I have an object Report that contains two integer fields: Month and Year.
I need to sort it by "date"
Report.desc(:year).desc(:month).each do |a|
puts a.year.to_s + " " + a.month.to_s
end
results:
2011 12
2011 11
2012 7
2012 6
2012 5
2012 4
2012 3
2012 2
2012 1
While I would think to get
2012 7
2012 6
2012 5
2012 4
2012 3
2012 2
2012 1
2011 12
2011 11
What am I doing wrong?
The Mongoid Criteria looks like:
irb(main):043:0> Report.desc(:year).desc(:month)
=> #<Mongoid::Criteria
selector: {},
options: {:sort=>{"year"=>-1, "month"=>-1}},
class: Report,
embedded: true>
The result you get is only sorted by month not by the whole date(thus the result you get). Maybe use the sort_by method with a body that takes into account both the year and the month?
Something like:
Report.sort_by{|t| [-t.year, -t.month]}
EDIT: I am using the negatives of both the year and the month to achieve decreasing order.
Related
I am trying to use the Date.parse method in Ruby but it is returning odd results when I try to parse early historical dates. For example:
Date.parse("16 January 27 BC")
=> #<Date: -0026-01-16 >
This returns the correct year (although it is -26 instead of -27 but that is normal I think).
But when I try:
Date.parse("19 august 1 BC")
=> #<Date: 2000-08-19 >
Here it assumes the year 2000 as a start date. Same thing for later years:
Date.parse("19 August 14")
=> #<Date: 2014-08-19 >
yearninetynine = Date.parse("19 august AD 99")
=> #<Date: 1999-08-19 >
(I have also tried to replace AD with "CE" but that makes no difference)
The year 100 is where it starts to work properly again:
Date.parse("19 august AD 100")
=> #<Date: 0100-08-19 >
To summarize: up to the year 100 it doesn't parse properly and treats it as years occurring in the 1990s and 2000s, negative years are one less than you would intuitively expect, negative years close to 0 are also not parsed properly.
I tried a few other methods like start=Date::ITALY, but these seem to make no difference. Is there a way to make it parse these years properly?
According to the docs that is exactly what the 2nd argument is for.
Date.parse("19 august 99") #=> Thu, 19 Aug 1999
Date.parse("19 august 99", false) #=> Mon, 19 Aug 0099
I cant get the next 15th day but not the working day.
DateTime.now.next_day(+15).strftime('%d %^B %Y')
how can i get the next 15th weekday?
You're just adding 15 days to the current date. What you want is to adjust the date:
date = DateTime.now
if (date.mday > 15)
date = date.next_month
end
date = date.next_day(15 - date.mday)
Where that adjusts to be the 15th of the next month if it's already past the 15th of the current month.
Now this can be extended to be an Enumerator:
def each_mday(mday, from: nil)
from ||= DateTime.now
Enumerator.new do |y|
loop do
if (from.mday > mday)
from = from.next_month
end
from = from.next_day(mday - from.mday)
y << from
from += 1
end
end
end
Which makes it possible to find the first day matching particular criteria, like being a weekday:
each_mday(15, from: Date.parse('2019-06-14')).find { |d| (1..5).include?(d.wday) }
Where that returns July 15th, as June 15th is a weekend.
The from argument is optional but useful for testing cases like this to ensure it's working correctly.
15.times.reduce(Date.civil 2019, 03, 24) do |acc, _|
begin
acc += 1
end while [0, 6].include? acc.wday
acc
end
#⇒ #<Date: 2019-04-12 ((2458586j,0s,0n),+0s,2299161j)>
So you want to add 15 business days from the current date. You can go with iGian or Aleksei vanilla ruby answers or use business_time gem:
15.business_days.from_now
If I understood correctly, you want to get next Monday if you hit Saturday or Sunday. Since wday gives you 0 for Sun and 6 for Sat, you can use it to as a conditional to add days towards Monday.
def date_add_next_week_day(days)
date = (Date.today + days)
date += 1 if date.wday == 6
date += 1 if date.wday == 0
date
end
date_add_next_week_day(15).strftime('%d %^B %Y')
If I get the point you need to find the 15th day after a specified date, skipping weekends.
One possible option is to define the skipping_weekend hash like this, considering Date.html#wday:
skip_weekend = { 6 => 2, 0 => 1}
skip_weekend.default = 0
Then:
next15 = DateTime.now.next_day(15)
next15_working = next15.next_day(skip_weekend[next15.wday]).strftime('%d %B %Y')
Now if next15 falls on a working day, next15_working is the same day (hash defaults to 0), otherwise it skips 2 days if Saturday (6th week day, hash maps to 2) or 1 day if Sunday (0th week day, hash maps to 1)
I assume that, given a starting date, ds (a Date object), and a positive integer n, the problem is determine a later date, dt, such that between ds+1 and dt, inclusive, there n weekdays.
