How would I write this reportCandidates' function better?
1
This is the input data in array named candidatesArray:
[
{"name":"george","languages":["php","javascript","java"],"age":19,"graduate_date":1044064800000,"phone":"32-991-511"},
{"name":"anna","languages":["java","javascript"],"age":23,"graduate_date":1391220000000,"phone":"32-991-512"},
{"name":"hailee","languages":["regex","javascript","perl","go","java"],"age":31,"graduate_date":1296525600000,"phone":"32-991-513"}
]
I need to transform in this collection as a result of the function:
{candidates: [
{name: "George", age: 19, phone: "32-991-511"},
{name: "Hailee", age: 31, phone: "32-991-513"},
{name: "Anna", age: 23, phone: "32-991-512"}
],
languages: [
{lang:"javascript",count:1},
{lang:"java", count:2},
{lang:"php", count:2},
{lang:"regex", count:1}
]}
The function repCandidates:
const reportCandidates = (candidatesArray) => {
return repObject}
- I need to write it in javascript ES6
- I shouldn't use loops(for, while, repeat) but foreach is allowed and it could be better if I use "reduce" function
- The candidates should be return by their name, age and phone organized by their graduate_date.
- The languages should be returned with their counter in alphabetic order .
Visit https://codepen.io/rillervincci/pen/NEyMoV?editors=0010 to see my code, please.
javascript arrays json sorting collections
edited Nov 22 '18 at 5:51
Mark Meyer
37.5k33159
asked Nov 22 '18 at 5:40
Riller VincciRiller Vincci
84
add a comment |
1
This is the input data in array named candidatesArray:
[
{"name":"george","languages":["php","javascript","java"],"age":19,"graduate_date":1044064800000,"phone":"32-991-511"},
{"name":"anna","languages":["java","javascript"],"age":23,"graduate_date":1391220000000,"phone":"32-991-512"},
{"name":"hailee","languages":["regex","javascript","perl","go","java"],"age":31,"graduate_date":1296525600000,"phone":"32-991-513"}
]
I need to transform in this collection as a result of the function:
{candidates: [
{name: "George", age: 19, phone: "32-991-511"},
{name: "Hailee", age: 31, phone: "32-991-513"},
{name: "Anna", age: 23, phone: "32-991-512"}
],
languages: [
{lang:"javascript",count:1},
{lang:"java", count:2},
{lang:"php", count:2},
{lang:"regex", count:1}
]}
The function repCandidates:
const reportCandidates = (candidatesArray) => {
return repObject}
- I need to write it in javascript ES6
- I shouldn't use loops(for, while, repeat) but foreach is allowed and it could be better if I use "reduce" function
- The candidates should be return by their name, age and phone organized by their graduate_date.
- The languages should be returned with their counter in alphabetic order .
Visit https://codepen.io/rillervincci/pen/NEyMoV?editors=0010 to see my code, please.
javascript arrays json sorting collections
edited Nov 22 '18 at 5:51
Mark Meyer
37.5k33159
asked Nov 22 '18 at 5:40
Riller VincciRiller Vincci
84
add a comment |
1
1
1
This is the input data in array named candidatesArray:
[
{"name":"george","languages":["php","javascript","java"],"age":19,"graduate_date":1044064800000,"phone":"32-991-511"},
{"name":"anna","languages":["java","javascript"],"age":23,"graduate_date":1391220000000,"phone":"32-991-512"},
{"name":"hailee","languages":["regex","javascript","perl","go","java"],"age":31,"graduate_date":1296525600000,"phone":"32-991-513"}
]
I need to transform in this collection as a result of the function:
{candidates: [
{name: "George", age: 19, phone: "32-991-511"},
{name: "Hailee", age: 31, phone: "32-991-513"},
{name: "Anna", age: 23, phone: "32-991-512"}
],
languages: [
{lang:"javascript",count:1},
{lang:"java", count:2},
{lang:"php", count:2},
{lang:"regex", count:1}
]}
The function repCandidates:
const reportCandidates = (candidatesArray) => {
return repObject}
- I need to write it in javascript ES6
- I shouldn't use loops(for, while, repeat) but foreach is allowed and it could be better if I use "reduce" function
- The candidates should be return by their name, age and phone organized by their graduate_date.
- The languages should be returned with their counter in alphabetic order .
Visit https://codepen.io/rillervincci/pen/NEyMoV?editors=0010 to see my code, please.
javascript arrays json sorting collections
edited Nov 22 '18 at 5:51
Mark Meyer
37.5k33159
asked Nov 22 '18 at 5:40
Riller VincciRiller Vincci
84
This is the input data in array named candidatesArray:
[
{"name":"george","languages":["php","javascript","java"],"age":19,"graduate_date":1044064800000,"phone":"32-991-511"},
{"name":"anna","languages":["java","javascript"],"age":23,"graduate_date":1391220000000,"phone":"32-991-512"},
{"name":"hailee","languages":["regex","javascript","perl","go","java"],"age":31,"graduate_date":1296525600000,"phone":"32-991-513"}
]
I need to transform in this collection as a result of the function:
{candidates: [
{name: "George", age: 19, phone: "32-991-511"},
{name: "Hailee", age: 31, phone: "32-991-513"},
{name: "Anna", age: 23, phone: "32-991-512"}
],
languages: [
{lang:"javascript",count:1},
{lang:"java", count:2},
{lang:"php", count:2},
{lang:"regex", count:1}
]}
The function repCandidates:
const reportCandidates = (candidatesArray) => {
return repObject}
- I need to write it in javascript ES6
- I shouldn't use loops(for, while, repeat) but foreach is allowed and it could be better if I use "reduce" function
- The candidates should be return by their name, age and phone organized by their graduate_date.
