Split an array of items into several arrays, given a conditionTransforming an array of arrays into an array...

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Split an array of items into several arrays, given a condition

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The question is very simple : I have an array of items, that all have a date. According to the date, I would like to split this array into several arrays.

In my case, I would like to split the array into 3 arrays : one that contains result of the last 24h, the other 48h, then the last one on the week.

Here is a minimal reproduction of the case, with the final solution.

const now = Date.now();

const day = 1000 * 3600 * 24;



const todayTime = now - day;

const yesterdayTime = now - day * 2;

const weekTime = now - day * 7;



const arr = [

  { id: 1, date: todayTime + 100 },

  { id: 2, date: yesterdayTime + 100 },

  { id: 3, date: weekTime + 100 },

];



const today = arr.filter(item => item.date - todayTime > 0);

const yesterday = arr.filter(item => item.date - yesterdayTime > 0 && item.date - todayTime < 0);

const week = arr.filter(item => item.date - weekTime > 0 && item.date - yesterdayTime < 0);



console.log(today, yesterday, week);

Is there a way to make it more concise ? I'm talking about code length : I would like to reduce it to the maximum, especially the definition of the 3 arrays. I tried with reduce, but the syntax is pretty ugly.

Any ideas ?

asked 13 hours ago

trichetriche

1114

New contributor

add a comment |

The question is very simple : I have an array of items, that all have a date. According to the date, I would like to split this array into several arrays.

In my case, I would like to split the array into 3 arrays : one that contains result of the last 24h, the other 48h, then the last one on the week.

Here is a minimal reproduction of the case, with the final solution.

const now = Date.now();

const day = 1000 * 3600 * 24;



const todayTime = now - day;

const yesterdayTime = now - day * 2;

const weekTime = now - day * 7;



const arr = [

  { id: 1, date: todayTime + 100 },

  { id: 2, date: yesterdayTime + 100 },

  { id: 3, date: weekTime + 100 },

];



const today = arr.filter(item => item.date - todayTime > 0);

const yesterday = arr.filter(item => item.date - yesterdayTime > 0 && item.date - todayTime < 0);

const week = arr.filter(item => item.date - weekTime > 0 && item.date - yesterdayTime < 0);



console.log(today, yesterday, week);

Any ideas ?

asked 13 hours ago

trichetriche

1114

New contributor

add a comment |

The question is very simple : I have an array of items, that all have a date. According to the date, I would like to split this array into several arrays.

In my case, I would like to split the array into 3 arrays : one that contains result of the last 24h, the other 48h, then the last one on the week.

Here is a minimal reproduction of the case, with the final solution.

const now = Date.now();

const day = 1000 * 3600 * 24;



const todayTime = now - day;

const yesterdayTime = now - day * 2;

const weekTime = now - day * 7;



const arr = [

  { id: 1, date: todayTime + 100 },

  { id: 2, date: yesterdayTime + 100 },

  { id: 3, date: weekTime + 100 },

];



const today = arr.filter(item => item.date - todayTime > 0);

const yesterday = arr.filter(item => item.date - yesterdayTime > 0 && item.date - todayTime < 0);

const week = arr.filter(item => item.date - weekTime > 0 && item.date - yesterdayTime < 0);



console.log(today, yesterday, week);

Any ideas ?

asked 13 hours ago

trichetriche

1114

New contributor

The question is very simple : I have an array of items, that all have a date. According to the date, I would like to split this array into several arrays.

In my case, I would like to split the array into 3 arrays : one that contains result of the last 24h, the other 48h, then the last one on the week.

