One line if statement in Ruby - ruby

I have following piece of code:
if day > 31
day -= 31
month = "April"
end
Can I write it in one line different than:
if day > 31 then day -= 31 and month = "April" end
?
I've tried it like:
if day > 31 {day -= 31; month = "April"}
But it doesn't work

(day -= 31; month = "April") if day > 31
Alternate way (As suggested by #mudasobwa in comments below) :
day, month = day - 31, "April" if day > 31

Related

Trying to Sync weekdays between 2 years

I'm trying to come up with something to Sync weekdays between years
So far I can do it for a 4 year gap just by subtracting 364 days
For example
Monday Feb 1 2021 - 364 days becomes Monday Feb 3rd 2020
Friday Feb 19 2021 - 364 days becomes Friday Feb 21nd 2020
Tuesday Mar 2 2021 - 364 days becomes Tuesday Mar 3rd 2020
notice how the weekday is in perfect sync (Monday to Monday, Tuesday to Tuesday etc)
and I can do this for 2 years just by using 728 days (364 * 2)
and so on for 3 and 4 years
my problem is after 4 years it stops working
if I do the same thing for 5 Years (364*5)
Monday Feb 1st 2021 becomes Monday Feb 8th 2016
however I would want it to be Monday Feb 1st 2016
I cant seem to crack how to deal with this for 5 years on
This is using Zeller’s Rule to get the day number, and then just modifying the actual date when subtracting 365 days to get the day name to match. I'm sure there are edge cases (e.g. I believe this would currently allow a result of something like "March 33"), but this should get you started:
const days = ['Sun', 'Mon', 'Tue', 'Wed', 'Thu', 'Fri', 'Sat']
const months = ["January", "February", "March", "April", "May", "June",
"July", "August", "September", "October", "November", "December"
];
const dayNum = (day, month, year) => {
if (month <= 2) {
year--;
month += 12;
}
month-=2;
const lastTwoDigitsOfYear = year % 100;
const firstTwoDigitsOfYear = (year - lastTwoDigitsOfYear) / 100
let F = day + Math.floor((13*month-1)/5) + lastTwoDigitsOfYear + Math.floor(lastTwoDigitsOfYear/4) +Math.floor(firstTwoDigitsOfYear/4)-2*firstTwoDigitsOfYear
let f = F >= 0 ? (F % 7) : 7 + (F % 7);
return f
}
// Ok, so we can get the day for any date.
const syncedYearAdd = (day, month, year, numYears) => {
const d1 = dayNum(day, month, year);
const d2 = dayNum(day, month, year + numYears);
if (d1 < d2) {
day -= (d2 - d1)
} else if (d1 > d2) {
day += (d1 - d2);
}
if (day < 0) {
day += 7;
}
console.log(days[d1], day, months[month-1], year + numYears)
}
// Monday Feb 1st 2021
for (let i = 0; i <= 5; i++) {
syncedYearAdd(1, 2, 2021, i * -1)
}

