I know there is strings.Index and strings.LastIndex, but they just find the first and last. Is there any function I can use, where I can specify the start index? Something like the last line in my example.
Example:
s := "go gopher, go"
fmt.Println(strings.Index(s, "go")) // Position 0
fmt.Println(strings.LastIndex(s, "go")) // Postion 11
fmt.Println(strings.Index(s, "go", 1)) // Position 3 - Start looking for "go" begining at index 1
It's an annoying oversight, you have to create your own function.
Something like:
func indexAt(s, sep string, n int) int {
idx := strings.Index(s[n:], sep)
if idx > -1 {
idx += n
}
return idx
}
No, but it might be simpler to apply strings.Index on a slice of the string
strings.Index(s[1:], "go")+1
strings.Index(s[n:], "go")+n
See example (for the case where the string isn't found, see OneOfOne's answer), but, as commented by Dewy Broto, one can simply test it with a 'if' statement including a simple statement:
(also called 'if' with an initialization statement)
if i := strings.Index(s[n:], sep) + n; i >= n {
...
}
Related
I am trying to optimize my stringpad library in Go. So far the only way I have found to fill a string (actually bytes.Buffer) with a known character value (ex. 0 or " ") is with a for loop.
the snippet of code is:
// PadLeft pads string on left side with p, c times
func PadLeft(s string, p string, c int) string {
var t bytes.Buffer
if c <= 0 {
return s
}
if len(p) < 1 {
return s
}
for i := 0; i < c; i++ {
t.WriteString(p)
}
t.WriteString(s)
return t.String()
}
The larger the string pad I believe there is more memory copies of the t buffer. Is there a more elegant way to make a known size buffer with a known value on initialization?
You can only use make() and new() to allocate buffers (byte slices or arrays) that are zeroed. You may use composite literals to obtain slices or arrays that initially contain non-zero values, but you can't describe the initial values dynamically (indices must be constants).
Take inspiration from the similar but very efficient strings.Repeat() function. It repeats the given string with given count:
func Repeat(s string, count int) string {
// Since we cannot return an error on overflow,
// we should panic if the repeat will generate
// an overflow.
// See Issue golang.org/issue/16237
if count < 0 {
panic("strings: negative Repeat count")
} else if count > 0 && len(s)*count/count != len(s) {
panic("strings: Repeat count causes overflow")
}
b := make([]byte, len(s)*count)
bp := copy(b, s)
for bp < len(b) {
copy(b[bp:], b[:bp])
bp *= 2
}
return string(b)
}
strings.Repeat() does a single allocation to obtain a working buffer (which will be a byte slice []byte), and uses the builtin copy() function to copy the repeatable string. One thing noteworthy is that it uses the working copy and attempts to copy the whole of it incrementally, meaning e.g. if the string has already been copied 4 times, copying this buffer will make it 8 times, etc. This will minimize the calls to copy(). Also the solution takes advantage of that copy() can copy bytes from a string without having to convert it to a byte slice.
What we want is something similar, but we want the result to be prepended to a string.
We can account for that, simply allocating a buffer that is used inside Repeat() plus the length of the string we're left-padding.
The result (without checking the count param):
func PadLeft(s, p string, count int) string {
ret := make([]byte, len(p)*count+len(s))
b := ret[:len(p)*count]
bp := copy(b, p)
for bp < len(b) {
copy(b[bp:], b[:bp])
bp *= 2
}
copy(ret[len(b):], s)
return string(ret)
}
Testing it:
fmt.Println(PadLeft("aa", "x", 1))
fmt.Println(PadLeft("aa", "x", 2))
fmt.Println(PadLeft("abc", "xy", 3))
Output (try it on the Go Playground):
xaa
xxaa
xyxyxyabc
See similar / related question: Is there analog of memset in go?
I know that for counting the occurrence of one substring I can use "strings.Count(, )". What if I want to count the number of occurrences of substring1 OR substring2? Is there a more elegant way than writing another new line with strings.count()?
Use a regular expression:
https://play.golang.org/p/xMsHIYKtkQ
aORb := regexp.MustCompile("A|B")
matches := aORb.FindAllStringIndex("A B C B A", -1)
fmt.Println(len(matches))
Another way to do substring matching is with the suffixarray package. Here is an example of matching multiple patterns:
package main
import (
"fmt"
"index/suffixarray"
"regexp"
)
func main() {
r := regexp.MustCompile("an")
index := suffixarray.New([]byte("banana"))
results := index.FindAllIndex(r, -1)
fmt.Println(len(results))
}
You can also match a single substring with the Lookup function.
