I'm trying to convert strings like "Sep 11, Oct 31, Feb 28" into DateTime instances as part of a screen-scraper. Using DateTime.parse() works fine apart from when the data goes across years, and it naively (and probably correctly) returns a date in the current year.
For example the following test case.
test "dateA convert next year" do
TimeService.stubs(:now).returns(Time.new(2013, 12, 30, 9, 30))
assert_equal(Date.new(2014, 1, 2), Extraction.dateA("Jan 2"))
end
I updated my method to look at what would be date with year + 1, and return the closest to 'now' - this works fine. However it feels a bit ugly, and I'm looking for a more elegant solution.
def Extraction.dateA(content)
return_value = DateTime.parse(content)
next_year = return_value.change(:year => return_value.year + 1)
now = TimeService.now.to_i
if (next_year.to_i - now).abs < (return_value.to_i - now).abs then
return_value = next_year
end
return_value
end
TimeService.now is just a utility to return current time to help with stubbing.
Excuse my ruby, I'm new to it.
I think this works as intended, allowing for closest date in previous year as well:
module Extraction
def Extraction.date_a(content)
parsed_date = DateTime.parse(content)
now = DateTime.now
dates = [ parsed_date, parsed_date.next_year, parsed_date.prev_year ]
dates.min_by { |d| ( d - now ).abs }
end
end
A few points:
Changed method name to date_a, just a Ruby convention that differs from Java.
I made use of some built-in methods next_year and prev_year on DateTime
I used a time difference metric and selected date with the minimal value of it from three candidate dates (this is what min_by does). This is simpler code than the conditional switching, especially with three dates to consider.
I forgot about min_by originally, I don't use it often, but it's a very good fit for this problem.
Note there is a pathological case - "29 Feb". If it appears correctly in text, by its nature it will define which year is valid, and it won't parse if current year is e.g. 2015.
Related
This thread provides an answer to find the first occurrence of a day (e.g. Thursday) after today's date.
I would like to get a more general function which gives the first occurrence of a day (e.g. Thursday) after any given date (e.g. November 1st, 2017).
The method would therefore take 2 arguments: the Date (either as a String, or as a Date object) and the day (as a String).
One way I found is to check each day after with the thursday? method (e.g. my_date_object.thursday?) but you would need a switch / if-else statement to check which day to check against, which makes the method rather bulky, as seen below.
def get_next_day_after_date(date, day)
days_in_week = 7
week = []
days_in_week.times { |day_after| week << (date + day_after) }
if day.casecmp('Monday').zero?
week.select(&:monday?)
elsif day.casecmp('Tuesday').zero?
week.select(&:tuesday?)
elsif day.casecmp('Wednesday').zero?
week.select(&:wednesday?)
# same code for Thursday to Sunday
end
end
Is there a cleaner way to achieve this in pure Ruby? (no use of active_support or rails-specific methods).
require "date"
def get_next_day_after_day(date, day)
date + (Date.strptime(day, "%A") - date) % 7
end
Objective
I am trying to calculate the distance in weeks since a given date without jumping through hoops. I'd prefer to do it in plain Ruby, but ActiveSupport is certainly an acceptable alternative.
My Code
I wrote the following, which seems to work but looks like the long way around to me.
require 'date'
DAYS_IN_WEEK = 7.0
def weeks_since date_string
date = Date.parse date_string
days = Date.today - date
weeks = days / DAYS_IN_WEEK
weeks.round 2
end
weeks_since '2015-06-15'
#=> 32.57
ActiveSupport's #weeks_since takes a number of weeks as its argument, so it doesn't fit this use case. Ruby's Date class doesn't seem to have anything relevant, either.
Alternatives?
