For the purpose of an event study, I would like to create lagging and leading variables in my data set. Unfortunately, my data is not in a balanced panel format.
My data set looks as:
clear
input id year month binary str7 implement
28845421 2007 3 0 2008-1
29118744 2018 10 1 2012-6
29118744 2016 7 1 2016-7
29183010 2019 3 1 2010-1
29320027 2013 3 0 2015-2
end
. list
+---------------------------------------------+
| id year month binary implem~t |
|---------------------------------------------|
1. | 2.88e+07 2007 3 0 2008-1 |
2. | 2.91e+07 2018 10 1 2012-6 |
3. | 2.91e+07 2016 7 1 2016-7 |
4. | 2.92e+07 2019 3 1 2010-1 |
5. | 2.93e+07 2013 3 0 2015-2 |
+---------------------------------------------+
The variable binary equals 1 at the point when its year and month combination has reached the implement date. Each observation is represented by an identifier id.
The goal is to create lagging and leading variables of the binary: binary-5, binary-4, ..., binary+1, binary+2, .... In other words, I want to shift implement by n years of increments / decrement to create the new binary variables.
How can I create such variables in Stata?
I just want to run the job twice per week. Every Sunday 11 PM and every Friday 11 pm I just want to trigger the job automatically. I successfully implemented for one scheduler but not sure how to use two in single .
Sunday scheduler :
H 11 * * 0
Friday scheduler:
H 11 * * 6
For scheduling the job below pattern need to be followed:-
0 - Sun Sunday
1 - Mon Monday
2 - Tue Tuesday
3 - Wed Wednesday
4 - Thu Thursday
5 - Fri Friday
6 - Sat Saturday
7 - Sun Sunday
For your case you can follow the below:-
0 23 * * 0,5
You better understanding of "H" in Jobs Scheduler follow this:
Maybe something like this. Note there's a couple changes from your example, changed 11 to 23, 11 is 11am, 23 is 11pm and using 5 instead of 6 for Friday.
H 23 * * 0,5
But note with the "H" it isn't going to run at exactly 11pm, from the Jenkins docs:
The H symbol can be thought of as a random value over a range, but it
actually is a hash of the job name, not a random function, so that the
value remains stable for any given project.
If you want it to run closer to 11pm, maybe something like this
H(1-5) 23 * * 0,5
I've got a dataset with repeated measures that looks roughly like this:
ID v1 v2 v3 v4
1 3 4 2 NA
1 2 NA 6 7
2 4 3 6 4
2 NA 2 7 9
. . . . .
n . . . .
What I want to know is how many NAs are there for each participants over the variables v1 - v4 (e.g. participant 1 is missing 2 of 8 responses)?
Missing Values are always displayed per Variable not per participant so how do I do this? Maybe there is a way using the AGGREGATE command with ID as BREAK?
Use COUNT to count the missing values as a new variable and then aggregate by the Id or split files by I'd and freq.
I have a dataset structured like below;
id contracthours13 contracthours14 contracthours13u contracthours14u
12 . 13 . 13
13 30 30 . .
14 . . 15 16
15 . 5 6 7
If contracthours13 is missing I want the value in contracthours14 to move across. If this is missing then I want contacthours13u to move across and the same then for contracthours14u if the previous 3 are all missing. I know this is fairly simple syntax but I just can't get my head around how to do it without having the run simpler syntax 3 times. If anyone could help it would be greatly appreciated.
Edit: below is what I would like my dataset to look like afterwards.
id contracthours13
12 13
13 30
14 15
15 5
Look up VECTOR / LOOP examples.
DATA LIST FREE / ID CH13 CH14 CH13U CH14U.
BEGIN DATA.
1 -1 13 -1 -1
2 30 30 -1 -1
3 -1 -1 15 16
4 -1 5 6 7
END DATA.
DATASET NAME DSRaw.
RECODE ALL (-1=SYSMIS).
VECTOR V= CH14 TO CH14U.
LOOP #i = 1 TO 3 IF (NVALID(CH13)=0).
COMPUTE CH13=V(#i).
END LOOP IF NVALID(V(#i))=1.
LIST.
EXE.
**List**
ID CH13 CH14 CH13U CH14U
1.00 13.00 13.00 . .
2.00 30.00 30.00 . .
3.00 15.00 . 15.00 16.00
4.00 5.00 5.00 6.00 7.00
Number of cases read: 4 Number of cases listed: 4
I need to calculate the number of business days between two dates. How can I pull that off using Ruby (or Rails...if there are Rails-specific helpers).
Likewise, I'd like to be able to add business days to a given date.
So if a date fell on a Thursday and I added 3 business days, it would return the next Tuesday.
Take a look at business_time. It can be used for both the things you're asking.
Calculating business days between two dates:
wednesday = Date.parse("October 17, 2018")
monday = Date.parse("October 22, 2018")
wednesday.business_days_until(monday) # => 3
Adding business days to a given date:
4.business_days.from_now
8.business_days.after(some_date)
Historical answer
When this question was originally asked, business_time didn't provide the business_days_until method so the method below was provided to answer the first part of the question.