If you want to count the number of matches in a large string, without allocating space for all the indices just to get the length and then throwing them away, you can use Regexp.FindStringIndex in a loop to match against successive substrings:
func countMatches(s string, re *regexp.Regexp) int {
total := 0
for start := 0; start < len(s); {
remaining := s[start:] // slicing the string is cheap
loc := re.FindStringIndex(remaining)
if loc == nil {
break
}
// loc[0] is the start index of the match,
// loc[1] is the end index (exclusive)
start += loc[1]
total++
}
return total
}
func main() {
s := "abracadabra"
fmt.Println(countMatches(s, regexp.MustCompile(`a|b`)))
}
runnable example at Go Playground
Here is the Go Playground code for this problem.
I am trying to write a golang string permutation using recursion.
The permutation function takes two arguments, prefix which is an empty string ("") and str which is "abc". The code is below
func main() {
str := "abc"
perm("", str)
}
func perm(prefix string, str string) {
n := len(str)
fmt.Println(n)
if n == 0 {
fmt.Println(prefix)
} else {
for i := 0; i < n; n++ {
perm(prefix+str[i:i+1], str[0:i]+str[(i+1):n])
}
}
}
When I run this code, I am shown 3,2,1,0 for the value of n as expected.
I successfully get "abc" but then I get a "panic: runtime error: slice bounds out of range" error.
It never shows the second round of 3,2,1,0 so it doesn't even get to the b combinations. I feel like if an error would occur is when it gets to the c portion of the string but since it doesn't even get to the b portion, I am not sure what is wrong.
Simple:
for i := 0; i < n; n++ {
^^^
?
Replace n by i, which is supposed to go from 0 to n-1.
Resulting playground.
This is just in case someone else is learning Golang and is wondering how to convert from a string to a string representation in binary.
Long story short, I have been looking at the standard library without being able to find the right call. So I started with something similar to the following:
func RuneToBinary(r rune) string {
var buf bytes.Buffer
b := []int64{128, 64, 32, 16, 8, 4, 2, 1}
v := int64(r)
for i := 0; i < len(b); i++ {
t := v-b[i]
if t >= 0 {
fmt.Fprintf(&buf, "1")
v = t
} else {
fmt.Fprintf(&buf, "0")
}
}
return buf.String()
}
This is all well and dandy, but after a couple of days looking around I found that I should have been using the fmt package instead and just format the rune with %b%:
var r rune
fmt.Printf("input: %b ", r)
Is there a better way to do this?
Thanks
Standard library support
fmt.Printf("%b", r) - this solution is already very compact and easy to write and understand. If you need the result as a string, you can use the analog Sprintf() function:
s := fmt.Sprintf("%b", r)
You can also use the strconv.FormatInt() function which takes a number of type int64 (so you first have to convert your rune) and a base where you can pass 2 to get the result in binary representation:
s := strconv.FormatInt(int64(r), 2)
Note that in Go rune is just an alias for int32, the 2 types are one and the same (just you may refer to it by 2 names).
Doing it manually ("Simple but Naive"):
If you'd want to do it "manually", there is a much simpler solution than your original. You can test the lowest bit with r & 0x01 == 0 and shift all bits with r >>= 1. Just "loop" over all bits and append either "1" or "0" depending on the bit:
Note this is just for demonstration, it is nowhere near optimal regarding performance (generates "redundant" strings):
func RuneToBin(r rune) (s string) {
if r == 0 {
return "0"
}
for digits := []string{"0", "1"}; r > 0; r >>= 1 {
s = digits[r&1] + s
}
return
}
Note: negative numbers are not handled by the function. If you also want to handle negative numbers, you can first check it and proceed with the positive value of it and start the return value with a minus '-' sign. This also applies the other manual solution below.