Is there a better built-in solution or well-known algorithm for calculating the number of weeks separating a pair of dates? I'm not trying to code-golf this, as readability trumps brevity, but simply to learn whether Ruby natively supports the type of date arithmetic I've coded by hand.
require 'date'
str = '2015-06-15'
Date.parse(str).step(Date.today, 7).count # => 33
Date.parse(str).upto(Date.today).count.fdiv(7).round(2) # => 32.71
Might be easier to convert the dates to time and then divide the time difference by a week. You can round it however you want or ceil.
def weeks_since(date_string)
time_in_past = Date.parse(date_string).to_time
now = Time.now
time_difference = now - time_in_past
(time_difference / 1.week).round(2)
end
in_weeks (Rails 6.1+)
Rails 6.1 introduces new ActiveSupport::Duration conversion methods like in_seconds, in_minutes, in_hours, in_days, in_weeks, in_months, and in_years.
As a result, now, your problem can be solved as:
date_1 = Time.parse('2020-10-18 00:00:00 UTC')
date_2 = Time.parse('2020-08-13 03:35:38 UTC')
(date_2 - date_1).seconds.in_weeks.to_i.abs
# => 9
Here is a link to the corresponding PR.
I have an issue where, I'm trying to work out if a certain alert on a webpage is calculating sums correctly. I'm using Capybara and Cucumber.
I have an alert that calculates records that expire within 30 days. When selecting this alert, the records are listed in a table and the date is presented in the following format, "1 feb 2016"
What I want to do is somehow take today's date, compare it to the date returned in the table and ensure that it's >= 30 days from the date in the alert.
I'm able to set today's date as the same format using Time.strftime etc.
When I try things like:
And(/^I can see record "([\d]*)" MOT is calculated due within 30 days$/) do |selection1|
today = Time.now.strftime('%l %b %Y')
thirty_days = (today + 30)
first_30day_mot = first('#clickable-rows > tbody > tr:nth-child(' + selection1 + ') > td:nth-child(3)')
if today + first_30day_mot <= thirty_days
puts 'alert correct'
else
(error handler here)
end
end
As you can see, this is quite a mess.
I keep getting the error TypeError: no implicit conversion of Fixnum into String
If anyone can think of a neater way to do this, please put me out of my misery.
Thanks
There are at least a couple of things wrong with your attempt.
You're converting dates to strings and then trying to compare lengths of time with strings. You should be converting strings to dates and then comparing them
#first returns the element in the page not the contents of the element
It's not 100% clear from your code what you're trying to do, but from the test naming I think you just want to make sure the date in the 3rd td cell (which is in the 1 feb 2016 format) of a given row is less than 30 days from now. If so the following should do what you want
mot_element = first("#clickable-rows > tbody > tr:nth-child(#{selection1}) > td:nth-child(3)")
date_of_mot = Date.parse(mot_element.text)
if (date_of_mot - Date.today) < 30
puts 'alert correct'
else
#error handler
end
Beyond that, I'm not sure why you're using #first with that selector since it seems like it should only ever match one element on the page, so you might want to swap that to #find instead, which would get you the benefits of Capybaras waiting behavior. If you do actually need #first, you might consider passing the minimum: 1 option to make sure it waits a bit for the matching element to appear on the page (if this is the first step after clicking a button to go to a new page for instance)
Convert selection1 to the string explicitly (or, better, use string interpolation):
first_30day_mot = first("#clickable-rows > tbody > tr:nth-child(#{selection1}) > td:nth-child(3)")
Also, I suspect that one line below it should be converted to integer to add it to today:
first_30day_mot.to_i <= 30
UPD OK, I finally got time to take a more thorough look at. You do not need all these voodoo magic with days calculus:
# today = Time.now.strftime('%l %b %Y') # today will be a string " 3 Feb 2016"
# thirty_days = (today + 30) this was causing an error
# correct:
# today = DateTime.now # correct, but not needed
# plus_30_days = today + 30.days # correct, but not needed
first_30day_mot = first("#clickable-rows > tbody > tr:nth-child(#{selection1}) > td:nth-child(3)")
if 30 > first_30day_mot.to_i
...
Hope it helps.
I'd strongly recommend not using Cucumber to do this sort of test. You'll find its:
Quite hard to set up
Has a high runtime cost
Doesn't give enough benefit to justify the setup/runtime costs
Instead consider writing a unit test of the thing that provides the date. Generally a good unit test can easily run 10 to 100 times faster than a scenario.
