Related
I have a function that takes input of packs which is a map of the pack size and the quantity and it takes a total quantity for an order.
I need to get all the divisibles for the pack sizes, remove all the under 1 values, and then pick the best divisible which is the lowest number remaining. This number is the key from the supplied packsizes
Note: I have a function further up the trace which eliminates any possibility of there not being a divisible.
Code:
func optimalDivisble(packs map[int]int, oq int) (int, error) {
divisables := make(map[int]float64)
for key := range packs {
divisables[key] = float64(oq) / float64(key)
}
// Remove zero divisibles
filteredDivisibles := make(map[int]float64)
for key, divisable := range divisables {
if divisable >= 1 {
filteredDivisibles[key] = divisable
}
}
// Get divisables
var divisableSlice []float64
for _, filteredDivisible := range filteredDivisibles {
divisableSlice = append(divisableSlice, filteredDivisible)
}
sort.Float64s(divisableSlice)
for key, filteredDivisible := range filteredDivisibles {
if filteredDivisible == divisableSlice[0] {
return key, nil
}
}
return 0, errors.New("Could not find a divisable for quantity")
}
Could someone help refactor this, as seeing 3 for loops doesn't seem ideal. What would be more idiomatic to Go?
You can process the packs, compute the min divisible and get the key for it in a single loop. You don't need the intermediate steps:
var minDiv float64
var minKey int
minSet:=false
for key := range packs {
divisable:=float64(oq) / float64(key)
if divisable>=1 {
if minDiv>divisable || !minSet {
minDiv=divisable
minKey=key
minSet=true
}
}
}
// minKey is what you need
The problem is: find the index of two numbers that nums[index1] + nums[index2] == target. Here is my attempt in golang (index starts from 1):
package main
import (
"fmt"
)
var nums = []int{0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 25182, 25184, 25186, 25188, 25190, 25192, 25194, 25196} // The number list is too long, I put the whole numbers in a gist: https://gist.github.com/nickleeh/8eedb39e008da8b47864
var target int = 16021
func twoSum(nums []int, target int) (int, int) {
if len(nums) <= 1 {
return 0, 0
}
hdict := make(map[int]int)
for i := 1; i < len(nums); i++ {
if val, ok := hdict[nums[i+1]]; ok {
return val, i + 1
} else {
hdict[target-nums[i+1]] = i + 1
}
}
return 0, 0
}
func main() {
fmt.Println(twoSum(nums, target))
}
The nums list is too long, I put it into a gist:
https://gist.github.com/nickleeh/8eedb39e008da8b47864
This code works fine, but I find the return 0,0 part is ugly, and it runs ten times slower than the Julia translation. I would like to know is there any part that is written terrible and affect the performance?
Edit:
Julia's translation:
function two_sum(nums, target)
if length(nums) <= 1
return false
end
hdict = Dict()
for i in 1:length(nums)
if haskey(hdict, nums[i])
return [hdict[nums[i]], i]
else
hdict[target - nums[i]] = i
end
end
end
In my opinion if no elements found adding up to target, best would be to return values which are invalid indices, e.g. -1. Although returning 0, 0 would be enough as a valid index pair can't be 2 equal indices, this is more convenient (because if you forget to check the return values and you attempt to use the invalid indices, you will immediately get a run-time panic, alerting you not to forget checking the validity of the return values). As so, in my solutions I will get rid of that i + 1 shifts as it makes no sense.
Benchmarking of different solutions can be found at the end of the answer.
If sorting allowed:
If the slice is big and not changing, and you have to call this twoSum() function many times, the most efficient solution would be to sort the numbers simply using sort.Ints() in advance:
sort.Ints(nums)
And then you don't have to build a map, you can use binary search implemented in sort.SearchInts():
func twoSumSorted(nums []int, target int) (int, int) {
for i, v := range nums {
v2 := target - v
if j := sort.SearchInts(nums, v2); v2 == nums[j] {
return i, j
}
}
return -1, -1
}
Note: Note that after sorting, the indices returned will be indices of values in the sorted slice. This may differ from indices in the original (unsorted) slice (which may or may not be a problem). If you do need indices from the original order (original, unsorted slice), you may store sorted and unsorted index mapping so you can get what the original index is. For details see this question:
Get the indices of the array after sorting in golang
If sorting is not allowed:
Here is your solution getting rid of that i + 1 shifts as it makes no sense. Slice and array indices are zero based in all languages. Also utilizing for ... range:
func twoSum(nums []int, target int) (int, int) {
if len(nums) <= 1 {
return -1, -1
}
m := make(map[int]int)
for i, v := range nums {
if j, ok := m[v]; ok {
return j, i
}
m[target-v] = i
}
return -1, -1
}
If the nums slice is big and the solution is not found fast (meaning the i index grows big) that means a lot of elements will be added to the map. Maps start with small capacity, and they are internally grown if additional space is required to host many elements (key-value pairs). An internal growing requires rehashing and rebuilding with the already added elements. This is "very" expensive.