- The languages should be returned with their counter in alphabetic order .
Visit https://codepen.io/rillervincci/pen/NEyMoV?editors=0010 to see my code, please.
javascript arrays json sorting collections
javascript arrays json sorting collections
edited Nov 22 '18 at 5:51
Mark Meyer
37.5k33159
asked Nov 22 '18 at 5:40
Riller VincciRiller Vincci
84
edited Nov 22 '18 at 5:51
Mark Meyer
37.5k33159
asked Nov 22 '18 at 5:40
Riller VincciRiller Vincci
84
edited Nov 22 '18 at 5:51
Mark Meyer
37.5k33159
edited Nov 22 '18 at 5:51
Mark Meyer
37.5k33159
edited Nov 22 '18 at 5:51
Mark Meyer
37.5k33159
37.5k33159
asked Nov 22 '18 at 5:40
Riller VincciRiller Vincci
84
asked Nov 22 '18 at 5:40
Riller VincciRiller Vincci
84
asked Nov 22 '18 at 5:40
Riller VincciRiller Vincci
84
84
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
1
One option would be to first reduce into the candidates subobject, while pushing the langauges of each to an array.
After iterating, sort the candidates and remove the graduate_date property from each candidate, then use reduce again to transform the languages array into one indexed by language, incrementing the count property each time:
const input = [{
"name": "george",
"languages": ["php", "javascript", "java"],
"age": 19,
"graduate_date": 1044064800000,
"phone": "32-991-511"
}, {
"name": "anna",
"languages": ["java", "javascript"],
"age": 23,
"graduate_date": 1391220000000,
"phone": "32-991-512"
}, {
"name": "hailee",
"languages": ["regex", "javascript", "perl", "go", "java"],
"age": 31,
"graduate_date": 1296525600000,
"phone": "32-991-513"
}];
const output = input.reduce((a, { languages, ...rest }) => {
a.candidates.push(rest);
a.languages.push(...languages);
return a;
}, { candidates: , languages: });
output.candidates.sort((a, b) => a.graduate_date - b.graduate_date);
output.candidates.forEach(candidate => delete candidate.graduate_date);
output.languages = Object.values(
output.languages.reduce((a, lang) => {
if (!a[lang]) a[lang] = { lang, count: 0 };
a[lang].count++;
return a;
}, {})
);
output.languages.sort((a, b) => a.lang.localeCompare(b.lang));
console.log(output);answered Nov 22 '18 at 5:47
CertainPerformanceCertainPerformance
79.4k143865
Wow that's amazing! I would never imagine do it like that. Very good... So how could I order the candidates by graduate date in desc and languages in alphabetic order
– Riller Vincci
Nov 22 '18 at 5:59
Thecandidatesare already being sorted, see thesortfunction, to sort the languages by language, uselocaleCompareon theirlangproperties
– CertainPerformance
Nov 22 '18 at 6:02
Can I use this to make first letter of the names like lowercase? name = name.charAt(0).toUpperCase() + name.slice(1); a.candidates.push({ name, age, phone });
– Riller Vincci
Nov 22 '18 at 6:20
You want to make each first character uppercase? Yep, that code looks like it would work, but note that you can use bracket notation instead ofcharAtfor less syntax noise, egname[0].toUpperCase()
– CertainPerformance
Nov 22 '18 at 6:21
Bracket notation? that's new to me
– Riller Vincci
Nov 22 '18 at 6:24
|
show 2 more comments
0
It's common practice to do everything in a reduce(), but sometimes it's easier to read if you break it up a bit. This creates a counter object as a helper to to track the language counts. map()s over the array to pull out the languages and personal info and then puts it all together:
let arr = [ {"name":"george","languages":["php","javascript","java"],"age":19,"graduate_date":1044064800000,"phone":"32-991-511"},{"name":"anna","languages":["java","javascript"],"age":23,"graduate_date":1391220000000,"phone":"32-991-512"},{"name":"hailee","languages":["regex","javascript","perl","go","java"],"age":31,"graduate_date":1296525600000,"phone":"32-991-513"}]
let lang_counter = {
// helper keeps counts of unique items
counts:{},
add(arr){
arr.forEach(item => this.counts[item] = this.counts[item] ? this.counts[item] + 1 : 1)
},
toarray(){
return Object.entries(this.counts).map(([key, val]) => ({[key]: val}))
}
}
// iterate over object to create candidates
let candidates = arr.map(row => {
let {languages, ...person} = row
lang_counter.add(languages) // side effect
return person
})
// put them together
console.log({candidates, languages:lang_counter.toarray()})answered Nov 22 '18 at 6:14
Mark MeyerMark Meyer
37.5k33159
Yeaah That's nice bro. I've changed some things to make all names receive the first letter in uppercase and organize all languages in alphabetical order. Thank you!!!