Here is a minimal reproduction of the case, with the final solution.

const now = Date.now();

const day = 1000 * 3600 * 24;



const todayTime = now - day;

const yesterdayTime = now - day * 2;

const weekTime = now - day * 7;



const arr = [

  { id: 1, date: todayTime + 100 },

  { id: 2, date: yesterdayTime + 100 },

  { id: 3, date: weekTime + 100 },

];



const today = arr.filter(item => item.date - todayTime > 0);

const yesterday = arr.filter(item => item.date - yesterdayTime > 0 && item.date - todayTime < 0);

const week = arr.filter(item => item.date - weekTime > 0 && item.date - yesterdayTime < 0);



console.log(today, yesterday, week);

Any ideas ?

const now = Date.now();

const day = 1000 * 3600 * 24;



const todayTime = now - day;

const yesterdayTime = now - day * 2;

const weekTime = now - day * 7;



const arr = [

  { id: 1, date: todayTime + 100 },

  { id: 2, date: yesterdayTime + 100 },

  { id: 3, date: weekTime + 100 },

];



const today = arr.filter(item => item.date - todayTime > 0);

const yesterday = arr.filter(item => item.date - yesterdayTime > 0 && item.date - todayTime < 0);

const week = arr.filter(item => item.date - weekTime > 0 && item.date - yesterdayTime < 0);



console.log(today, yesterday, week);

const now = Date.now();

const day = 1000 * 3600 * 24;



const todayTime = now - day;

const yesterdayTime = now - day * 2;

const weekTime = now - day * 7;



const arr = [

  { id: 1, date: todayTime + 100 },

  { id: 2, date: yesterdayTime + 100 },

  { id: 3, date: weekTime + 100 },

];



const today = arr.filter(item => item.date - todayTime > 0);

const yesterday = arr.filter(item => item.date - yesterdayTime > 0 && item.date - todayTime < 0);

const week = arr.filter(item => item.date - weekTime > 0 && item.date - yesterdayTime < 0);



console.log(today, yesterday, week);

javascript array

asked 13 hours ago

trichetriche

1114

New contributor

asked 13 hours ago

trichetriche

1114

New contributor

asked 13 hours ago

trichetriche

1114

New contributor

asked 13 hours ago

trichetriche

1114

asked 13 hours ago

trichetriche

1114

New contributor

trichetriche is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment |

2 Answers
2

active

oldest

votes

item.date - todayTime > 0 is the same as item.date > todayTime.

Your less-than condition should be less-than-or-equal, otherwise anything falling on a day boundary won't match any category.

const now = Date.now(),

     day = 1000 * 3600 * 24,

     arr = [

            { id: 1, date: now - day + 100 },

            { id: 2, date: now - day*2 + 100 },

            { id: 3, date: now - day*7 + 100 },

           ],

    daysAgo=[ 1, 2, 7 ],

    filtered=daysAgo.map( 

        (days,i) => arr.filter( 

            j => j.date > now-day*days && j.date <= now-day*( i ? daysAgo[i-1] : -1 )        

        ) 

    );



console.log(filtered);

answered 12 hours ago

Oh My Goodness

58529

$begingroup$
As said on your previous comment, I just wrote that fast and didn't really cared about all that. I was more focused on the array operators
$endgroup$
– trichetriche
11 hours ago

$begingroup$
well you can factor out some of the calculations with an array of days - see my code. It's only a little shorter than your version, but it will scale up to many intervals with hardly any increase in code length.
$endgroup$
– Oh My Goodness
11 hours ago

$begingroup$
I have posted this question on SOF and they made the same statement, to which I replied that my case is very specific and I actually don't need genericity, because the data that is one week old gets deleted. I have and will always have 3 arrays, so I don't need to make it generic at all. But your answer is nicely thought indeed, so have my upvote !
$endgroup$
– trichetriche
10 hours ago

add a comment |

Always create a function

Its not really code if its not a function, you may as well just output the 3 arrays directly. (my personal view on global inline code).

You may say, "This is just an example". No it can not be just an example, a function has special powers and is written differently than inline code. Plus examples are boarder line off topic here, this question may get closed.

Part of a functions power is that it makes you think about how you solve the problem differently from inline code.

Look for change

When you write a function you look for the parts of the logic and data that change. You pass that to the function as arguments. The function uses these argument to process the data and return the desired results.