It does not seem significant but it really is. Since you know the max elements that will end up in the map (worst case is len(nums)), you can create a map with a big-enough capacity to hold all elements for the worst case. The gain will be that no internal growing and rehashing will be required. You can provide the initial capacity as the second argument to make() when creating the map. This speeds up twoSum2() big time if nums is big:
func twoSum2(nums []int, target int) (int, int) {
if len(nums) <= 1 {
return -1, -1
}
m := make(map[int]int, len(nums))
for i, v := range nums {
if j, ok := m[v]; ok {
return j, i
}
m[target-v] = i
}
return -1, -1
}
Benchmarking
Here's a little benchmarking code to test execution speed of the 3 solutions with the input nums and target you provided. Note that in order to test twoSumSorted(), you first have to sort the nums slice.
Save this into a file named xx_test.go and run it with go test -bench .:
package main
import (
"sort"
"testing"
)
func BenchmarkTwoSum(b *testing.B) {
for i := 0; i < b.N; i++ {
twoSum(nums, target)
}
}
func BenchmarkTwoSum2(b *testing.B) {
for i := 0; i < b.N; i++ {
twoSum2(nums, target)
}
}
func BenchmarkTwoSumSorted(b *testing.B) {
sort.Ints(nums)
b.ResetTimer()
for i := 0; i < b.N; i++ {
twoSumSorted(nums, target)
}
}
Output:
BenchmarkTwoSum-4 1000 1405542 ns/op
BenchmarkTwoSum2-4 2000 722661 ns/op
BenchmarkTwoSumSorted-4 10000000 133 ns/op
As you can see, making a map with big enough capacity speeds up: it runs twice as fast.
And as mentioned, if nums can be sorted in advance, that is ~10,000 times faster!
If nums is always sorted, you can do a binary search to see if the complement to whichever number you're on is also in the slice.
func binary(haystack []int, needle, startsAt int) int {
pivot := len(haystack) / 2
switch {
case haystack[pivot] == needle:
return pivot + startsAt
case len(haystack) <= 1:
return -1
case needle > haystack[pivot]:
return binary(haystack[pivot+1:], needle, startsAt+pivot+1)
case needle < haystack[pivot]:
return binary(haystack[:pivot], needle, startsAt)
}
return -1 // code can never fall off here, but the compiler complains
// if you don't have any returns out of conditionals.
}
func twoSum(nums []int, target int) (int, int) {
for i, num := range nums {
adjusted := target - num
if j := binary(nums, adjusted, 0); j != -1 {
return i, j
}
}
return 0, 0
}
playground example
Or you can use sort.SearchInts which implements binary searching.
func twoSum(nums []int, target int) (int, int) {
for i, num := range nums {
adjusted := target - num
if j := sort.SearchInts(nums, adjusted); nums[j] == adjusted {
// sort.SearchInts returns the index where the searched number
// would be if it was there. If it's not, then nums[j] != adjusted.
return i, j
}
}
return 0, 0
}
I imported the math library in my program, and I was trying to find the minimum of three numbers in the following way:
v1[j+1] = math.Min(v1[j]+1, math.Min(v0[j+1]+1, v0[j]+cost))
where v1 is declared as:
t := "stackoverflow"
v1 := make([]int, len(t)+1)
However, when I run my program I get the following error:
./levenshtein_distance.go:36: cannot use int(v0[j + 1] + 1) (type int) as type float64 in argument to math.Min
I thought it was weird because I have another program where I write
fmt.Println(math.Min(2,3))
and that program outputs 2 without complaining.
so I ended up casting the values as float64, so that math.Min could work:
v1[j+1] = math.Min(float64(v1[j]+1), math.Min(float64(v0[j+1]+1), float64(v0[j]+cost)))
With this approach, I got the following error:
./levenshtein_distance.go:36: cannot use math.Min(int(v1[j] + 1), math.Min(int(v0[j + 1] + 1), int(v0[j] + cost))) (type float64) as type int in assignment
so to get rid of the problem, I just casted the result back to int
I thought this was extremely inefficient and hard to read:
v1[j+1] = int(math.Min(float64(v1[j]+1), math.Min(float64(v0[j+1]+1), float64(v0[j]+cost))))
I also wrote a small minInt function, but I think this should be unnecessary because the other programs that make use of math.Min work just fine when taking integers, so I concluded this has to be a problem of my program and not the library per se.
