Golang Check for existing items in slice of structs

user_2882913 0
Viewed 726

Let's say I've got the following structs:

type A struct {
    Field1 string
    Field2 string
}

type B struct {
    Field3 string
    Field4 int
}

and one more struct which uses the previous two:

type C struct {
    A
    MyList []B
}

Now, I've got some data that I want to group and map into a C struct and return a slice of C:

var results []C

I've got a slice of structs that looks like (since it's a slice if could happen that we have repeated A):

type X struct {
    A
    B
}

So I want to group the results by A. To do so I'm gonna iterate through a slice of X that my method receive as parameter:

var results []C

// Y is an slice of Xs 

for _, elem := range Y {
    // If elem.A exists then append elem.B into C.A
    // If not then add elem.A and elem.B into C
}

How can achieve the stuff stated in the comments above? I mean, how could I check if a struct A exists already in the slice of structs C?

goland go slice struct
1 Answers

Usually when you want to aggregate something by some value, you use a map. The key will be the value you want to aggregate by, type of A in your case.

This is easiest here too, this only requires the key type to be comparable.

You may simply gather the data like this:

// Example data:
xs := []X{
    {A{"a1", "a2"}, B{"b1", 2}},
    {A{"a1", "a2"}, B{"b11", 22}},
    {A{"a1", "a3"}, B{"b4", 5}},
}
m := map[A][]B{}

for _, x := range xs {
    m[x.A] = append(m[x.A], x.B)
}

fmt.Println(m)

This will output:

map[{a1 a2}:[{b1 2} {b11 22}] {a1 a3}:[{b4 5}]]

If you do need the result as []C, iterate over the map and populate it:

for a, bs := range m {
    results = append(results, C{a, bs})
}
fmt.Println(results)

This will output:

[{{a1 a2} [{b1 2} {b11 22}]} {{a1 a3} [{b4 5}]}]

Try the examples on the Go Playground.

If by any chance A wouldn't be comparable, you would have to resort to using loops instead of map lookups where you would have to manually check for equivalence, and if you find a match, you'd perform the append operation.

user_17055980
Related