In this case you have

the array of items to split.

The number of days old to split the data, eg 1,2,and 7 days.

Thus we can have something like the following. The function is declared inside another so that all it needs is safely encapsulated and outside of unrelated scopes.

const splitByDaysOld = (() => {

    const MS_IN_DAY = 8.64e7;

    const sorter = (a, b) => a - b;

    return function(array, periods) {

        const now = Date.now();

        var start = 0;

        return periods.sort(sorter).map(day => {

            const res = array.filter(item => {

                const daysOld = (item.date - now) / MS_IN_DAY;

                return daysOld >= start && daysOld < day;

            });

            start = day;

            return res;

        });

    };

})();

It returns an array of arrays, one for each period.

Example below has 3 periods 0 to < 1, 1 to < 2, and 2 to < 7 days old.

const [today, yesterday, week] = splitByDaysOld(arr, [1, 2, 7]);

Note that the periods array is sorted from most recent to oldest and that the result will be in the same order

answered 11 hours ago

Blindman67

8,1951521

$begingroup$
The function is called listAllPercos and it exists. So yes, it is just an example. Didn't think about making it like this though. Question totally out of context, but is this considerd a factory ? And I'm going to try to implement it with what I need and don't, and keep you updated. Anyway, have my upvote too !
$endgroup$
– trichetriche
10 hours ago

add a comment |

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

item.date - todayTime > 0 is the same as item.date > todayTime.

Your less-than condition should be less-than-or-equal, otherwise anything falling on a day boundary won't match any category.

const now = Date.now(),

     day = 1000 * 3600 * 24,

     arr = [

            { id: 1, date: now - day + 100 },

            { id: 2, date: now - day*2 + 100 },

            { id: 3, date: now - day*7 + 100 },

           ],

    daysAgo=[ 1, 2, 7 ],

    filtered=daysAgo.map( 

        (days,i) => arr.filter( 

            j => j.date > now-day*days && j.date <= now-day*( i ? daysAgo[i-1] : -1 )        

        ) 

    );



console.log(filtered);

answered 12 hours ago

Oh My Goodness

58529

$begingroup$
As said on your previous comment, I just wrote that fast and didn't really cared about all that. I was more focused on the array operators
$endgroup$
– trichetriche
11 hours ago

$begingroup$
well you can factor out some of the calculations with an array of days - see my code. It's only a little shorter than your version, but it will scale up to many intervals with hardly any increase in code length.
$endgroup$
– Oh My Goodness
11 hours ago

$begingroup$
I have posted this question on SOF and they made the same statement, to which I replied that my case is very specific and I actually don't need genericity, because the data that is one week old gets deleted. I have and will always have 3 arrays, so I don't need to make it generic at all. But your answer is nicely thought indeed, so have my upvote !
$endgroup$
– trichetriche
10 hours ago

add a comment |

item.date - todayTime > 0 is the same as item.date > todayTime.

Your less-than condition should be less-than-or-equal, otherwise anything falling on a day boundary won't match any category.

const now = Date.now(),

     day = 1000 * 3600 * 24,

     arr = [

            { id: 1, date: now - day + 100 },

            { id: 2, date: now - day*2 + 100 },

            { id: 3, date: now - day*7 + 100 },

           ],

    daysAgo=[ 1, 2, 7 ],

    filtered=daysAgo.map( 

        (days,i) => arr.filter( 

            j => j.date > now-day*days && j.date <= now-day*( i ? daysAgo[i-1] : -1 )        

        ) 

    );



console.log(filtered);

answered 12 hours ago

Oh My Goodness

58529

$begingroup$
As said on your previous comment, I just wrote that fast and didn't really cared about all that. I was more focused on the array operators
$endgroup$
– trichetriche
11 hours ago

$begingroup$
well you can factor out some of the calculations with an array of days - see my code. It's only a little shorter than your version, but it will scale up to many intervals with hardly any increase in code length.
$endgroup$
– Oh My Goodness
11 hours ago

$begingroup$
I have posted this question on SOF and they made the same statement, to which I replied that my case is very specific and I actually don't need genericity, because the data that is one week old gets deleted. I have and will always have 3 arrays, so I don't need to make it generic at all. But your answer is nicely thought indeed, so have my upvote !
$endgroup$
– trichetriche
10 hours ago

add a comment |

item.date - todayTime > 0 is the same as item.date > todayTime.