Is there anything that I'm doing terrible wrong?
Here's a program that you can use to reproduce the issues above, line 36 specifically:
package main
import (
"math"
)
func main() {
LevenshteinDistance("stackoverflow", "stackexchange")
}
func LevenshteinDistance(s string, t string) int {
if s == t {
return 0
}
if len(s) == 0 {
return len(t)
}
if len(t) == 0 {
return len(s)
}
v0 := make([]int, len(t)+1)
v1 := make([]int, len(t)+1)
for i := 0; i < len(v0); i++ {
v0[i] = i
}
for i := 0; i < len(s); i++ {
v1[0] = i + 1
for j := 0; j < len(t); j++ {
cost := 0
if s[i] != t[j] {
cost = 1
}
v1[j+1] = int(math.Min(float64(v1[j]+1), math.Min(float64(v0[j+1]+1), float64(v0[j]+cost))))
}
for j := 0; j < len(v0); j++ {
v0[j] = v1[j]
}
}
return v1[len(t)]
}
Until Go 1.18 a one-off function was the standard way; for example, the stdlib's sort.go does it near the top of the file:
func min(a, b int) int {
if a < b {
return a
}
return b
}
You might still want or need to use this approach so your code works on Go versions below 1.18!
Starting with Go 1.18, you can write a generic min function which is just as efficient at run time as the hand-coded single-type version, but works with any type with < and > operators:
func min[T constraints.Ordered](a, b T) T {
if a < b {
return a
}
return b
}
func main() {
fmt.Println(min(1, 2))
fmt.Println(min(1.5, 2.5))
fmt.Println(min("Hello", "世界"))
}
There's been discussion of updating the stdlib to add generic versions of existing functions, but if that happens it won't be until a later version.
math.Min(2, 3) happened to work because numeric constants in Go are untyped. Beware of treating float64s as a universal number type in general, though, since integers above 2^53 will get rounded if converted to float64.
There is no built-in min or max function for integers, but it’s simple to write your own. Thanks to support for variadic functions we can even compare more integers with just one call:
func MinOf(vars ...int) int {
min := vars[0]
for _, i := range vars {
if min > i {
min = i
}
}
return min
}
Usage:
MinOf(3, 9, 6, 2)
Similarly here is the max function:
func MaxOf(vars ...int) int {
max := vars[0]
for _, i := range vars {
if max < i {
max = i
}
}
return max
}
For example,
package main
import "fmt"
func min(x, y int) int {
if x < y {
return x
}
return y
}
func main() {
t := "stackoverflow"
v0 := make([]int, len(t)+1)
v1 := make([]int, len(t)+1)
cost := 1
j := 0
v1[j+1] = min(v1[j]+1, min(v0[j+1]+1, v0[j]+cost))
fmt.Println(v1[j+1])
}
Output:
1
Though the question is quite old, maybe my package imath can be helpful for someone who does not like reinventing a bicycle. There are few functions, finding minimal of two integers: ix.Min (for int), i8.Min (for int8), ux.Min (for uint) and so on. The package can be obtained with go get, imported in your project by URL and functions referred as typeabbreviation.FuncName, for example:
package main
import (
"fmt"
"<Full URL>/go-imath/ix"
)
func main() {
a, b := 45, -42
fmt.Println(ix.Min(a, b)) // Output: -42
}
As the accepted answer states, with the introduction of generics in go 1.18 it's now possible to write a generic function that provides min/max for different numeric types (there is not one built into the language). And with variadic arguments we can support comparing 2 elements or a longer list of elements.
func Min[T constraints.Ordered](args ...T) T {
min := args[0]
for _, x := range args {
if x < min {
min = x
}
}
return min
}
func Max[T constraints.Ordered](args ...T) T {
max := args[0]
for _, x := range args {
if x > max {
max = x
}
}
return max
}
example calls:
Max(1, 2) // 2
Max(4, 5, 3, 1, 2) // 5
Could use https://github.com/pkg/math:
import (
"fmt"
"github.com/pkg/math"
)
func main() {
a, b := 45, -42
fmt.Println(math.Min(a, b)) // Output: -42
}
Since the issue has already been resolved, I would like to add a few words. Always remember that the math package in Golang operates on float64. You can use type conversion to cast int into a float64. Keep in mind to account for type ranges. For example, you cannot fit a float64 into an int16 if the number exceeds the limit for int16 which is 32767. Last but not least, if you convert a float into an int in Golang, the decimal points get truncated without any rounding.
