Associative array/Merging
- Task
Define two associative arrays, where one represents the following "base" data:
| Key | Value |
| "name" | "Rocket Skates" |
| "price" | 12.75 |
| "color" | "yellow" |
And the other represents "update" data:
| Key | Value |
| "price" | 15.25 |
| "color" | "red" |
| "year" | 1974 |
Merge these into a new associative array that contains every key found in either of the source ones. Each key should map to the value in the second (update) table if that exists, or else to the value in the first (base) table. If possible, do this in a way that does not mutate the original two associative arrays. Obviously this should be done in a way that would work for any data, not just the specific data given here, but in this example the result should be:
| Key | Value |
| "name" | "Rocket Skates" |
| "price" | 15.25 |
| "color" | "red" |
| "year" | 1974 |
Factor
The assoc-union word does this. assoc-union! is a variant that mutates the first associative array.
<lang factor>USING: assocs prettyprint ;
{ { "name" "Rocket Skates" } { "price" 12.75 } { "color" "yellow" } } { { "price" 15.25 } { "color" "red" } { "year" 1974 } } assoc-union .</lang>
- Output:
V{
{ "name" "Rocket Skates" }
{ "price" 15.25 }
{ "color" "red" }
{ "year" 1974 }
}
Go
<lang go>package main
import "fmt"
type assoc map[string]interface{}
func merge(base, update assoc) assoc {
result := make(assoc)
for k, v := range base {
result[k] = v
}
for k, v := range update {
result[k] = v
}
return result
}
func main() {
base := assoc{"name": "Rocket Skates", "price": 12.75, "color": "yellow"}
update := assoc{"price": 15.25, "color": "red", "year": 1974}
result := merge(base, update)
fmt.Println(result)
}</lang>
- Output:
map[color:red name:Rocket Skates price:15.25 year:1974]
JavaScript
<lang javascript>(() => {
'use strict';
console.log(JSON.stringify(
Object.assign({}, // Fresh dictionary.
{ // Base.
"name": "Rocket Skates",
"price": 12.75,
"color": "yellow"
}, { // Update.
"price": 15.25,
"color": "red",
"year": 1974
}
),
null, 2
))
})();</lang>
- Output:
{
"name": "Rocket Skates",
"price": 15.25,
"color": "red",
"year": 1974
}
Lua
<lang Lua>base = {name="Rocket Skates", price=12.75, color="yellow"} update = {price=15.25, color="red", year=1974}
--clone the base data -- result = {} for key,val in pairs(base) do
result[key] = val
end
--copy in the update data -- for key,val in pairs(update) do
result[key] = val
end
--print the result -- for key,val in pairs(result) do
print(string.format("%s: %s", key, val))
end</lang>
- Output:
price: 15.25 color: red year: 1974 name: Rocket Skates
MiniScript
MiniScript supports merging maps with the + operator. <lang MiniScript>base = {"name":"Rocket Skates", "price":12.75, "color":"yellow"} update = {"price":15.25, "color":"red", "year":1974}
result = base + update
print result</lang>
- Output:
{"color": "red", "name": "Rocket Skates", "price": 15.25, "year": 1974}
Perl 6
I must say I somewhat disagree with the terminology. The requested operation is an update not a merge. Demonstrate both an update and a merge. Associative arrays are commonly called hashes in Perl 6.
<lang perl6># Show original hashes say my %base = :name('Rocket Skates'), :price<12.75>, :color<yellow>; say my %update = :price<15.25>, :color<red>, :year<1974>;
- Need to assign to anonymous hash to get the desired results and avoid mutating
- TIMTOWTDI
say "\nUpdate:\n", join "\n", sort %=%base, %update;
- Same
say "\nUpdate:\n", {%base, %update}.sort.join: "\n";
say "\nMerge:\n", join "\n", sort ((%=%base).push: %update)».join: ', ';
- Same
say "\nMerge:\n", ({%base}.push: %update)».join(', ').sort.join: "\n";
- Demonstrate unmutated hashes
say "\n", %base, "\n", %update;</lang>
- Output:
{color => yellow, name => Rocket Skates, price => 12.75}
{color => red, price => 15.25, year => 1974}
Update:
color red
name Rocket Skates
price 15.25
year 1974
Update:
color red
name Rocket Skates
price 15.25
year 1974
Merge:
color yellow, red
name Rocket Skates
price 12.75, 15.25
year 1974
Merge:
color yellow, red
name Rocket Skates
price 12.75, 15.25
year 1974
{color => yellow, name => Rocket Skates, price => 12.75}
{color => red, price => 15.25, year => 1974}
Python
As of Python 3.5, this can be solved with the dictionary unpacking operator. <lang Python>base = {"name":"Rocket Skates", "price":12.75, "color":"yellow"} update = {"price":15.25, "color":"red", "year":1974}
result = {**base, **update}
print(result)</lang>
- Output:
{'name': 'Rocket Skates', 'price': 15.25, 'color': 'red', 'year': 1974}
REXX
The REXX language has no native support to acquire the keys of an associate array, so it must be performed at definition time. <lang rexx>/*REXX program merges two associative arrays (requiring an external list of indices). */ $.= /*define default value(s) for arrays. */ call defAA 'base', "name Rocket Skates", 'price 12.75', "color yellow" call defAA 'update', "price 15.25", "color red", 'year 1974' call show 'base' call show 'update' call show 'new' exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ defAA: procedure expose $.; parse arg AAn; new= 'new'
do j=2 to arg(); parse value arg(j) with key val
$.AAn.key=val; if wordpos(key, $.AAN.?keys)==0 then $.AAn.?keys=$.AAn.?keys key
$.new.key=val; if wordpos(key, $.new.?keys)==0 then $.new.?keys=$.new.?keys key
end /*j*/
return
/*──────────────────────────────────────────────────────────────────────────────────────*/ show: procedure expose $.; parse arg AAn; say; _= '═'
do j=1 for words($.AAn.?keys)
if j==1 then say center('associate array', 21, _) center('key' , 9, _),
center('value', 13, _)
key= word($.AAn.?keys, j); say center(AAn, 21) right(key, 9) $.AAn.key
end /*j*/
return</lang>
- output when using the internal default inputs:
═══associate array═══ ═══key═══ ════value════
base name Rocket Skates
base price 12.75
base color yellow
═══associate array═══ ═══key═══ ════value════
update price 15.25
update color red
update year 1974
═══associate array═══ ═══key═══ ════value════
new name Rocket Skates
new price 15.25
new color red
new year 1974
zkl
<lang zkl>base:=Dictionary(
"name", "Rocket Skates", "price", 12.75, "color", "yellow",);
update:=Dictionary(
"price", 15.25, "color", "red", "year", 1974,);
update.pump( new:=base.copy() );
new.pump(Void,fcn([(k,v)]){ println("%s\t%s".fmt(k,v)) });</lang>
- Output:
price 15.25 color red year 1974 name Rocket Skates