Your less-than condition should be less-than-or-equal, otherwise anything falling on a day boundary won't match any category.

const now = Date.now(),

     day = 1000 * 3600 * 24,

     arr = [

            { id: 1, date: now - day + 100 },

            { id: 2, date: now - day*2 + 100 },

            { id: 3, date: now - day*7 + 100 },

           ],

    daysAgo=[ 1, 2, 7 ],

    filtered=daysAgo.map( 

        (days,i) => arr.filter( 

            j => j.date > now-day*days && j.date <= now-day*( i ? daysAgo[i-1] : -1 )        

        ) 

    );



console.log(filtered);

answered 12 hours ago

Oh My Goodness

58529

item.date - todayTime > 0 is the same as item.date > todayTime.

Your less-than condition should be less-than-or-equal, otherwise anything falling on a day boundary won't match any category.

const now = Date.now(),

     day = 1000 * 3600 * 24,

     arr = [

            { id: 1, date: now - day + 100 },

            { id: 2, date: now - day*2 + 100 },

            { id: 3, date: now - day*7 + 100 },

           ],

    daysAgo=[ 1, 2, 7 ],

    filtered=daysAgo.map( 

        (days,i) => arr.filter( 

            j => j.date > now-day*days && j.date <= now-day*( i ? daysAgo[i-1] : -1 )        

        ) 

    );



console.log(filtered);

const now = Date.now(),

     day = 1000 * 3600 * 24,

     arr = [

            { id: 1, date: now - day + 100 },

            { id: 2, date: now - day*2 + 100 },

            { id: 3, date: now - day*7 + 100 },

           ],

    daysAgo=[ 1, 2, 7 ],

    filtered=daysAgo.map( 

        (days,i) => arr.filter( 

            j => j.date > now-day*days && j.date <= now-day*( i ? daysAgo[i-1] : -1 )        

        ) 

    );



console.log(filtered);

const now = Date.now(),

     day = 1000 * 3600 * 24,

     arr = [

            { id: 1, date: now - day + 100 },

            { id: 2, date: now - day*2 + 100 },

            { id: 3, date: now - day*7 + 100 },

           ],

    daysAgo=[ 1, 2, 7 ],

    filtered=daysAgo.map( 

        (days,i) => arr.filter( 

            j => j.date > now-day*days && j.date <= now-day*( i ? daysAgo[i-1] : -1 )        

        ) 

    );



console.log(filtered);

answered 12 hours ago

Oh My Goodness

58529

answered 12 hours ago

Oh My Goodness

58529

answered 12 hours ago

Oh My Goodness

58529

answered 12 hours ago

Oh My Goodness

58529

$begingroup$
As said on your previous comment, I just wrote that fast and didn't really cared about all that. I was more focused on the array operators
$endgroup$
– trichetriche
11 hours ago

$begingroup$
well you can factor out some of the calculations with an array of days - see my code. It's only a little shorter than your version, but it will scale up to many intervals with hardly any increase in code length.
$endgroup$
– Oh My Goodness
11 hours ago

$begingroup$
I have posted this question on SOF and they made the same statement, to which I replied that my case is very specific and I actually don't need genericity, because the data that is one week old gets deleted. I have and will always have 3 arrays, so I don't need to make it generic at all. But your answer is nicely thought indeed, so have my upvote !
$endgroup$
– trichetriche
10 hours ago

add a comment |

$begingroup$
As said on your previous comment, I just wrote that fast and didn't really cared about all that. I was more focused on the array operators
$endgroup$
– trichetriche
11 hours ago