require 'date'
def given_date_plus_week_days(dt, week_days)
wday = dt.wday
weeks, days = (week_days + {0=>4, 6=>4}.fetch(wday, wday-1)).divmod(5)
dt - (wday.zero? ? 6 : (wday - 1)) + 7*weeks + days
end
The variable wday is assigned to the day of week for the start date, dt. The start date is moved back to the previous Monday, unless it falls on a Monday, in which case it is not changed. That is reflected in the expression
wday.zero? ? 6 : (wday - 1)
which is subtracted from dt. The number of week days is correspondingly adjusted to
week_days + { 0=>4, 6=>4 }.fetch(wday, wday-1)
The remaining calculations are straightforward.
def display(start_str, week_days)
start = Date.parse(start_str)
7.times.map { |i| start + i }.each do |ds|
de = given_date_plus_week_days(ds, week_days)
puts "#{ds.strftime("%a, %b %d, %Y")} + #{week_days} -> #{de.strftime("%a, %b %d, %Y")}"
end
end
display("April 8", 15)
Mon, Apr 08, 2019 + 15 -> Mon, Apr 29, 2019
Tue, Apr 09, 2019 + 15 -> Tue, Apr 30, 2019
Wed, Apr 10, 2019 + 15 -> Wed, May 01, 2019
Thu, Apr 11, 2019 + 15 -> Thu, May 02, 2019
Fri, Apr 12, 2019 + 15 -> Fri, May 03, 2019
Sat, Apr 13, 2019 + 15 -> Fri, May 03, 2019
Sun, Apr 14, 2019 + 15 -> Fri, May 03, 2019
display("April 8", 17)
Mon, Apr 08, 2019 + 17 -> Wed, May 01, 2019
Tue, Apr 09, 2019 + 17 -> Thu, May 02, 2019
Wed, Apr 10, 2019 + 17 -> Fri, May 03, 2019
Thu, Apr 11, 2019 + 17 -> Mon, May 06, 2019
Fri, Apr 12, 2019 + 17 -> Tue, May 07, 2019
Sat, Apr 13, 2019 + 17 -> Tue, May 07, 2019
Sun, Apr 14, 2019 + 17 -> Tue, May 07, 2019
Let's suppose I have the following data on ElasticSearch:
#timestamp; userId; currentPoints
August 7th 2017, 00:30:37.319; myUserName; 4
August 7th 2017, 00:43:22.121; myUserName; 10
August 7th 2017, 00:54:29.177; myUserName; 7
August 7th 2017, 01:10:29.352; myUserName; 4
August 7th 2017, 00:32:37.319; myOtherUserName; 12
August 7th 2017, 00:44:22.121; myOtherUserName; 17
August 7th 2017, 00:56:29.177; myOtherUserName; 8
August 7th 2017, 01:18:29.352; myOtherUserName; 11
I'm looking to draw a date histogram that will show me the sum of all max:currentPoints per username per hour, which whould generate the following data to plot:
August 7th 2017, 00; SumOfMaxCurrentPoints -> 27 (max from hour 00h from both users 10 + 17)
August 7th 2017, 00; SumOfMaxCurrentPoints -> 15 (max from hour 01h from both users 4 + 11)
This would usually be done with a subquery, extracting the max(currentPoints) for each hour, user and then sum the results and aggregate per hour.
Is this possible with Kibana Timelion for instance? I can't find a way to achieve this using the documentation.
Thanks
Alex
While working on another project, I've stumpled upon the answer to do this in Kibana/Elasticsearch without using Timelion.
The feature is called Sibling Pipeline Aggregation, and in this case you use the Sum Bucket. You can use it with any recent Kibana/Elastic visualization (I'm using version 5.5).
For a dataset such as:
#timestamp; userId; currentPoints
August 7th 2017, 00:30:37.319; myUserName; 4
August 7th 2017, 00:43:22.121; myUserName; 10
August 7th 2017, 00:54:29.177; myUserName; 7
August 7th 2017, 01:10:29.352; myUserName; 4
August 7th 2017, 00:32:37.319; myOtherUserName; 12
August 7th 2017, 00:44:22.121; myOtherUserName; 17
August 7th 2017, 00:56:29.177; myOtherUserName; 8
August 7th 2017, 01:18:29.352; myOtherUserName; 11
Where you want an hourly SUM of(currentPoints) all MAXs(currentPoints) per userId, resulting in:
August 7th 2017, 00; SumOfMaxCurrentPoints -> 27 (max from hour 00h from both users 10 + 17)
August 7th 2017, 00; SumOfMaxCurrentPoints -> 15 (max from hour 01h from both users 4 + 11)
You can do:
Metric
Aggregation: Sibling Pipeline Aggregation (Sum Bucket)
Bucket Aggregation Type: Terms
Bucket Field: userId
Bucket Size: Comfortable value above the # of users if you want total precision
Metric Aggregation: Max
Metric Field: currentPoints
Bucket
Buckets type: Split Rows
Bucket Aggregation: Date Histogram
Histogram Field: #timestamp
Histogram Interval: Hourly
Trying to convert strings like 9 weeks ago, 1 year, 6 months ago, 20 hours ago into a ruby time object like Tue, 10 Mar 2015 12:06:15 PDT -07:00.