Ruby - Get time at start of next minute

I'm looking for a concise way to get a Ruby Time object representing the top of the next minute (and hour/day/month/year, if possible). I want this to work in a pure Ruby environment, so the Rails function Time.change or similar doesn't fit the bill.
At first this seems simple - just add 1 to Time.now, but there are edge cases where if, for example, you try to instantiate a Time object with Time.now.min + 1 when the current minute is 59, you get an ArgumentError: min out of range. This goes for hour, day, and month as well.
I have some lengthy code that does the job. It's ugly, but I'm just experimenting:
def add_minute
now = Time.local
year = now.year
month = now.month
day = now.day
hour = now.hour
min = now.min
if now.min == 59
if now.hour == 23
if now.day == Date.civil(now.year, now.month, -1).day
if month == 12
year = year + 1
month = 1
day = 1
hour = 0
min = 0
else
month = now.month + 1
day = 1
hour = 0
min = 0
end
else
day = now.day + 1
hour = 0
min = 0
end
else
hour = now.hour + 1
min = 0
end
else
min = now.min + 1
end
Time.local year, month, day, hour, min, 0
end
This seems absurdly verbose for what seems like it should be a simple or built-in task, but I haven't found a native Ruby solution. Does one exist?
You could convert the Time object to UNIX epoch time (seconds since 1970) using #to_i, add 60 s, and then convert back to a Time object.
time_unix = Time.now.to_i
time_unix_one_min_later = time_unix + 60
time_one_min_later = t = Time.at(time_unix_one_min_later)
time_one_min_later_rounded_down = Time.new(t.year, t.month, t.day, t.hour, t.min)
EDIT: Even shorter - you can just add integer seconds to Time.now directly:
time_one_min_later = t = Time.now + 60
time_one_min_later_rounded_down = Time.new(t.year, t.month, t.day, t.hour, t.min)
EDIT 2: One-liner - just subtract Time.now.sec:
time_one_min_later_rounded_down = Time.now + 60 - Time.now.sec
Other option, given one second to midnight:
require 'time'
now = Time.strptime('2018-12-31 23:59:59', '%Y-%m-%d %H:%M:%S')
Within one minute:
Time.at(now + 60) #=> 2019-01-01 00:00:59 +0100
Time.at(now + 60 - now.sec) #=> 2019-01-01 00:00:00 +0100
You get: # HAPPY NEW YEAR!
Ruby has built in methods for adding months (>>) and days (+). A year is 12 months, and an hour is 1/24th of a day.
require 'date'
def add_time(time, year: 0 ,month: 0, day: 0, hour: 0, minute: 0)
time >>= 12*year
time >>= month
time += day
time += Rational(hour,24) # or (hour/24.0) if you dislike rationals
time += Rational(minute, 24*60) # (minute/24.0*60) if you dislike rationals
end
p t = DateTime.now
p add_time(t, year: 1, minute: 30)
Not that clean without ActiveSupport:
new_date = (DateTime.now + 1.to_f / (60*24))
DateTime.new(new_date.year, new_date.month, new_date.day, new_date.hour, new_date.minute)
We can make this calculation easier to understand by getting the current number of seconds we are through the day. (optional)
DateTime.current.to_i gives us the number of seconds since 1970
DateTime.current.to_i - DateTime.current.beginning_of_day.to_i gives us the number of seconds since the start of the day.
(((number_of_seconds_through_the_day + 60)/60) * 60) gives us the number of seconds we will be at when the next minute starts
Then we subtract the two to give us the number of seconds until the top of the next minute.
If we want the exact time at start of the next minute then we can do:
DateTime.current + seconds_until_start_of_the_next_minute.seconds
def seconds_until_start_of_the_next_minute
number_of_seconds_through_the_day = DateTime.current.to_i - DateTime.current.beginning_of_day.to_i
number_of_seconds_through_the_day_at_next_minute = (((number_of_seconds_through_the_day + 60)/60) * 60)
seconds_until_next_minute_starts = number_of_seconds_through_the_day_at_next_minute - number_of_seconds_through_the_day
return seconds_until_next_minute_starts
end

How do I convert seconds since Epoch to current date and time?