– Riller Vincci
Nov 22 '18 at 6:40
add a comment |
0
You can use Array.reduce and Object.values like below
let arr = [{"name":"george","languages":["php","javascript","java"],"age":19,"graduate_date":1044064800000,"phone":"32-991-511"},{"name":"anna","languages":["java","javascript"],"age":23,"graduate_date":1391220000000,"phone":"32-991-512"},{"name":"hailee","languages":["regex","javascript","perl","go","java"],"age":31,"graduate_date":1296525600000,"phone":"32-991-513"}]
let res = arr.reduce((o, {name, age, phone, graduate_date, languages}) => {
o.candidates.push({name, age, phone, graduate_date})
languages.forEach(l => {
o.languages[l] = o.languages[l] || { lang:l, count: 0 }
o.languages[l].count++
})
return o
}
, { candidates: , languages: {}})
res.candidates = res.candidates.sort((a,b) => a.graduate_date - b.graduate_date)
.map(({ graduate_date, ...rest }) => rest)
res.languages = Object.values(res.languages).sort((a,b) => a.lang.localeCompare(b.lang))
console.log(res)answered Nov 22 '18 at 6:05
Nitish NarangNitish Narang
2,948815
1
It's very interesting You've used map function to organize candidates Woow Congratulations... Thank you
– Riller Vincci
Nov 22 '18 at 6:44
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
One option would be to first reduce into the candidates subobject, while pushing the langauges of each to an array.
After iterating, sort the candidates and remove the graduate_date property from each candidate, then use reduce again to transform the languages array into one indexed by language, incrementing the count property each time:
const input = [{
"name": "george",
"languages": ["php", "javascript", "java"],
"age": 19,
"graduate_date": 1044064800000,
"phone": "32-991-511"
}, {
"name": "anna",
"languages": ["java", "javascript"],
"age": 23,
"graduate_date": 1391220000000,
"phone": "32-991-512"
}, {
"name": "hailee",
"languages": ["regex", "javascript", "perl", "go", "java"],
"age": 31,
"graduate_date": 1296525600000,
"phone": "32-991-513"
}];
const output = input.reduce((a, { languages, ...rest }) => {
a.candidates.push(rest);
a.languages.push(...languages);
return a;
}, { candidates: , languages: });
output.candidates.sort((a, b) => a.graduate_date - b.graduate_date);
output.candidates.forEach(candidate => delete candidate.graduate_date);
output.languages = Object.values(
output.languages.reduce((a, lang) => {
if (!a[lang]) a[lang] = { lang, count: 0 };
a[lang].count++;
return a;
}, {})
);
output.languages.sort((a, b) => a.lang.localeCompare(b.lang));
console.log(output);answered Nov 22 '18 at 5:47
CertainPerformanceCertainPerformance
79.4k143865
Wow that's amazing! I would never imagine do it like that. Very good... So how could I order the candidates by graduate date in desc and languages in alphabetic order
– Riller Vincci
Nov 22 '18 at 5:59
Thecandidatesare already being sorted, see thesortfunction, to sort the languages by language, uselocaleCompareon theirlangproperties
– CertainPerformance
Nov 22 '18 at 6:02
Can I use this to make first letter of the names like lowercase? name = name.charAt(0).toUpperCase() + name.slice(1); a.candidates.push({ name, age, phone });
– Riller Vincci
Nov 22 '18 at 6:20
You want to make each first character uppercase? Yep, that code looks like it would work, but note that you can use bracket notation instead ofcharAtfor less syntax noise, egname[0].toUpperCase()
– CertainPerformance
Nov 22 '18 at 6:21
Bracket notation? that's new to me
– Riller Vincci
Nov 22 '18 at 6:24
|
show 2 more comments
1
One option would be to first reduce into the candidates subobject, while pushing the langauges of each to an array.
After iterating, sort the candidates and remove the graduate_date property from each candidate, then use reduce again to transform the languages array into one indexed by language, incrementing the count property each time:
const input = [{
"name": "george",
"languages": ["php", "javascript", "java"],
"age": 19,
"graduate_date": 1044064800000,
"phone": "32-991-511"
}, {
"name": "anna",
"languages": ["java", "javascript"],
"age": 23,
"graduate_date": 1391220000000,
"phone": "32-991-512"
}, {
"name": "hailee",
"languages": ["regex", "javascript", "perl", "go", "java"],
"age": 31,
"graduate_date": 1296525600000,
"phone": "32-991-513"
}];
const output = input.reduce((a, { languages, ...rest }) => {
a.candidates.push(rest);
a.languages.push(...languages);
return a;
}, { candidates: , languages: });
output.candidates.sort((a, b) => a.graduate_date - b.graduate_date);
output.candidates.forEach(candidate => delete candidate.graduate_date);
output.languages = Object.values(
output.languages.reduce((a, lang) => {
if (!a[lang]) a[lang] = { lang, count: 0 };
a[lang].count++;
return a;
}, {})
);
output.languages.sort((a, b) => a.lang.localeCompare(b.lang));
console.log(output);answered Nov 22 '18 at 5:47
CertainPerformanceCertainPerformance
79.4k143865
Wow that's amazing! I would never imagine do it like that. Very good... So how could I order the candidates by graduate date in desc and languages in alphabetic order
– Riller Vincci
Nov 22 '18 at 5:59
Thecandidatesare already being sorted, see thesortfunction, to sort the languages by language, uselocaleCompareon theirlangproperties
– CertainPerformance
Nov 22 '18 at 6:02
Can I use this to make first letter of the names like lowercase? name = name.charAt(0).toUpperCase() + name.slice(1); a.candidates.push({ name, age, phone });
– Riller Vincci
Nov 22 '18 at 6:20
You want to make each first character uppercase? Yep, that code looks like it would work, but note that you can use bracket notation instead ofcharAtfor less syntax noise, egname[0].toUpperCase()
– CertainPerformance
Nov 22 '18 at 6:21
Bracket notation? that's new to me
– Riller Vincci
Nov 22 '18 at 6:24
|
show 2 more comments
1
1
1
One option would be to first reduce into the candidates subobject, while pushing the langauges of each to an array.
