Group and concatenate a 2d array by a 1d array

Question

I have the following arrays:

array_order = [ 'SiteA', 'SiteC', 'SiteB' ]

string_array = [
                [ 'Site C', 'StringC1_' ] ,
                [ 'Site A', 'StringA1_' ] ,
                [ 'Site C', 'StringC2_' ] ,
                [ 'Site B', 'StringB1_' ] ,
                [ 'Site B', 'StringB2_' ] ,
                [ 'Site B', 'StringC3_' ] ,
                [ 'Site B', 'StringA2_' ]
               ]

I want to group the strings by site and output them as a concatenated string in order of array_order, so something like this:

output = [ 'StringA1_StringA2_', 'StringC1_StringC2_StringC3_', 'StringB1_StringB2_' ]

The concatenation order doesn't matter, for example 'StringA1StringA2' is just as fine as 'StringA2StringA1' so I don't need to worry about sorting them.

I was initially thinking I could convert string_array to an object, then somehow concatenate by group, then map it something like array_order.map(key => obj[key]) but I am struggling with making an object to be iterated and concatenated.

Is StringC3_ supposed to be under "Site B"?, as you grouping by the letter C found in the string or by the "Site"? — Nick Parsons
– Nick Parsons, Commented Jun 25, 2021 at 11:28
In array_order you do not have space in group name, but in string_array you do — Yuriy Yakym
– Yuriy Yakym, Commented Jun 25, 2021 at 11:34

Sergio Belevskij · Accepted Answer · 2021-06-25 11:46:00Z

2

Yet another way:

const siteOrder = ['Site A', 'Site C', 'Site B']

const strings = [
  ['Site C', 'StringC1_'],
  ['Site A', 'StringA1_'],
  ['Site C', 'StringC2_'],
  ['Site B', 'StringB1_'],
  ['Site B', 'StringB2_'],
  ['Site B', 'StringC3_'],
  ['Site B', 'StringA2_'],
]

const result = siteOrder.reduce((acc, site) => {
  acc[site] = ''
  return acc
}, {})

strings.forEach((siteString) => {
  const [site, string] = siteString
  result[site] += string
})
console.log({ result })

answered Jun 25, 2021 at 11:46

Sergio Belevskij

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1 Comment

ikhvjs Over a year ago

@RajdeepDebnath, I think mine is cleaner, but this is a good answer too : )

ikhvjs · Accepted Answer · 2021-06-25 11:56:46Z

2

Just use simple map and reduce

array = ["Site A", "Site C", "Site B", "Site X"];

data = [
  ["Site C", "StringC1_"],
  ["Site A", "StringA1_"],
  ["Site C", "StringC2_"],
  ["Site B", "StringB1_"],
  ["Site B", "StringB2_"],
  ["Site B", "StringB3_"],
  ["Site B", "StringB4_"],
];

const output = array.map(e =>
  data.reduce((a, b) => (a += b[0] === e ? b[1] : ""), "")
);

console.log(output);

answered Jun 25, 2021 at 11:56

ikhvjs

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1 Comment

Rajdeep D Over a year ago

This one is amazing mate :)

Terry Lennox · Accepted Answer · 2021-06-25 12:02:00Z

1

You can use Array.reduce on string_array to build your output array, we'll find the correct output index from running Array.findIndex on the array_order input array.

If we don't find the item, we'll skip adding it to the output:

array_order = [ 'SiteA', 'SiteX', 'SiteC', 'SiteB']

string_array = [
                [ 'SiteC', 'StringC1_' ] ,
                [ 'SiteA', 'StringA1_' ] ,
                [ 'SiteC', 'StringC2_' ] ,
                [ 'SiteB', 'StringB1_' ] ,
                [ 'SiteB', 'StringB2_' ] ,
                [ 'SiteC', 'StringC3_' ] ,
                [ 'SiteA', 'StringA2_' ]
               ]
               
 let result = string_array.reduce((acc, [site, value], arr) => { 
      let index = array_order.findIndex(el => el === site);
      if (index >= 0) { 
         acc[index] = (acc[index] || "") + value;
      }
      return acc;
 }, array_order.map(el => ""))
 
 
 console.log('Result:', result);

And another way of doing this with Array.map, Array.filter and Array.join:

array_order = [ 'SiteA', 'SiteX', 'SiteC', 'SiteB']

string_array = [
                [ 'SiteC', 'StringC1_' ] ,
                [ 'SiteA', 'StringA1_' ] ,
                [ 'SiteC', 'StringC2_' ] ,
                [ 'SiteB', 'StringB1_' ] ,
                [ 'SiteB', 'StringB2_' ] ,
                [ 'SiteC', 'StringC3_' ] ,
                [ 'SiteA', 'StringA2_' ]
               ]
     
let result = array_order
 .map(el => string_array.filter(([site, val]) => site === el)
 .map(([site, val]) => val)
 .join(""));
 
console.log("Result:",result);

edited Jun 25, 2021 at 12:02

answered Jun 25, 2021 at 11:43

Terry Lennox

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2 Comments

redditor Over a year ago

how would I edit this to have a null value if there is are no matches, rather than skip it? eg Result: [ "StringA1_StringA2_", "", "StringB1_StringB2_"]

Terry Lennox Over a year ago

I've updated to leave an empty string if no matches are found.

Yuriy Yakym · Accepted Answer · 2021-06-25 11:52:36Z

1

First yo need to group the items, then you can map each group concatenating the strings:

const grouped_arrays = string_array.reduce((result, entry) => {
  const [groupName, value] = entry;
  result[groupName] = result[groupName] || [];
  result[groupName].push(value);
  return result;
}, {});

const result = array_order.map(groupName => {
  return (grouped_arrays[groupName] || []).join('');
});

Also note that your array_order group names do not have space: 'Site C' !== 'SiteC'

edited Jun 25, 2021 at 11:52

answered Jun 25, 2021 at 11:27

Yuriy Yakym

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1 Comment

redditor Over a year ago

thanks, I made a couple of typos in the question. in real life, Site C is SiteC. will try shortly.

Rajdeep D · Accepted Answer · 2021-06-25 11:54:27Z

You can use reduce for this

array_order = [ 'Site A', 'Site C', 'Site B' ]

string_array = [
                [ 'Site C', 'StringC1_' ] ,
                [ 'Site A', 'StringA1_' ] ,
                [ 'Site C', 'StringC2_' ] ,
                [ 'Site B', 'StringB1_' ] ,
                [ 'Site B', 'StringB2_' ] ,
                [ 'Site B', 'StringC3_' ] ,
                [ 'Site B', 'StringA2_' ]
               ]
               
let result = [...string_array.reduce((acc,[k,v]) => {
                    acc.has(k)? acc.set(k, acc.get(k)+v) : acc.set(k,v);
                    return acc;
                 }
               , new Map())]
      .sort((a,b) => array_order.indexOf(a[0])-array_order.indexOf(b[0]))
      .map(i => i[1])

console.log(result);

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