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
Related
I'm trying to replicate this algorithm for finding duplicates in an array in Golang. Here's the javascript version:
function hasDuplicateValue(array) {
let existingNumbers = [];
for(let i = 0; i < array.length; i++) {
if(existingNumbers[array[i]] === 1) {
return true;
} else {
existingNumbers[array[i]] = 1;
}
}
return false;
}
On line 2, the algorithm creates an empty array of unknown length, and then adds 1 to an index in the array corresponding with each number that it finds (e.g. if it finds the number 3 in the array, it will add a 1 to index 3 in existing numbers.
I'm wondering — how do I replicate this in Golang (since we need to have slots allocated in the slice before reading it). Would I first need to find the max value in the array and then declare the existingNumbers slice to be of that same size?
Or is there a more efficient way of doing this (instead of searching through the array and finding the max value before constructing the slice).
Thanks!
Edit:
I realized that I can't do this with a slice because I can't read from an empty value. However, as #icza suggested, it will work with a map:
func findDuplicates(list []int)(bool) {
temp := make(map[int]int)
for _, elem := range list {
if temp[elem] == 1 {
return true
} else {
temp[elem] = 1
}
}
return false
}
As comments, I would also suggest using a map to keep the state of the duplications, but we can use map[int]struct{} because empty structs are not consumed any memory in Go.
And also I have simplified the code a bit and it is as follows.
func findDuplicates(list []int) bool {
temp := make(map[int]struct{})
for _, elem := range list {
if _, ok := temp[elem]; ok {
return true
}
temp[elem] = struct{}{}
}
return false
}
Full code can be executed here
I have to deal with a lot of maps with int keys which contain pointers to different datatypes.
I need a function (and not 10 functions for each map type) to range over those maps and get the maximum and minimum key values.
Use the reflect package to operate on maps with integer keys and arbitrary value types:
func getMaxKey(inout interface{}) int {
keys := reflect.ValueOf(inout).MapKeys()
if len(keys) == 0 {
return 0
}
max := keys[0].Int()
for _, key := range keys[1:] {
n := key.Int()
if n > max {
max = n
}
}
return int(max)
}
Run it on the playground.
This is what I came up with. It might also work for other map types:
https://play.golang.org/p/-T8s-bPCNm4
It allows to have passed in any map of type map[int]*somepointer
No type assertion needed (in this case)
-
func getMaxKey(inout interface{}) int {
auxMap:= make(map[int]string)
body, _ := json.Marshal(inout)
json.Unmarshal(body, &auxMap)
maxKey := 0
for key,_ := range auxMap {
if key > maxKey {
maxKey = key
}
}
return maxKey
}
func getMinKey(inout interface{}) int {
auxMap:= make(map[int]string)
body, _ := json.Marshal(inout)
json.Unmarshal(body, &auxMap)
minKey := 0
for key,_ := range auxMap {
if key < minKey || minKey == 0 {
minKey = key
}
}
return minKey
}
In the language there is a minimum function https://golang.org/pkg/math/#Min But what if I have more than 2 numbers? I must to write a manual comparison in a for loop, or is there another way? The numbers are in the slice.
No, there isn't any better way than looping. Not only is it cleaner than any other approach, it's also the fastest.
values := []int{4, 20, 0, -11, -10}
min := values[0]
for _, v := range values {
if (v < min) {
min = v
}
}
fmt.Println(min)
EDIT
Since there has been some discussion in the comments about error handling and how to handle empty slices, here is a basic function that determines the minimum value. Remember to import errors.
func Min(values []int) (min int, e error) {
if len(values) == 0 {
return 0, errors.New("Cannot detect a minimum value in an empty slice")
}
min = values[0]
for _, v := range values {
if (v < min) {
min = v
}
}
return min, nil
}
General answer is: "Yes, you must use a loop, if you do not know exact number of items to compare".
In this package Min functions are implemented like:
// For 2 values
func Min(value_0, value_1 int) int {
if value_0 < value_1 {
return value_0
}
return value_1
}
// For 1+ values
func Mins(value int, values ...int) int {
for _, v := range values {
if v < value {
value = v
}
}
return value
}
You should write a loop. It does not make sense to create dozens of function in standard library to find min/max/count/count_if/all_of/any_of/none_of etc. like in C++ (most of them in 4 flavours according arguments).
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