I know I posted this a while ago, but I figured out the solution. I wrote this code for a game called Roblox, but I'm just posting the code here in case anyone else who has this same problem needs a solution. Anyways, here's the code:
outputTime = true -- true: will print the current time to output window. false: won't print time
createVariable = true -- true: creates variables under game.Lighting. false: won't create variables
-----------------------------------------------------------------------------------------------
--DO NOT EDIT BELOW----------------------------------------------------------------------------
-----------------------------------------------------------------------------------------------
if(createVariable) then
yearVar = Instance.new("IntValue", game.Lighting)
yearVar.Name = "Year"
yearVar.Value = 0
monthVar = Instance.new("IntValue", game.Lighting)
monthVar.Name = "Month"
monthVar.Value = 0
dayVar = Instance.new("IntValue", game.Lighting)
dayVar.Name = "Day"
dayVar.Value = 0
hourVar = Instance.new("IntValue", game.Lighting)
hourVar.Name = "Hour"
hourVar.Value = 0
minuteVar = Instance.new("IntValue", game.Lighting)
minuteVar.Name = "Minute"
minuteVar.Value = 0
secondVar = Instance.new("IntValue", game.Lighting)
secondVar.Name = "Second"
secondVar.Value = 0
dayOfWeek = Instance.new("StringValue", game.Lighting)
dayOfWeek.Name = "DayOfWeek"
dayOfWeek.Value = "Thursday"
end
function giveZero(data)
if string.len(data) <= 1 then
return "0" .. data
else
return data
end
end
function hasDecimal(value)
if not(value == math.floor(value)) then
return true
else
return false
end
end
function isLeapYear(year)
if(not hasDecimal(year / 4)) then
if(hasDecimal(year / 100)) then
return true
else
if(not hasDecimal(year / 400)) then
return true
else
return false
end
end
else
return false
end
end
local eYear = 1970
local timeStampDayOfWeak = 5
local secondsInHour = 3600
local secondsInDay = 86400
local secondsInYear = 31536000
local secondsInLeapYear = 31622400
local monthWith28 = 2419200
local monthWith29 = 2505600
local monthWith30 = 2592000
local monthWith31 = 2678400
local monthsWith30 = {4, 6, 9, 11}
local monthsWith31 = {1, 3, 5, 7, 8, 10, 12}
local daysSinceEpoch = 0
local DOWAssociates = {"Tursday", "Friday", "Saturday", "Sunday", "Monday", "Tuesday", "Wednesday"}
while(true) do
now = tick()
year = 1970
secs = 0
daysSinceEpoch = 0
while((secs + secondsInLeapYear) < now or (secs + secondsInYear) < now) do
if(isLeapYear(year+1)) then
if((secs + secondsInLeapYear) < now) then
secs = secs + secondsInLeapYear
year = year + 1
daysSinceEpoch = daysSinceEpoch + 366
end
else
if((secs + secondsInYear) < now) then
secs = secs + secondsInYear
year = year + 1
daysSinceEpoch = daysSinceEpoch + 365
end
end
end
secondsRemaining = now - secs
monthSecs = 0
yearIsLeapYear = isLeapYear(year)
month = 1 -- January
while((monthSecs + monthWith28) < secondsRemaining or (monthSecs + monthWith30) < secondsRemaining or (monthSecs + monthWith31) < secondsRemaining) do
if(month == 1) then
if((monthSecs + monthWith31) < secondsRemaining) then
month = 2
monthSecs = monthSecs + monthWith31
daysSinceEpoch = daysSinceEpoch + 31
else
break
end
end
if(month == 2) then
if(not yearIsLeapYear) then
if((monthSecs + monthWith28) < secondsRemaining) then
month = 3
monthSecs = monthSecs + monthWith28
daysSinceEpoch = daysSinceEpoch + 28
else
break
end
else
if((monthSecs + monthWith29) < secondsRemaining) then
month = 3
monthSecs = monthSecs + monthWith29
daysSinceEpoch = daysSinceEpoch + 29
else
break
end
end
end
if(month == 3) then
if((monthSecs + monthWith31) < secondsRemaining) then
month = 4
monthSecs = monthSecs + monthWith31
daysSinceEpoch = daysSinceEpoch + 31
else
break
end
end
if(month == 4) then
if((monthSecs + monthWith30) < secondsRemaining) then
month = 5
monthSecs = monthSecs + monthWith30
daysSinceEpoch = daysSinceEpoch + 30