$begingroup$
well you can factor out some of the calculations with an array of days - see my code. It's only a little shorter than your version, but it will scale up to many intervals with hardly any increase in code length.
$endgroup$
– Oh My Goodness
11 hours ago

$begingroup$
I have posted this question on SOF and they made the same statement, to which I replied that my case is very specific and I actually don't need genericity, because the data that is one week old gets deleted. I have and will always have 3 arrays, so I don't need to make it generic at all. But your answer is nicely thought indeed, so have my upvote !
$endgroup$
– trichetriche
10 hours ago

As said on your previous comment, I just wrote that fast and didn't really cared about all that. I was more focused on the array operators

– trichetriche
11 hours ago

well you can factor out some of the calculations with an array of days - see my code. It's only a little shorter than your version, but it will scale up to many intervals with hardly any increase in code length.

– Oh My Goodness
11 hours ago

I have posted this question on SOF and they made the same statement, to which I replied that my case is very specific and I actually don't need genericity, because the data that is one week old gets deleted. I have and will always have 3 arrays, so I don't need to make it generic at all. But your answer is nicely thought indeed, so have my upvote !

– trichetriche
10 hours ago

add a comment |

Always create a function

Its not really code if its not a function, you may as well just output the 3 arrays directly. (my personal view on global inline code).

Part of a functions power is that it makes you think about how you solve the problem differently from inline code.

Look for change

In this case you have

the array of items to split.

The number of days old to split the data, eg 1,2,and 7 days.

Thus we can have something like the following. The function is declared inside another so that all it needs is safely encapsulated and outside of unrelated scopes.

const splitByDaysOld = (() => {

    const MS_IN_DAY = 8.64e7;

    const sorter = (a, b) => a - b;

    return function(array, periods) {

        const now = Date.now();

        var start = 0;

        return periods.sort(sorter).map(day => {

            const res = array.filter(item => {

                const daysOld = (item.date - now) / MS_IN_DAY;

                return daysOld >= start && daysOld < day;

            });

            start = day;

            return res;

        });

    };

})();

It returns an array of arrays, one for each period.

Example below has 3 periods 0 to < 1, 1 to < 2, and 2 to < 7 days old.

const [today, yesterday, week] = splitByDaysOld(arr, [1, 2, 7]);

Note that the periods array is sorted from most recent to oldest and that the result will be in the same order

answered 11 hours ago

Blindman67

8,1951521

$begingroup$
The function is called listAllPercos and it exists. So yes, it is just an example. Didn't think about making it like this though. Question totally out of context, but is this considerd a factory ? And I'm going to try to implement it with what I need and don't, and keep you updated. Anyway, have my upvote too !
$endgroup$
– trichetriche
10 hours ago

add a comment |

Always create a function

Its not really code if its not a function, you may as well just output the 3 arrays directly. (my personal view on global inline code).

Part of a functions power is that it makes you think about how you solve the problem differently from inline code.

Look for change

In this case you have

the array of items to split.

The number of days old to split the data, eg 1,2,and 7 days.

Thus we can have something like the following. The function is declared inside another so that all it needs is safely encapsulated and outside of unrelated scopes.

const splitByDaysOld = (() => {

    const MS_IN_DAY = 8.64e7;

    const sorter = (a, b) => a - b;

    return function(array, periods) {

        const now = Date.now();

        var start = 0;

        return periods.sort(sorter).map(day => {

            const res = array.filter(item => {

                const daysOld = (item.date - now) / MS_IN_DAY;

                return daysOld >= start && daysOld < day;

            });

            start = day;

            return res;

        });

    };

})();

It returns an array of arrays, one for each period.