After iterating, sort the candidates and remove the graduate_date property from each candidate, then use reduce again to transform the languages array into one indexed by language, incrementing the count property each time:
const input = [{
"name": "george",
"languages": ["php", "javascript", "java"],
"age": 19,
"graduate_date": 1044064800000,
"phone": "32-991-511"
}, {
"name": "anna",
"languages": ["java", "javascript"],
"age": 23,
"graduate_date": 1391220000000,
"phone": "32-991-512"
}, {
"name": "hailee",
"languages": ["regex", "javascript", "perl", "go", "java"],
"age": 31,
"graduate_date": 1296525600000,
"phone": "32-991-513"
}];
const output = input.reduce((a, { languages, ...rest }) => {
a.candidates.push(rest);
a.languages.push(...languages);
return a;
}, { candidates: , languages: });
output.candidates.sort((a, b) => a.graduate_date - b.graduate_date);
output.candidates.forEach(candidate => delete candidate.graduate_date);
output.languages = Object.values(
output.languages.reduce((a, lang) => {
if (!a[lang]) a[lang] = { lang, count: 0 };
a[lang].count++;
return a;
}, {})
);
output.languages.sort((a, b) => a.lang.localeCompare(b.lang));
console.log(output);answered Nov 22 '18 at 5:47
CertainPerformanceCertainPerformance
79.4k143865
One option would be to first reduce into the candidates subobject, while pushing the langauges of each to an array.
After iterating, sort the candidates and remove the graduate_date property from each candidate, then use reduce again to transform the languages array into one indexed by language, incrementing the count property each time:
const input = [{
"name": "george",
"languages": ["php", "javascript", "java"],
"age": 19,
"graduate_date": 1044064800000,
"phone": "32-991-511"
}, {
"name": "anna",
"languages": ["java", "javascript"],
"age": 23,
"graduate_date": 1391220000000,
"phone": "32-991-512"
}, {
"name": "hailee",
"languages": ["regex", "javascript", "perl", "go", "java"],
"age": 31,
"graduate_date": 1296525600000,
"phone": "32-991-513"
}];
const output = input.reduce((a, { languages, ...rest }) => {
a.candidates.push(rest);
a.languages.push(...languages);
return a;
}, { candidates: , languages: });
output.candidates.sort((a, b) => a.graduate_date - b.graduate_date);
output.candidates.forEach(candidate => delete candidate.graduate_date);
output.languages = Object.values(
output.languages.reduce((a, lang) => {
if (!a[lang]) a[lang] = { lang, count: 0 };
a[lang].count++;
return a;
}, {})
);
output.languages.sort((a, b) => a.lang.localeCompare(b.lang));
console.log(output);const input = [{
"name": "george",
"languages": ["php", "javascript", "java"],
"age": 19,
"graduate_date": 1044064800000,
"phone": "32-991-511"
}, {
"name": "anna",
"languages": ["java", "javascript"],
"age": 23,
"graduate_date": 1391220000000,
"phone": "32-991-512"
}, {
"name": "hailee",
"languages": ["regex", "javascript", "perl", "go", "java"],
"age": 31,
"graduate_date": 1296525600000,
"phone": "32-991-513"
}];
const output = input.reduce((a, { languages, ...rest }) => {
a.candidates.push(rest);
a.languages.push(...languages);
return a;
}, { candidates: , languages: });
output.candidates.sort((a, b) => a.graduate_date - b.graduate_date);
output.candidates.forEach(candidate => delete candidate.graduate_date);
output.languages = Object.values(
output.languages.reduce((a, lang) => {
if (!a[lang]) a[lang] = { lang, count: 0 };
a[lang].count++;
return a;
}, {})
);
output.languages.sort((a, b) => a.lang.localeCompare(b.lang));
console.log(output);const input = [{
"name": "george",
"languages": ["php", "javascript", "java"],
"age": 19,
"graduate_date": 1044064800000,
"phone": "32-991-511"
}, {
"name": "anna",
"languages": ["java", "javascript"],
"age": 23,
"graduate_date": 1391220000000,
"phone": "32-991-512"
}, {
"name": "hailee",
"languages": ["regex", "javascript", "perl", "go", "java"],
"age": 31,
"graduate_date": 1296525600000,
"phone": "32-991-513"
}];
const output = input.reduce((a, { languages, ...rest }) => {
a.candidates.push(rest);
a.languages.push(...languages);
return a;
}, { candidates: , languages: });
output.candidates.sort((a, b) => a.graduate_date - b.graduate_date);
output.candidates.forEach(candidate => delete candidate.graduate_date);
output.languages = Object.values(
output.languages.reduce((a, lang) => {
if (!a[lang]) a[lang] = { lang, count: 0 };
a[lang].count++;
return a;
}, {})
);
output.languages.sort((a, b) => a.lang.localeCompare(b.lang));
console.log(output);answered Nov 22 '18 at 5:47
CertainPerformanceCertainPerformance
79.4k143865
edited Nov 22 '18 at 6:02
answered Nov 22 '18 at 5:47
CertainPerformanceCertainPerformance
79.4k143865
answered Nov 22 '18 at 5:47
CertainPerformanceCertainPerformance
79.4k143865
answered Nov 22 '18 at 5:47
CertainPerformanceCertainPerformance
79.4k143865
79.4k143865