else
break
end
end
if(month == 5) then
if((monthSecs + monthWith31) < secondsRemaining) then
month = 6
monthSecs = monthSecs + monthWith31
daysSinceEpoch = daysSinceEpoch + 31
else
break
end
end
if(month == 6) then
if((monthSecs + monthWith30) < secondsRemaining) then
month = 7
monthSecs = monthSecs + monthWith30
daysSinceEpoch = daysSinceEpoch + 30
else
break
end
end
if(month == 7) then
if((monthSecs + monthWith31) < secondsRemaining) then
month = 8
monthSecs = monthSecs + monthWith31
daysSinceEpoch = daysSinceEpoch + 31
else
break
end
end
if(month == 8) then
if((monthSecs + monthWith31) < secondsRemaining) then
month = 9
monthSecs = monthSecs + monthWith31
daysSinceEpoch = daysSinceEpoch + 31
else
break
end
end
if(month == 9) then
if((monthSecs + monthWith30) < secondsRemaining) then
month = 10
monthSecs = monthSecs + monthWith30
daysSinceEpoch = daysSinceEpoch + 30
else
break
end
end
if(month == 10) then
if((monthSecs + monthWith31) < secondsRemaining) then
month = 11
monthSecs = monthSecs + monthWith31
daysSinceEpoch = daysSinceEpoch + 31
else
break
end
end
if(month == 11) then
if((monthSecs + monthWith30) < secondsRemaining) then
month = 12
monthSecs = monthSecs + monthWith30
daysSinceEpoch = daysSinceEpoch + 30
else
break
end
end
end
day = 1 -- 1st
daySecs = 0
daySecsRemaining = secondsRemaining - monthSecs
while((daySecs + secondsInDay) < daySecsRemaining) do
day = day + 1
daySecs = daySecs + secondsInDay
daysSinceEpoch = daysSinceEpoch + 1
end
hour = 0 -- Midnight
hourSecs = 0
hourSecsRemaining = daySecsRemaining - daySecs
while((hourSecs + secondsInHour) < hourSecsRemaining) do
hour = hour + 1
hourSecs = hourSecs + secondsInHour
end
minute = 0 -- Midnight
minuteSecs = 0
minuteSecsRemaining = hourSecsRemaining - hourSecs
while((minuteSecs + 60) < minuteSecsRemaining) do
minute = minute + 1
minuteSecs = minuteSecs + 60
end
second = math.floor(now % 60)
year = giveZero(year)
month = giveZero(month)
day = giveZero(day)
hour = giveZero(hour)
minute = giveZero(minute)
second = giveZero(second)
remanderForDOW = daysSinceEpoch % 7
DOW = DOWAssociates[remanderForDOW + 1]
if(createVariable) then
yearVar.Value = year
monthVar.Value = month
dayVar.Value = day
hourVar.Value = hour
minuteVar.Value = minute
secondVar.Value = second
dayOfWeek.Value = DOW
end
if(outputTime) then
str = "Year: " .. year .. ", Month: " .. month .. ", Day: " .. day .. ", Hour: " .. hour .. ", Minute: " .. minute .. ", Second: ".. second .. ", Day of Week: " .. DOW
print(str)
end
wait(1)
end
----ORIGINAL POST----
What are the formulas for calculating the following given no resources except the seconds since Epoch?
Here's a list of what I need:
Current Month of year Ex: 7
Current day of month Ex: 25
Current day of week Ex: Thursday (1-7 would be acceptable)
Current hour of day Ex: 22
Current minute of hour Ex: 34
Current second of minute: 07
Here is some Lua code adapted from some C code found by Google. It does not handle timezones or Daylight Saving Time and so the outputs refers to Universal Coordinated Time (UTC).
-- based on http://www.ethernut.de/api/gmtime_8c_source.html
local floor=math.floor
local DSEC=24*60*60 -- secs in a day
local YSEC=365*DSEC -- secs in a year
local LSEC=YSEC+DSEC -- secs in a leap year
local FSEC=4*YSEC+DSEC -- secs in a 4-year interval
local BASE_DOW=4 -- 1970-01-01 was a Thursday
local BASE_YEAR=1970 -- 1970 is the base year
local _days={
-1, 30, 58, 89, 119, 150, 180, 211, 242, 272, 303, 333, 364
}
local _lpdays={}
for i=1,2 do _lpdays[i]=_days[i] end
for i=3,13 do _lpdays[i]=_days[i]+1 end
function gmtime(t)
print(os.date("!\n%c\t%j",t),t)
local y,j,m,d,w,h,n,s
local mdays=_days
s=t
-- First calculate the number of four-year-interval, so calculation
-- of leap year will be simple. Btw, because 2000 IS a leap year and
-- 2100 is out of range, this formula is so simple.
y=floor(s/FSEC)
s=s-y*FSEC