If you want the minimum of a set of N integers you can use (assuming N > 0):
import "sort"
func min(set []int) int {
sort.Slice(set, func(i, j int) bool {
return set[i] < set[j]
})
return set[0]
}
Where the second argument to min function is your less function, that is, the function that decides when an element i of the passed slice is less than an element j
Check it out here in Go Playground: https://go.dev/play/p/lyQYlkwKrsA
Is there a way to check slices/maps for the presence of a value?
I would like to add a value to a slice only if it does not exist in the slice.
This works, but it seems verbose. Is there a better way to do this?
orgSlice := []int{1, 2, 3}
newSlice := []int{}
newInt := 2
newSlice = append(newSlice, newInt)
for _, v := range orgSlice {
if v != newInt {
newSlice = append(newSlice, v)
}
}
newSlice == [2 1 3]
Your approach would take linear time for each insertion. A better way would be to use a map[int]struct{}. Alternatively, you could also use a map[int]bool or something similar, but the empty struct{} has the advantage that it doesn't occupy any additional space. Therefore map[int]struct{} is a popular choice for a set of integers.
Example:
set := make(map[int]struct{})
set[1] = struct{}{}
set[2] = struct{}{}
set[1] = struct{}{}
// ...
for key := range(set) {
fmt.Println(key)
}
// each value will be printed only once, in no particular order
// you can use the ,ok idiom to check for existing keys
if _, ok := set[1]; ok {
fmt.Println("element found")
} else {
fmt.Println("element not found")
}
Most efficient is likely to be iterating over the slice and appending if you don't find it.
func AppendIfMissing(slice []int, i int) []int {
for _, ele := range slice {
if ele == i {
return slice
}
}
return append(slice, i)
}
It's simple and obvious and will be fast for small lists.
Further, it will always be faster than your current map-based solution. The map-based solution iterates over the whole slice no matter what; this solution returns immediately when it finds that the new value is already present. Both solutions compare elements as they iterate. (Each map assignment statement certainly does at least one map key comparison internally.) A map would only be useful if you could maintain it across many insertions. If you rebuild it on every insertion, then all advantage is lost.
If you truly need to efficiently handle large lists, consider maintaining the lists in sorted order. (I suspect the order doesn't matter to you because your first solution appended at the beginning of the list and your latest solution appends at the end.) If you always keep the lists sorted then you you can use the sort.Search function to do efficient binary insertions.
Another option:
package main
import "golang.org/x/tools/container/intsets"
func main() {
var (
a intsets.Sparse
b bool
)
b = a.Insert(9)
println(b) // true
b = a.Insert(9)
println(b) // false
}
https://pkg.go.dev/golang.org/x/tools/container/intsets
This option if the number of missing numbers is unknown
AppendIfMissing := func(sl []int, n ...int) []int {
cache := make(map[int]int)
for _, elem := range sl {
cache[elem] = elem
}
for _, elem := range n {
if _, ok := cache[elem]; !ok {
sl = append(sl, elem)
}
}
return sl
}
distincting a array of a struct :
func distinctObjects(objs []ObjectType) (distinctedObjs [] ObjectType){
var output []ObjectType
for i:= range objs{
if output==nil || len(output)==0{
output=append(output,objs[i])
} else {
founded:=false
for j:= range output{
if output[j].fieldname1==objs[i].fieldname1 && output[j].fieldname2==objs[i].fieldname2 &&......... {
founded=true
}
}
if !founded{
output=append(output,objs[i])
}
}
}
return output
}
where the struct here is something like :
type ObjectType struct {
fieldname1 string
fieldname2 string
.........
}
the object will distinct by checked fields here :
if output[j].fieldname1==objs[i].fieldname1 && output[j].fieldname2==objs[i].fieldname2 &&......... {
How would you implement the deleteRecords function in the code below:
Example:
type Record struct {
id int
name string
}
type RecordList []*Record
func deleteRecords( l *RecordList, ids []int ) {
// Assume the RecordList can contain several 100 entries.
// and the number of the of the records to be removed is about 10.
// What is the fastest and cleanest ways to remove the records that match
// the id specified in the records list.
}
I did some micro-benchmarking on my machine, trying out most of the approaches given in the replies here, and this code comes out fastest when you've got up to about 40 elements in the ids list:
func deleteRecords(data []*Record, ids []int) []*Record {
w := 0 // write index
loop:
for _, x := range data {
for _, id := range ids {
if id == x.id {
continue loop
}
}
data[w] = x
w++
}
return data[:w]
}
You didn't say whether it's important to preserve the order of records in the list. If you don't then this function is faster than the above and still fairly clean.
func reorder(data []*Record, ids []int) []*Record {
n := len(data)
i := 0
loop:
for i < n {
r := data[i]
for _, id := range ids {
if id == r.id {
data[i] = data[n-1]
n--
continue loop
}
}
i++
}
return data[0:n]
}
As the number of ids rises, so does the cost of the linear search. At around 50 elements, using a map or doing a binary search to look up the id becomes more efficient, as long as you can avoid rebuilding the map (or resorting the list) every time. At several hundred ids, it becomes more efficient to use a map or a binary search even if you have to rebuild it every time.