Wow that's amazing! I would never imagine do it like that. Very good... So how could I order the candidates by graduate date in desc and languages in alphabetic order
– Riller Vincci
Nov 22 '18 at 5:59
Thecandidatesare already being sorted, see thesortfunction, to sort the languages by language, uselocaleCompareon theirlangproperties
– CertainPerformance
Nov 22 '18 at 6:02
Can I use this to make first letter of the names like lowercase? name = name.charAt(0).toUpperCase() + name.slice(1); a.candidates.push({ name, age, phone });
– Riller Vincci
Nov 22 '18 at 6:20
You want to make each first character uppercase? Yep, that code looks like it would work, but note that you can use bracket notation instead ofcharAtfor less syntax noise, egname[0].toUpperCase()
– CertainPerformance
Nov 22 '18 at 6:21
Bracket notation? that's new to me
– Riller Vincci
Nov 22 '18 at 6:24
|
show 2 more comments
Wow that's amazing! I would never imagine do it like that. Very good... So how could I order the candidates by graduate date in desc and languages in alphabetic order
– Riller Vincci
Nov 22 '18 at 5:59
Thecandidatesare already being sorted, see thesortfunction, to sort the languages by language, uselocaleCompareon theirlangproperties
– CertainPerformance
Nov 22 '18 at 6:02
Can I use this to make first letter of the names like lowercase? name = name.charAt(0).toUpperCase() + name.slice(1); a.candidates.push({ name, age, phone });
– Riller Vincci
Nov 22 '18 at 6:20
You want to make each first character uppercase? Yep, that code looks like it would work, but note that you can use bracket notation instead ofcharAtfor less syntax noise, egname[0].toUpperCase()
– CertainPerformance
Nov 22 '18 at 6:21
Bracket notation? that's new to me
– Riller Vincci
Nov 22 '18 at 6:24
Wow that's amazing! I would never imagine do it like that. Very good... So how could I order the candidates by graduate date in desc and languages in alphabetic order
– Riller Vincci
Nov 22 '18 at 5:59
Wow that's amazing! I would never imagine do it like that. Very good... So how could I order the candidates by graduate date in desc and languages in alphabetic order
– Riller Vincci
Nov 22 '18 at 5:59
The
candidates are already being sorted, see the sort function, to sort the languages by language, use localeCompare on their lang properties– CertainPerformance
Nov 22 '18 at 6:02
The
candidates are already being sorted, see the sort function, to sort the languages by language, use localeCompare on their lang properties– CertainPerformance
Nov 22 '18 at 6:02
Can I use this to make first letter of the names like lowercase? name = name.charAt(0).toUpperCase() + name.slice(1); a.candidates.push({ name, age, phone });
– Riller Vincci
Nov 22 '18 at 6:20
Can I use this to make first letter of the names like lowercase? name = name.charAt(0).toUpperCase() + name.slice(1); a.candidates.push({ name, age, phone });
– Riller Vincci
Nov 22 '18 at 6:20
You want to make each first character uppercase? Yep, that code looks like it would work, but note that you can use bracket notation instead of
charAt for less syntax noise, eg name[0].toUpperCase()– CertainPerformance
Nov 22 '18 at 6:21
You want to make each first character uppercase? Yep, that code looks like it would work, but note that you can use bracket notation instead of
charAt for less syntax noise, eg name[0].toUpperCase()– CertainPerformance
Nov 22 '18 at 6:21
Bracket notation? that's new to me
– Riller Vincci
Nov 22 '18 at 6:24
Bracket notation? that's new to me
– Riller Vincci
Nov 22 '18 at 6:24
|
show 2 more comments
0
It's common practice to do everything in a reduce(), but sometimes it's easier to read if you break it up a bit. This creates a counter object as a helper to to track the language counts. map()s over the array to pull out the languages and personal info and then puts it all together:
let arr = [ {"name":"george","languages":["php","javascript","java"],"age":19,"graduate_date":1044064800000,"phone":"32-991-511"},{"name":"anna","languages":["java","javascript"],"age":23,"graduate_date":1391220000000,"phone":"32-991-512"},{"name":"hailee","languages":["regex","javascript","perl","go","java"],"age":31,"graduate_date":1296525600000,"phone":"32-991-513"}]
let lang_counter = {
// helper keeps counts of unique items
counts:{},
add(arr){
arr.forEach(item => this.counts[item] = this.counts[item] ? this.counts[item] + 1 : 1)
},
toarray(){
return Object.entries(this.counts).map(([key, val]) => ({[key]: val}))
}
}
// iterate over object to create candidates
let candidates = arr.map(row => {
let {languages, ...person} = row
lang_counter.add(languages) // side effect
return person
})
// put them together
console.log({candidates, languages:lang_counter.toarray()})answered Nov 22 '18 at 6:14
Mark MeyerMark Meyer
37.5k33159
Yeaah That's nice bro. I've changed some things to make all names receive the first letter in uppercase and organize all languages in alphabetical order. Thank you!!!