y=y*4+BASE_YEAR -- 1970, 1974, 1978, ...
if s>=YSEC then
y=y+1 -- 1971, 1975, 1979,...
s=s-YSEC
if s>=YSEC then
y=y+1 -- 1972, 1976, 1980,... (leap years!)
s=s-YSEC
if s>=LSEC then
y=y+1 -- 1971, 1975, 1979,...
s=s-LSEC
else -- leap year
mdays=_lpdays
end
end
end
j=floor(s/DSEC)
s=s-j*DSEC
local m=1
while mdays[m]<j do m=m+1 end
m=m-1
local d=j-mdays[m]
-- Calculate day of week. Sunday is 0
w=(floor(t/DSEC)+BASE_DOW)%7
-- Calculate the time of day from the remaining seconds
h=floor(s/3600)
s=s-h*3600
n=floor(s/60)
s=s-n*60
print("y","j","m","d","w","h","n","s")
print(y,j+1,m,d,w,h,n,s)
end
local t=os.time()
gmtime(t)
t=os.time{year=1970, month=1, day=1, hour=0} gmtime(t)
t=os.time{year=1970, month=1, day=3, hour=0} gmtime(t)
t=os.time{year=1970, month=1, day=2, hour=23-3, min=59, sec=59} gmtime(t)
The formula is not simple for a few reasons, especially leap years. You should probably use the date function on this page rather than trying to calculate it yourself.
You could use luatz
x = 1234567890
t = require "luatz.timetable".new_from_timestamp ( x )
print(t.year,t.month,t.day,t.hour,t.min,t.sec,t.yday,t.wday)
-- Or just
print(t:rfc_3339())
This is how I do it.
> time0=os.time()
> time0
1571439964
> os.date("%Y%m%d%H%M%S",time0)
20191019120604
>
os.date is a standard Lua function, if passing the first argument as "%t", it will return a table containing the following fields: year (four digits), month (1--12), day (1--31), hour (0--23), min (0--59), sec (0--61), wday (weekday, Sunday is 1), yday (day of the year), and isdst (daylight saving flag, a boolean).
Give it a test:
time = os.time()
print("time since epoch: " .. time)
date = os.date("*t", time)
print("year: " .. date.year)
print("month: " .. date.month)
print("day: " .. date.day)
print("hour: " .. date.hour)
print("minute: " .. date.min)
print("second: " .. date.sec)
print("weekday: " .. date.wday)
Output:
time since epoch: 1374826427
year: 2013
month: 7
day: 26
hour: 16
minute: 13
second: 47
weekday: 6
A much faster solution would be to use my method, which I haven't really seen anyone else use because most have access to os.date()
Since I do not have access to os.date(), here is my solution:
local tabIndexOverflow = function(seed, table)
-- This subtracts values from the table from seed until an overflow
-- This can be used for probability :D
for i = 1, #table do
if seed - table[i] <= 0 then
return i, seed
end
seed = seed - table[i]
end
end
local getDate = function(unix)
-- Given unix date, return string date
assert(unix == nil or type(unix) == "number" or unix:find("/Date%((%d+)"), "Please input a valid number to \"getDate\"")
local unix = (type(unix) == "string" and unix:match("/Date%((%d+)") / 1000 or unix or os.time()) -- This is for a certain JSON compatability. It works the same even if you don't need it
local dayCount, year, days, month = function(yr) return (yr % 4 == 0 and (yr % 100 ~= 0 or yr % 400 == 0)) and 366 or 365 end, 1970, math.ceil(unix/86400)
while days >= dayCount(year) do days = days - dayCount(year) year = year + 1 end -- Calculate year and days into that year
month, days = tabIndexOverflow(days, {31,(dayCount(year) == 366 and 29 or 28),31,30,31,30,31,31,30,31,30,31}) -- Subtract from days to find current month and leftover days
-- hours = hours > 12 and hours - 12 or hours == 0 and 12 or hours -- Change to proper am or pm time
-- local period = hours > 12 and "pm" or "am"
-- Formats for you!
-- string.format("%d/%d/%04d", month, days, year)
-- string.format("%02d:%02d:%02d %s", hours, minutes, seconds, period)
return {Month = month, day = days, year = year, hours = math.floor(unix / 3600 % 24), minutes = math.floor(unix / 60 % 60), seconds = math.floor(unix % 60)}
end
You are, however, on your own when it comes to finding the day of the week. I never cared to find whether it be the day of Thor or the day of Frige.

