	float := math.Float64frombits(bits)
	return float
}

func Float64bytes(float float64) []byte {
	bits := math.Float64bits(float)
	bytes := make([]byte, 8)
	binary.LittleEndian.PutUint64(bytes, bits)
	return bytes
}

func main() {
	bytes := Float64bytes(math.Pi)
	fmt.Println(bytes)
	float := Float64frombytes(bytes)
	fmt.Println(float)
}

Output:

[24 45 68 84 251 33 9 64]
3.141592653589793

I think this example from Go documentation is what you are looking for:
http://golang.org/pkg/encoding/binary/#example_Read

var pi float64
b := []byte{0x18, 0x2d, 0x44, 0x54, 0xfb, 0x21, 0x09, 0x40}
buf := bytes.NewReader(b)
err := binary.Read(buf, binary.LittleEndian, &pi)
if err != nil {
  fmt.Println("binary.Read failed:", err)
}
fmt.Print(pi)

Prints 3.141592653589793

As the comments read, it all depends on what kind of data you have in your []uint8 slice.

If it is bytes representing an IEEE 754 floating-point value in Little Endian order, then use Kluyg's or peterSo's (better performance without use of reflection) answer.

If it is a textual representation in Latin-1/UTF-8 encoding, then you should be able to do what you just did:

package main

import (
	"fmt"
	"strconv"
)

func main() {
	var f float64
	text := []uint8("1.23") // A decimal value represented as Latin-1 text
	
	f, err := strconv.ParseFloat(string(text), 64)
	if err != nil {
		panic(err)
	}
	
	fmt.Println(f)
}

Result:

>1.23

Playground: http://play.golang.org/p/-7iKRDG_ZM

I hope this hack helps. The purpose of it is to convert the long stream of binary numbers to float.

For example:
0110111100010010100000111100000011001010001000010000100111000000 -> -3.1415

func binFloat(bin string) float64 {

    var s1 []byte
    var result float64

    if len(bin) % 8 == 0 {

            for i := 0; i < len(bin) / 8; i++ {

                    //Chop the strings into a segment with a length of 8.
                    //Convert the string to Integer and to byte

                    num, _ := strconv.ParseInt(bin[8*i: 8*(i + 1)], 2, 64) 
                    //Store the byte into a slice s1
                    s1 = append(s1, byte(num)) 
            }

    }
    
    //convert the byte slice to a float64. 
    //The algorithm below are copied from golang binary examples. 

    buf := bytes.NewReader(s1)

    //You can also change binary.LittleEndian to binary.BigEndian
    //For the details of Endianness, please google Endianness
  
    err := binary.Read(buf, binary.LittleEndian, &result) 

    if err != nil {

            panic(err)
            fmt.Println("Length of the binary is not in the length of 8")
    }

    return result
}