– Riller Vincci
Nov 22 '18 at 6:40
add a comment |
0
It's common practice to do everything in a reduce(), but sometimes it's easier to read if you break it up a bit. This creates a counter object as a helper to to track the language counts. map()s over the array to pull out the languages and personal info and then puts it all together:
let arr = [ {"name":"george","languages":["php","javascript","java"],"age":19,"graduate_date":1044064800000,"phone":"32-991-511"},{"name":"anna","languages":["java","javascript"],"age":23,"graduate_date":1391220000000,"phone":"32-991-512"},{"name":"hailee","languages":["regex","javascript","perl","go","java"],"age":31,"graduate_date":1296525600000,"phone":"32-991-513"}]
let lang_counter = {
// helper keeps counts of unique items
counts:{},
add(arr){
arr.forEach(item => this.counts[item] = this.counts[item] ? this.counts[item] + 1 : 1)
},
toarray(){
return Object.entries(this.counts).map(([key, val]) => ({[key]: val}))
}
}
// iterate over object to create candidates
let candidates = arr.map(row => {
let {languages, ...person} = row
lang_counter.add(languages) // side effect
return person
})
// put them together
console.log({candidates, languages:lang_counter.toarray()})answered Nov 22 '18 at 6:14
Mark MeyerMark Meyer
37.5k33159
Yeaah That's nice bro. I've changed some things to make all names receive the first letter in uppercase and organize all languages in alphabetical order. Thank you!!!
– Riller Vincci
Nov 22 '18 at 6:40
add a comment |
0
0
0
It's common practice to do everything in a reduce(), but sometimes it's easier to read if you break it up a bit. This creates a counter object as a helper to to track the language counts. map()s over the array to pull out the languages and personal info and then puts it all together:
let arr = [ {"name":"george","languages":["php","javascript","java"],"age":19,"graduate_date":1044064800000,"phone":"32-991-511"},{"name":"anna","languages":["java","javascript"],"age":23,"graduate_date":1391220000000,"phone":"32-991-512"},{"name":"hailee","languages":["regex","javascript","perl","go","java"],"age":31,"graduate_date":1296525600000,"phone":"32-991-513"}]
let lang_counter = {
// helper keeps counts of unique items
counts:{},
add(arr){
arr.forEach(item => this.counts[item] = this.counts[item] ? this.counts[item] + 1 : 1)
},
toarray(){
return Object.entries(this.counts).map(([key, val]) => ({[key]: val}))
}
}
// iterate over object to create candidates
let candidates = arr.map(row => {
let {languages, ...person} = row
lang_counter.add(languages) // side effect
return person
})
// put them together
console.log({candidates, languages:lang_counter.toarray()})answered Nov 22 '18 at 6:14
Mark MeyerMark Meyer
37.5k33159
It's common practice to do everything in a reduce(), but sometimes it's easier to read if you break it up a bit. This creates a counter object as a helper to to track the language counts. map()s over the array to pull out the languages and personal info and then puts it all together:
let arr = [ {"name":"george","languages":["php","javascript","java"],"age":19,"graduate_date":1044064800000,"phone":"32-991-511"},{"name":"anna","languages":["java","javascript"],"age":23,"graduate_date":1391220000000,"phone":"32-991-512"},{"name":"hailee","languages":["regex","javascript","perl","go","java"],"age":31,"graduate_date":1296525600000,"phone":"32-991-513"}]
let lang_counter = {
// helper keeps counts of unique items
counts:{},
add(arr){
arr.forEach(item => this.counts[item] = this.counts[item] ? this.counts[item] + 1 : 1)
},
toarray(){
return Object.entries(this.counts).map(([key, val]) => ({[key]: val}))
}
}
// iterate over object to create candidates
let candidates = arr.map(row => {
let {languages, ...person} = row
lang_counter.add(languages) // side effect
return person
})
// put them together
console.log({candidates, languages:lang_counter.toarray()})let arr = [ {"name":"george","languages":["php","javascript","java"],"age":19,"graduate_date":1044064800000,"phone":"32-991-511"},{"name":"anna","languages":["java","javascript"],"age":23,"graduate_date":1391220000000,"phone":"32-991-512"},{"name":"hailee","languages":["regex","javascript","perl","go","java"],"age":31,"graduate_date":1296525600000,"phone":"32-991-513"}]
let lang_counter = {
// helper keeps counts of unique items
counts:{},
add(arr){
arr.forEach(item => this.counts[item] = this.counts[item] ? this.counts[item] + 1 : 1)
},
toarray(){
return Object.entries(this.counts).map(([key, val]) => ({[key]: val}))
}
}
// iterate over object to create candidates