Converting date to string in vb script

I have date like 6/24/2013 , i want to get only month name and date like June 24th as output in vb script.
Use two functions to deal with the two (sub) problems - name of month, ordinal of number - separately:
Option Explicit
Dim n
For n = -2 To 10
WScript.Echo fmtDate(DateAdd("d", n, Date))
Next
Function fmtDate(dtX)
fmtDate = MonthName(Month(dtX)) & " " & ordinal(Day(dtX))
End Function
' !! http://stackoverflow.com/a/4011232/603855
Function ordinal(n)
Select Case n Mod 10
case 1 : ordinal = "st"
case 2 : ordinal = "nd"
case 3 : ordinal = "rd"
case Else : ordinal = "th"
End Select
ordinal = n & ordinal
End Function
output:
June 22nd
June 23rd
June 24th
June 25th
June 26th
June 27th
June 28th
June 29th
June 30th
July 1st
July 2nd
July 3rd
July 4th
Update:
(Hopefully) improved version of ordinal():
Function ordinal(n)
Select Case n Mod 31
case 1, 21, 31 : ordinal = "st"
case 2, 22 : ordinal = "nd"
case 3, 23 : ordinal = "rd"
case Else : ordinal = "th"
End Select
ordinal = n & ordinal
End Function

Ruby Determine Season (Fall, Winter, Spring or Summer)

I am working on a script that is supposed to determine the "season" of the year based on date ranges:
For Example:
January 1 - April 1: Winter
April 2 - June 30: Spring
July 1 - September 31: Summer
October 1 - December 31: Fall
I am not sure how the best way (or the best ruby way) to go about doing this. Anyone else run across how to do this?
31 September?
As leifg suggested, here it is in code:
require 'Date'
class Date
def season
# Not sure if there's a neater expression. yday is out due to leap years
day_hash = month * 100 + mday
case day_hash
when 101..401 then :winter
when 402..630 then :spring
when 701..930 then :summer
when 1001..1231 then :fall
end
end
end
Once defined, call it e.g. like this:
d = Date.today
d.season
You could try with ranges and Date objects:
http://www.tutorialspoint.com/ruby/ruby_ranges.htm
without ranges.
require 'date'
def season
year_day = Date.today.yday().to_i
year = Date.today.year.to_i
is_leap_year = year % 4 == 0 && year % 100 != 0 || year % 400 == 0
if is_leap_year and year_day > 60
# if is leap year and date > 28 february
year_day = year_day - 1
end
if year_day >= 355 or year_day < 81
result = :winter
elsif year_day >= 81 and year_day < 173
result = :spring
elsif year_day >= 173 and year_day < 266
result = :summer
elsif year_day >= 266 and year_day < 355
result = :autumn
end
return result
end
Neil Slater's answer's approach is great but for me those dates aren't quite correct. They show fall ending on December 31st which isn't the case in any scenario I can think of.
Using the northern meteorological seasons:
Spring runs from March 1 to May 31;
Summer runs from June 1 to August 31;
Fall (autumn) runs from September 1 to November 30; and
Winter runs from December 1 to February 28 (February 29 in a leap year).
The code would need to be updated to:
require "date"
class Date
def season
day_hash = month * 100 + mday
case day_hash
when 101..300 then :winter
when 301..531 then :spring
when 601..831 then :summer
when 901..1130 then :fall
when 1201..1231 then :winter
end
end
end

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