This could still be useful to someone who didn't need any of the other functionality from business_time and wanted to avoid adding an additional dependency.
def business_days_between(date1, date2)
business_days = 0
date = date2
while date > date1
business_days = business_days + 1 unless date.saturday? or date.sunday?
date = date - 1.day
end
business_days
end
This can also be fine tuned to handle the cases that Tipx mentions in the way that you would like.
We used to use the algorithm suggested in the mikej's answer and discovered that calculating 25,000 ranges of several years each takes 340 seconds.
Here's another algorithm with asymptotic complexity O(1). It does the same calculations in 0.41 seconds.
# Calculates the number of business days in range (start_date, end_date]
#
# #param start_date [Date]
# #param end_date [Date]
#
# #return [Fixnum]
def business_days_between(start_date, end_date)
days_between = (end_date - start_date).to_i
return 0 unless days_between > 0
# Assuming we need to calculate days from 9th to 25th, 10-23 are covered
# by whole weeks, and 24-25 are extra days.
#
# Su Mo Tu We Th Fr Sa # Su Mo Tu We Th Fr Sa
# 1 2 3 4 5 # 1 2 3 4 5
# 6 7 8 9 10 11 12 # 6 7 8 9 ww ww ww
# 13 14 15 16 17 18 19 # ww ww ww ww ww ww ww
# 20 21 22 23 24 25 26 # ww ww ww ww ed ed 26
# 27 28 29 30 31 # 27 28 29 30 31
whole_weeks, extra_days = days_between.divmod(7)
unless extra_days.zero?
# Extra days start from the week day next to start_day,
# and end on end_date's week date. The position of the
# start date in a week can be either before (the left calendar)
# or after (the right one) the end date.
#
# Su Mo Tu We Th Fr Sa # Su Mo Tu We Th Fr Sa
# 1 2 3 4 5 # 1 2 3 4 5
# 6 7 8 9 10 11 12 # 6 7 8 9 10 11 12
# ## ## ## ## 17 18 19 # 13 14 15 16 ## ## ##
# 20 21 22 23 24 25 26 # ## 21 22 23 24 25 26
# 27 28 29 30 31 # 27 28 29 30 31
#
# If some of the extra_days fall on a weekend, they need to be subtracted.
# In the first case only corner days can be days off,
# and in the second case there are indeed two such days.
extra_days -= if start_date.tomorrow.wday <= end_date.wday
[start_date.tomorrow.sunday?, end_date.saturday?].count(true)
else
2
end
end
(whole_weeks * 5) + extra_days
end
business_time has all the functionallity you want.
From the readme:
#you can also calculate business duration between two dates
friday = Date.parse("December 24, 2010")
monday = Date.parse("December 27, 2010")
friday.business_days_until(monday) #=> 1
Adding business days to a given date:
some_date = Date.parse("August 4th, 1969")
8.business_days.after(some_date) #=> 14 Aug 1969
Here is my (non gem and non holiday) weekday count example:
first_date = Date.new(2016,1,5)
second_date = Date.new(2016,1,12)
count = 0
(first_date...second_date).each{|d| count+=1 if (1..5).include?(d.wday)}
count
Take a look at Workpattern. It alows you to specify working and resting periods and can add/subtract durations to/from a date as well as calculate the minutes between two dates.
You can set up workpatterns for different scenarios such as mon-fri working or sun-thu and you can have holidays and whole or part days.
I wrote this as away to learn Ruby. Still need to make it more Ruby-ish.
Based on #mikej's answer. But this also takes into account holidays, and returns a fraction of a day (up to the hour accurancy):
def num_days hi, lo
num_hours = 0
while hi > lo
num_hours += 1 if hi.workday? and !hi.holiday?
hi -= 1.hour
end
num_hours.to_f / 24
end
This uses the holidays and business_time gems.
Simple script to calculate total number of working days
require 'date'
(DateTime.parse('2016-01-01')...DateTime.parse('2017-01-01')).
inject({}) do |s,e|
s[e.month]||=0
if((1..5).include?(e.wday))
s[e.month]+=1
end
s
end
# => {1=>21, 2=>21, 3=>23, 4=>21, 5=>22, 6=>22, 7=>21, 8=>23, 9=>22, 10=>21, 11=>22, 12=>22}
There are two problems with the most popular solutions listed above:
They involve loops to count every single day between each date (meaning that performance gets worse the further apart the dates are.
They are unclear about whether they count from the beginning of the day or the end. If you count from the morning, there is one weekday between Friday and Saturday. If you count from the night, there are zero weekdays between Friday and Saturday.
After stewing over it, I propose this solution that addresses both problems. The below takes a reference date and an other date and calculates the number of weekdays between them (returning a negative number if other is before the reference date). The argument eod_base controls whether counting is done from end of day (eod) or start of day. It could be written more compactly but hopefully it's relatively easy to understand and it doesn't require gems or rails.
require 'date'
def weekdays_between(ref,otr,eod_base=true)
dates = [ref,otr].sort
return 0 if dates[0] == dates[1]
full_weeks = ((dates[1]-dates[0])/7).floor
dates[eod_base ? 0 : 1] += (eod_base ? 1 : -1)
part_week = Range.new(dates[0],dates[1])
.inject(0){|m,v| (v.wday >=1 && v.wday <= 5) ? (m+1) : m }
return (otr <=> ref) * (full_weeks*5 + part_week)
end