let candidates = arr.map(row => {
let {languages, ...person} = row
lang_counter.add(languages) // side effect
return person
})
// put them together
console.log({candidates, languages:lang_counter.toarray()})let arr = [ {"name":"george","languages":["php","javascript","java"],"age":19,"graduate_date":1044064800000,"phone":"32-991-511"},{"name":"anna","languages":["java","javascript"],"age":23,"graduate_date":1391220000000,"phone":"32-991-512"},{"name":"hailee","languages":["regex","javascript","perl","go","java"],"age":31,"graduate_date":1296525600000,"phone":"32-991-513"}]
let lang_counter = {
// helper keeps counts of unique items
counts:{},
add(arr){
arr.forEach(item => this.counts[item] = this.counts[item] ? this.counts[item] + 1 : 1)
},
toarray(){
return Object.entries(this.counts).map(([key, val]) => ({[key]: val}))
}
}
// iterate over object to create candidates
let candidates = arr.map(row => {
let {languages, ...person} = row
lang_counter.add(languages) // side effect
return person
})
// put them together
console.log({candidates, languages:lang_counter.toarray()})answered Nov 22 '18 at 6:14
Mark MeyerMark Meyer
37.5k33159
answered Nov 22 '18 at 6:14
Mark MeyerMark Meyer
37.5k33159
answered Nov 22 '18 at 6:14
Mark MeyerMark Meyer
37.5k33159
answered Nov 22 '18 at 6:14
Mark MeyerMark Meyer
37.5k33159
37.5k33159
Yeaah That's nice bro. I've changed some things to make all names receive the first letter in uppercase and organize all languages in alphabetical order. Thank you!!!
– Riller Vincci
Nov 22 '18 at 6:40
add a comment |
Yeaah That's nice bro. I've changed some things to make all names receive the first letter in uppercase and organize all languages in alphabetical order. Thank you!!!
– Riller Vincci
Nov 22 '18 at 6:40
Yeaah That's nice bro. I've changed some things to make all names receive the first letter in uppercase and organize all languages in alphabetical order. Thank you!!!
– Riller Vincci
Nov 22 '18 at 6:40
Yeaah That's nice bro. I've changed some things to make all names receive the first letter in uppercase and organize all languages in alphabetical order. Thank you!!!
– Riller Vincci
Nov 22 '18 at 6:40
add a comment |
0
You can use Array.reduce and Object.values like below
let arr = [{"name":"george","languages":["php","javascript","java"],"age":19,"graduate_date":1044064800000,"phone":"32-991-511"},{"name":"anna","languages":["java","javascript"],"age":23,"graduate_date":1391220000000,"phone":"32-991-512"},{"name":"hailee","languages":["regex","javascript","perl","go","java"],"age":31,"graduate_date":1296525600000,"phone":"32-991-513"}]
let res = arr.reduce((o, {name, age, phone, graduate_date, languages}) => {
o.candidates.push({name, age, phone, graduate_date})
languages.forEach(l => {
o.languages[l] = o.languages[l] || { lang:l, count: 0 }
o.languages[l].count++
})
return o
}
, { candidates: , languages: {}})
res.candidates = res.candidates.sort((a,b) => a.graduate_date - b.graduate_date)
.map(({ graduate_date, ...rest }) => rest)
res.languages = Object.values(res.languages).sort((a,b) => a.lang.localeCompare(b.lang))
console.log(res)answered Nov 22 '18 at 6:05
Nitish NarangNitish Narang
2,948815
1
It's very interesting You've used map function to organize candidates Woow Congratulations... Thank you
– Riller Vincci
Nov 22 '18 at 6:44
add a comment |
0
You can use Array.reduce and Object.values like below
let arr = [{"name":"george","languages":["php","javascript","java"],"age":19,"graduate_date":1044064800000,"phone":"32-991-511"},{"name":"anna","languages":["java","javascript"],"age":23,"graduate_date":1391220000000,"phone":"32-991-512"},{"name":"hailee","languages":["regex","javascript","perl","go","java"],"age":31,"graduate_date":1296525600000,"phone":"32-991-513"}]
let res = arr.reduce((o, {name, age, phone, graduate_date, languages}) => {
o.candidates.push({name, age, phone, graduate_date})
languages.forEach(l => {
o.languages[l] = o.languages[l] || { lang:l, count: 0 }
o.languages[l].count++
})
return o
}
, { candidates: , languages: {}})
res.candidates = res.candidates.sort((a,b) => a.graduate_date - b.graduate_date)
.map(({ graduate_date, ...rest }) => rest)
res.languages = Object.values(res.languages).sort((a,b) => a.lang.localeCompare(b.lang))
console.log(res)answered Nov 22 '18 at 6:05
Nitish NarangNitish Narang
2,948815
1
It's very interesting You've used map function to organize candidates Woow Congratulations... Thank you
– Riller Vincci
Nov 22 '18 at 6:44
add a comment |
0
0
0
You can use Array.reduce and Object.values like below