Example below has 3 periods 0 to < 1, 1 to < 2, and 2 to < 7 days old.

const [today, yesterday, week] = splitByDaysOld(arr, [1, 2, 7]);

Note that the periods array is sorted from most recent to oldest and that the result will be in the same order

answered 11 hours ago

Blindman67

8,1951521

$begingroup$
The function is called listAllPercos and it exists. So yes, it is just an example. Didn't think about making it like this though. Question totally out of context, but is this considerd a factory ? And I'm going to try to implement it with what I need and don't, and keep you updated. Anyway, have my upvote too !
$endgroup$
– trichetriche
10 hours ago

add a comment |

Always create a function

Its not really code if its not a function, you may as well just output the 3 arrays directly. (my personal view on global inline code).

Part of a functions power is that it makes you think about how you solve the problem differently from inline code.

Look for change

In this case you have

the array of items to split.

The number of days old to split the data, eg 1,2,and 7 days.

Thus we can have something like the following. The function is declared inside another so that all it needs is safely encapsulated and outside of unrelated scopes.

const splitByDaysOld = (() => {

    const MS_IN_DAY = 8.64e7;

    const sorter = (a, b) => a - b;

    return function(array, periods) {

        const now = Date.now();

        var start = 0;

        return periods.sort(sorter).map(day => {

            const res = array.filter(item => {

                const daysOld = (item.date - now) / MS_IN_DAY;

                return daysOld >= start && daysOld < day;

            });

            start = day;

            return res;

        });

    };

})();

It returns an array of arrays, one for each period.

Example below has 3 periods 0 to < 1, 1 to < 2, and 2 to < 7 days old.

const [today, yesterday, week] = splitByDaysOld(arr, [1, 2, 7]);

Note that the periods array is sorted from most recent to oldest and that the result will be in the same order

answered 11 hours ago

Blindman67

8,1951521

Always create a function

Its not really code if its not a function, you may as well just output the 3 arrays directly. (my personal view on global inline code).

Part of a functions power is that it makes you think about how you solve the problem differently from inline code.

Look for change

In this case you have

the array of items to split.

The number of days old to split the data, eg 1,2,and 7 days.

Thus we can have something like the following. The function is declared inside another so that all it needs is safely encapsulated and outside of unrelated scopes.

const splitByDaysOld = (() => {

    const MS_IN_DAY = 8.64e7;

    const sorter = (a, b) => a - b;

    return function(array, periods) {

        const now = Date.now();

        var start = 0;

        return periods.sort(sorter).map(day => {

            const res = array.filter(item => {

                const daysOld = (item.date - now) / MS_IN_DAY;

                return daysOld >= start && daysOld < day;

            });

            start = day;

            return res;

        });

    };

})();

It returns an array of arrays, one for each period.

Example below has 3 periods 0 to < 1, 1 to < 2, and 2 to < 7 days old.

const [today, yesterday, week] = splitByDaysOld(arr, [1, 2, 7]);

Note that the periods array is sorted from most recent to oldest and that the result will be in the same order

answered 11 hours ago

Blindman67

8,1951521

answered 11 hours ago

Blindman67

8,1951521

answered 11 hours ago

Blindman67

8,1951521

answered 11 hours ago

Blindman67

8,1951521

$begingroup$
The function is called listAllPercos and it exists. So yes, it is just an example. Didn't think about making it like this though. Question totally out of context, but is this considerd a factory ? And I'm going to try to implement it with what I need and don't, and keep you updated. Anyway, have my upvote too !
$endgroup$
– trichetriche
10 hours ago

add a comment |

$begingroup$
The function is called listAllPercos and it exists. So yes, it is just an example. Didn't think about making it like this though. Question totally out of context, but is this considerd a factory ? And I'm going to try to implement it with what I need and don't, and keep you updated. Anyway, have my upvote too !
$endgroup$
– trichetriche
10 hours ago

The function is called listAllPercos and it exists. So yes, it is just an example. Didn't think about making it like this though. Question totally out of context, but is this considerd a factory ? And I'm going to try to implement it with what I need and don't, and keep you updated. Anyway, have my upvote too !

– trichetriche
10 hours ago

add a comment |

trichetriche is a new contributor. Be nice, and check out our Code of Conduct.

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