Manual Performance-wise solution:
For a fast solution we shouldn't append strings. Since strings in Go are just byte slices encoded using UTF-8, appending a digit is just appending the byte value of the rune '0' or '1' which is just one byte (not multi). So we can allocate a big enough buffer/array (rune is 32 bits so max 32 binary digits), and fill it backwards so we won't even have to reverse it at the end. And return the used part of the array converted to string at the end. Note that I don't even call the built-in append function to append the binary digits, I just set the respective element of the array in which I build the result:
func RuneToBinFast(r rune) string {
if r == 0 {
return "0"
}
b, i := [32]byte{}, 31
for ; r > 0; r, i = r>>1, i-1 {
if r&1 == 0 {
b[i] = '0'
} else {
b[i] = '1'
}
}
return string(b[i+1:])
}
Let's say for example that I have one string, like this:
<h1>Hello World!</h1>
What Go code would be able to extract Hello World! from that string? I'm still relatively new to Go. Any help is greatly appreciated!
If the string looks like whatever;START;extract;END;whatever you can use this which will get the string in between:
// GetStringInBetween Returns empty string if no start string found
func GetStringInBetween(str string, start string, end string) (result string) {
s := strings.Index(str, start)
if s == -1 {
return
}
s += len(start)
e := strings.Index(str[s:], end)
if e == -1 {
return
}
e += s + e - 1
return str[s:e]
}
What happens here is it will find first index of START, adds length of START string and returns all that exists from there until first index of END.
There are lots of ways to split strings in all programming languages.
Since I don't know what you are especially asking for I provide a sample way to get the output
you want from your sample.
package main
import "strings"
import "fmt"
func main() {
initial := "<h1>Hello World!</h1>"
out := strings.TrimLeft(strings.TrimRight(initial,"</h1>"),"<h1>")
fmt.Println(out)
}
In the above code you trim <h1> from the left of the string and </h1> from the right.
As I said there are hundreds of ways to split specific strings and this is only a sample to get you started.
Hope it helps, Good luck with Golang :)
DB
I improved the Jan Kardaš`s answer.
now you can find string with more than 1 character at the start and end.
func GetStringInBetweenTwoString(str string, startS string, endS string) (result string,found bool) {
s := strings.Index(str, startS)
if s == -1 {
return result,false
}
newS := str[s+len(startS):]
e := strings.Index(newS, endS)
if e == -1 {
return result,false
}
result = newS[:e]
return result,true
}
Here is my answer using regex. Not sure why no one suggested this safest approach
package main
import (
"fmt"
"regexp"
)
func main() {
content := "<h1>Hello World!</h1>"
re := regexp.MustCompile(`<h1>(.*)</h1>`)
match := re.FindStringSubmatch(content)
if len(match) > 1 {
fmt.Println("match found -", match[1])
} else {
fmt.Println("match not found")
}
}
Playground - https://play.golang.org/p/Yc61x1cbZOJ
In the strings pkg you can use the Replacer to great affect.
r := strings.NewReplacer("<h1>", "", "</h1>", "")
fmt.Println(r.Replace("<h1>Hello World!</h1>"))
Go play!
func findInString(str, start, end string) ([]byte, error) {
var match []byte
index := strings.Index(str, start)
if index == -1 {
return match, errors.New("Not found")
}
index += len(start)
for {
char := str[index]
if strings.HasPrefix(str[index:index+len(match)], end) {
break
}
match = append(match, char)
index++
}
return match, nil
}
Read up on the strings package. Have a look into the SplitAfter function which can do something like this:
var sample = "[this][is my][string]"
t := strings.SplitAfter(sample, "[")
That should produce a slice something like: "[", "this][", "is my][", "string]". Using further functions for Trimming you should get your solution. Best of luck.
func Split(str, before, after string) string {
a := strings.SplitAfterN(str, before, 2)
b := strings.SplitAfterN(a[len(a)-1], after, 2)
if 1 == len(b) {
return b[0]
}
return b[0][0:len(b[0])-len(after)]
}
the first call of SplitAfterN will split the original string into array of 2 parts divided by the first found after string, or it will produce array containing 1 part equal to the original string.
second call of SplitAfterN uses a[len(a)-1] as input, as it is "the last item of array a". so either string after after or the original string str. the input will be split into array of 2 parts divided by the first found before string, or it will produce array containing 1 part equal to the input.
if after was not found than we can simply return b[0] as it is equal to a[len(a)-1]
if after is found, it will be included at the end of b[0] string, therefore you have to trim it via b[0][0:len(b[0])-len(after)]
all strings are case sensitive