Whilst with a single scenario you won't experience that much pain, once you have alot of scenarios like this the pain will accumulate. Part of the art of using Cucumber is to get plenty of bang for each scenario you write.
Can anyone help me make sense of this, please?
I am getting a very weird behaviour (reverse logic), when I am trying to use the following code.
require 'active_support/all'
c = {
id: 5,
years_of_experience: 4,
github_points: 293,
languages: ['C', 'Ruby', 'Python', 'Clojure'],
date_applied: 5.days.ago.to_date,
age: 26
}
c["date_applied"] > 15.days.ago.to_date - #works
c["date_applied"] < 15.days.ago.to_date - #doesnt work
c["date_applied"] gives a date value stored in a hash.
The latter makes more logical sense, but the first returns the right answer.
The code's behavior is correct, but I think I understand the confusion.
You're reading
c["date_applied"] > 15.days.ago
as:
Is the date applied more than 15 days ago?
and
c["date_applied"] < 15.days.ago
as:
Is the date applied less than 15 days ago?
and it's giving you the reverse of the answer you expect, right?
If that's the case, you should take a moment to understand how time comparisons operate. When you type date1 > date2, you're actually saying,
If I plot date1 and date2 on a number line with time increasing from left to right,
is date1 to the right of date2?
This is the same as when you type 2 > 1. It means,
If I plot 1 and 2 on a number line with the numbers increasing from left to right,
is 2 to the right of 1?
Given that this is how time comparisons operate, let's reexamine your code.
require 'active_support/all'
c = { date_applied: 5.days.ago.to_date }
c[:date_applied] > 15.days.ago.to_date
Correctly interpreted, this says
Is the date 5 days ago further rightward on a left-to-right timeline than the date 15 days ago?
and the answer is yes, or true.
If, on the other hand, you were to incorrectly interpret this as
Is 5 days ago more than 15 days ago?
you would get (or expect to get) the mistaken answer of no, or false.
The correct way to think about the task in English is to reframe the question of
Is date d more than n days ago?
and instead think of it as
Is date d earlier than the date n days ago?
and the correct code becomes apparent:
d.to_date < n.days.ago.to_date
If I understood your question correctly, this should explain it.
irb ## ruby-1.9.3-p448
require 'active_support/time'
c = {
id: 5,
years_of_experience: 4,
github_points: 293,
languages: ['C', 'Ruby', 'Python', 'Clojure'],
date_applied: 5.days.ago.to_date,
age: 26
}
(c[:date_applied] > 15.days.ago.to_date) - #true
(c[:date_applied] < 15.days.ago.to_date) - #false
###or you can try it by adding your own private methods###
class Fixnum
def days
self * 60 * 60 * 24 # we store seconds in a day
end
def ago
Time.now - self
end
end
I need to loop through all of the days and months for the past couple decades numerically as well as to have the name of the month and day for each date. Obviously a few series of loops can accomplish this, but I wanted to know the most concise ruby-like way to accomplish this.
Essentially I'd need output like this for each day over the past X years:
3 January 2011 and 1/3/2011
What's the cleanest approach?
Dates can work as a range, so it's fairly easy to iterate over a range. The only real trick is how to output them as a formatted string, which can be found in the Date#strftime method, which is documented here.
from_date = Date.new(2011, 1, 1)
to_date = Date.new(2011, 1, 10)
(from_date..to_date).each { |d| puts d.strftime("%-d %B %Y and %-m/%-d/%Y") }
# => 1 January 2011 and 1/1/2011
# => 2 January 2011 and 1/2/2011
# => ...
# => 9 January 2011 and 1/9/2011
# => 10 January 2011 and 1/10/2011
(Note: I recall having some bad luck a ways back with unpadded percent formats like %-d in Windows, but if the above doesn't work and you want them unpadded in that environment you can remove the dash and employ your own workarounds.)
Given start_date & end_date:
(start_date..end_date).each do |date|
# do things with date
end
as David said, this is possible because of Date#succ. You can use Date#strftime to get the date in any format you'd like.
See if you can construct a Range where the min and max are Date objects, then call .each on the range. If the Date object supports the succ method this should work.