let arr = [{"name":"george","languages":["php","javascript","java"],"age":19,"graduate_date":1044064800000,"phone":"32-991-511"},{"name":"anna","languages":["java","javascript"],"age":23,"graduate_date":1391220000000,"phone":"32-991-512"},{"name":"hailee","languages":["regex","javascript","perl","go","java"],"age":31,"graduate_date":1296525600000,"phone":"32-991-513"}]
let res = arr.reduce((o, {name, age, phone, graduate_date, languages}) => {
o.candidates.push({name, age, phone, graduate_date})
languages.forEach(l => {
o.languages[l] = o.languages[l] || { lang:l, count: 0 }
o.languages[l].count++
})
return o
}
, { candidates: , languages: {}})
res.candidates = res.candidates.sort((a,b) => a.graduate_date - b.graduate_date)
.map(({ graduate_date, ...rest }) => rest)
res.languages = Object.values(res.languages).sort((a,b) => a.lang.localeCompare(b.lang))
console.log(res)answered Nov 22 '18 at 6:05
Nitish NarangNitish Narang
2,948815
You can use Array.reduce and Object.values like below
let arr = [{"name":"george","languages":["php","javascript","java"],"age":19,"graduate_date":1044064800000,"phone":"32-991-511"},{"name":"anna","languages":["java","javascript"],"age":23,"graduate_date":1391220000000,"phone":"32-991-512"},{"name":"hailee","languages":["regex","javascript","perl","go","java"],"age":31,"graduate_date":1296525600000,"phone":"32-991-513"}]
let res = arr.reduce((o, {name, age, phone, graduate_date, languages}) => {
o.candidates.push({name, age, phone, graduate_date})
languages.forEach(l => {
o.languages[l] = o.languages[l] || { lang:l, count: 0 }
o.languages[l].count++
})
return o
}
, { candidates: , languages: {}})
res.candidates = res.candidates.sort((a,b) => a.graduate_date - b.graduate_date)
.map(({ graduate_date, ...rest }) => rest)
res.languages = Object.values(res.languages).sort((a,b) => a.lang.localeCompare(b.lang))
console.log(res)let arr = [{"name":"george","languages":["php","javascript","java"],"age":19,"graduate_date":1044064800000,"phone":"32-991-511"},{"name":"anna","languages":["java","javascript"],"age":23,"graduate_date":1391220000000,"phone":"32-991-512"},{"name":"hailee","languages":["regex","javascript","perl","go","java"],"age":31,"graduate_date":1296525600000,"phone":"32-991-513"}]
let res = arr.reduce((o, {name, age, phone, graduate_date, languages}) => {
o.candidates.push({name, age, phone, graduate_date})
languages.forEach(l => {
o.languages[l] = o.languages[l] || { lang:l, count: 0 }
o.languages[l].count++
})
return o
}
, { candidates: , languages: {}})
res.candidates = res.candidates.sort((a,b) => a.graduate_date - b.graduate_date)
.map(({ graduate_date, ...rest }) => rest)
res.languages = Object.values(res.languages).sort((a,b) => a.lang.localeCompare(b.lang))
console.log(res)let arr = [{"name":"george","languages":["php","javascript","java"],"age":19,"graduate_date":1044064800000,"phone":"32-991-511"},{"name":"anna","languages":["java","javascript"],"age":23,"graduate_date":1391220000000,"phone":"32-991-512"},{"name":"hailee","languages":["regex","javascript","perl","go","java"],"age":31,"graduate_date":1296525600000,"phone":"32-991-513"}]
let res = arr.reduce((o, {name, age, phone, graduate_date, languages}) => {
o.candidates.push({name, age, phone, graduate_date})
languages.forEach(l => {
o.languages[l] = o.languages[l] || { lang:l, count: 0 }
o.languages[l].count++
})
return o
}
, { candidates: , languages: {}})
res.candidates = res.candidates.sort((a,b) => a.graduate_date - b.graduate_date)
.map(({ graduate_date, ...rest }) => rest)
res.languages = Object.values(res.languages).sort((a,b) => a.lang.localeCompare(b.lang))
console.log(res)answered Nov 22 '18 at 6:05
Nitish NarangNitish Narang
2,948815
edited Nov 22 '18 at 6:40
answered Nov 22 '18 at 6:05
Nitish NarangNitish Narang
2,948815
answered Nov 22 '18 at 6:05
Nitish NarangNitish Narang
2,948815
answered Nov 22 '18 at 6:05
Nitish NarangNitish Narang
2,948815
2,948815
1
It's very interesting You've used map function to organize candidates Woow Congratulations... Thank you
– Riller Vincci
Nov 22 '18 at 6:44
add a comment |
1
It's very interesting You've used map function to organize candidates Woow Congratulations... Thank you
– Riller Vincci
Nov 22 '18 at 6:44
1
1
It's very interesting You've used map function to organize candidates Woow Congratulations... Thank you
– Riller Vincci
Nov 22 '18 at 6:44
It's very interesting You've used map function to organize candidates Woow Congratulations... Thank you
– Riller Vincci
Nov 22 '18 at 6:44
add a comment |
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