If you wish to preserve original contents of the slice, something like this is more appropriate:
func deletePreserve(data []*Record, ids []int) []*Record {
wdata := make([]*Record, len(data))
w := 0
loop:
for _, x := range data {
for _, id := range ids {
if id == x.id {
continue loop
}
}
wdata[w] = x
w++
}
return wdata[0:w]
}
For a personal project, I did something like this:
func filter(sl []int, fn func(int) bool) []int {
result := make([]int, 0, len(sl))
last := 0
for i, v := range sl {
if fn(v) {
result = append(result, sl[last:i]...)
last = i + 1
}
}
return append(result, sl[last:]...)
}
It doesn't mutate the original, but should be relatively efficient.
It's probably better to just do:
func filter(sl []int, fn func(int) bool) (result []int) {
for _, v := range sl {
if !fn(v) {
result = append(result, v)
}
}
return
}
Simpler and cleaner.
If you want to do it in-place, you probably want something like:
func filter(sl []int, fn func(int) bool) []int {
outi := 0
res := sl
for _, v := range sl {
if !fn(v) {
res[outi] = v
outi++
}
}
return res[0:outi]
}
You can optimize this to use copy to copy ranges of elements, but that's twice
the code and probably not worth it.
So, in this specific case, I'd probably do something like:
func deleteRecords(l []*Record, ids []int) []*Record {
outi := 0
L:
for _, v := range l {
for _, id := range ids {
if v.id == id {
continue L
}
}
l[outi] = v
outi++
}
return l[0:outi]
}
(Note: untested.)
No allocations, nothing fancy, and assuming the rough size of the list of Records and the list of ids you presented, a simple linear search is likely to do as well as fancier things but without any overhead. I realize that my version mutates the slice and returns a new slice, but that's not un-idiomatic in Go, and it avoids forcing the slice at the callsite to be heap allocated.
For the case you described, where len(ids) is approximately 10 and len(*l) is in the several hundreds, this should be relatively fast, since it minimizes memory allocations by updating in place.
package main
import (
"fmt"
"strconv"
)
type Record struct {
id int
name string
}
type RecordList []*Record
func deleteRecords(l *RecordList, ids []int) {
rl := *l
for i := 0; i < len(rl); i++ {
rid := rl[i].id
for j := 0; j < len(ids); j++ {
if rid == ids[j] {
copy(rl[i:len(*l)-1], rl[i+1:])
rl[len(rl)-1] = nil
rl = rl[:len(rl)-1]
break
}
}
}
*l = rl
}
func main() {
l := make(RecordList, 777)
for i := range l {
l[i] = &Record{int(i), "name #" + strconv.Itoa(i)}
}
ids := []int{0, 1, 2, 4, 8, len(l) - 1, len(l)}
fmt.Println(ids, len(l), cap(l), *l[0], *l[1], *l[len(l)-1])
deleteRecords(&l, ids)
fmt.Println(ids, len(l), cap(l), *l[0], *l[1], *l[len(l)-1])
}
Output:
[0 1 2 4 8 776 777] 777 777 {0 name #0} {1 name #1} {776 name #776}
[0 1 2 4 8 776 777] 772 777 {1 name #1} {3 name #3} {775 name #775}
Instead of repeatedly searching ids, you could use a map. This code preallocates the full size of the map, and then just moves array elements in place. There are no other allocations.
func deleteRecords(l *RecordList, ids []int) {
m := make(map[int]bool, len(ids))
for _, id := range ids {
m[id] = true
}
s, x := *l, 0
for _, r := range s {
if !m[r.id] {
s[x] = r
x++
}
}
*l = s[0:x]
}
Use the vector package's Delete method as a guide, or just use a Vector instead of a slice.
Here is one option but I would hope there are cleaner/faster more functional looking ones:
func deleteRecords( l *RecordList, ids []int ) *RecordList {
var newList RecordList
for _, rec := range l {
toRemove := false
for _, id := range ids {
if rec.id == id {
toRemove = true
}
if !toRemove {
newList = append(newList, rec)
}
}
return newList
}
With large enough l and ids it will be more effective to Sort() both lists first and then do a single loop over them instead of two nested loops