I've been doing this:
eval("10 days ago".gsub(' ', '.'))
This works fine, but for strings like 1 year, 6 months ago blows up.
I just need to do comparisons like:
eval("10 days ago".gsub(' ', '.')) < (Time.now - 7.days)
I'm using sinatra so no fancy rails helpers.
Please never use eval in production code..
Converting from timeago notation would be quite complex and resource intensive.
However, this way seems the least error prone: It will convert a string like "5 seconds ago" to "5S" and use mapping to find what it means in time, after which it will subtract that time from the current time.
The parse string is dynamically built so it can accomodate most every timeago notation.
require('date')
mapping = {"D"=> "%d","W"=>"%U","H"=>"%T","Y"=>"%Y","M"=>"%m","S"=>"%S"}
timerel = "1 year, 6 months ago".split(",").map { |n| n.gsub(/\s+/, "").upcase()[0,2].split('')}
Date.strptime(
timerel.map {|n| n[0]}.join(" "),
timerel.map {|n| mapping[n[1]]}.join(" ")
)
date = Date.new(0) + (Date.today - Date.strptime(timerel.map {|n| n[0]}.join(" "), timerel.map {|n| mapping[n[1]]}.join(" ")))
=> #<Date: 2014-10-10 ((2456941j,0s,0n),+0s,2299161j)>
It goes without saying that is very error prone. Use at your own risk:
def parse(date:)
eval(date.gsub(/ ?(,|and) ?/, '+').tr(' ', '.').gsub(/^(.*)(\.ago)$/, '(\1)\2'))
end
parse(date: '1 year, 6 months ago') # => Wed, 10 Sep 2014 21:29:11 BST +01:00
parse(date: '1 year, 6 months, 3 weeks, 6 days, 9 hours and 12 seconds ago')
# => Thu, 14 Aug 2014 12:33:07 BST +01:00
The idea is to convert the original string to:
'(1.year+6.months).ago'
I can only find algorithm for getting ISO 8601 week (week starts on a Monday).
However, the iCal spec says
A week is defined as a seven day period, starting on the day of the
week defined to be the week start (see WKST). Week number one of the
calendar year is the first week that contains at least four (4) days
in that calendar year.
Therefore, it is more complex than ISO 8601 since the start of week can be any day of the week.
Is there an algorithm to determine what is the week number of a date, with a custom start day of week?
or... is there a function in iCal4j that does this? Determine a weekno from a date?
Thanks!
p.s. Limitation: I'm using a JVM language that cannot extend a Java class, but I can invoke Java methods or instantiate Java classes.
if (input_date < firstDateOfTheYear(WKST, year))
{
return ((isLeapYear(year-1))?53:52);
}
else
{
return ((dayOfYear(input_date) - firstDateOfTheYear(WKST, year).day)/7 + 1);
}
firstDateOfTheYear returns the first calendar date given a start of week(WKST) and the year, e.g. if WKST = Thursday, year = 2012, then it returns Jan 5th.
dayOfYear returns sequencial numerical day of the year, e.g. Feb 1st = 32
Example #1: Jan 18th, 2012, start of week is Monday
dayOfYear(Jan 18th, 2012) = 18
firstDateOfTheYear(Monday, 2012) = Jan 2nd, 2012
(18 - 2)/7 + 1 = 3
Answer Week no. 3
Example #2: Jan 18th, 2012, start of week is Thursday
dayOfYear(Jan 18th, 2012) = 18
firstDateOfTheYear(Thursday, 2012) = Jan 5th, 2012
(18 - 5)/7 + 1 = 2
Answer Week no. 2
Example #3: Jan 1st, 2012, start of week is Monday
firstDateOfTheYear(Monday, 2012) = Jan 2nd, 2012
IsLeapYear(2012-1) = false
Jan 1st, 2012 < Jan 2nd, 2012
Answer Week no. 52
Let daysInFirstWeek be the number of days on the first week of the year that are in January. Week starts on a WKST day. (e.g. if Jan 1st is a WKST day, return 7)
Set dayOfYear to the n-th days of the input date's year (e.g. Feb 1st = 32)
If dayOfYear is less than or equal to daysInFirstWeek
3.1. if daysInFirstWeek is greater than or equal to 4, weekNo is 1, skip to step 5.
3.2. Let daysInFirstWeekOfLastYear be the number of days on the first week of the previous year that are in January. Week starts on a WKST day.
3.3. if daysInFirstWeekOfLastYear is 4 or last year is Leap year and daysInFirstWeekOfLastYear is 5, weekNo is 53, otherwise weekNo is 52, skip to step 5.
Set weekNo to ceiling((dayOfYear - daysInFirstWeek) / 7)
4.1. if daysInFirstWeek greater than or equal to 4, increment weekNo by 1
4.2. if daysInFirstWeek equal 53 and count of days on the first week (starting from WKST) of January in the year of inputDate's year + 1 is greater than or equal to 4, set weekNo to 1
return weekNo