Create Week Variable Based on Date Time Variable - sas

Simple question, I think.
I have a checkin_date_time variable in a database with thousands of unique records.
Database
ID checkin_date_time
1 January 01, 2019 11:36:50
2 January 01, 2019 11:36:55
....
60000 December 31, 2019 11:36:50
60001 December 31, 2019 11:36:55
I would like to create a 'week' variable based on the checkin_date_time variable. So for example 'January 01, 2019 11:36:55' would equal week 1 and 'December 31, 2019 15:16:57' would equal week 52.
Desired Output
ID datetime Week
1 January 01, 2019 11:36:50 1
2 January 01, 2019 11:36:55 1
....
60000 December 31, 2019 11:36:50 52
60001 December 31, 2019 11:36:55 52
I tried using the following code but its saying my
data testl;
set ed_tat;
week=week(checkin_date_time);
run;
NOTE: Missing values were generated as a result of performing an operation on missing values.
Each place is given by: (Number of times) at (Line):(Column).


Week operates on a date variable, use DATEPART() to get the date first and then determine the week.
week = week(datepart(checkin_date_time));

Related

display each day's date python from today

I was able to display the week that starts every Saturday by:
today = now().date()
sat_offset = (today.weekday() - 5) % 7
week_start = today - datetime.timedelta(days=sat_offset)
This will display the week from last Saturday but how would I show the dates of each day forward as well? So if the week: Oct. 27, 2018 is display it should say:
Saturday : Oct. 27, 2018
Sunday: Oct. 28, 2018
Monday: Oct. 29, 2018
Tuesday: Oct. 30, 2018
Wednesday: Oct. 31, 2018
Thursday: Nov. 01, 2018
Friday: Nov. 02, 2018
Thank you for your help.

You can iterate through the days of the week using range and time delta like so:
for i in range(7):
week_start += datetime.timedelta(days=1)
print(week_start.strftime("%A %d. %B %Y"))
This will produce a dates like:
Monday : Oct. 28, 2018
Tuesday : Oct. 29, 2018
Wednesday : Oct. 30, 2018
Thursday : Oct. 31, 2018
Friday : Nov. 01, 2018
Saturday : Nov. 02, 2018
Sunday : Nov. 03, 2018
You can format the string how ever you want. Here is some info on dates in python.

Trouble rounding times in Stata

I am attempting to round times to the nearest 15 minute interval in Stata, so for instance Dec 31, 2017 23:58 would become Jan 01, 2018 00:00. I have time stored (based on my understanding of the documentation) as the number of milliseconds since the start of 1960. So I thought this would do it:
gen round = round(datetime, 60000*15)
However, this doesn't quite work. For instance Nov 03, 2017 19:45:27 becomes Nov 03, 2017 19:46:01, when I think I should become 19:45:00. Does anyone know what I'm missing here?

Let's show a worked example illustrating my comment that you need to store datetime values as double rather than float.
. clear
. set obs 1
number of observations (_N) was 0, now 1
. gen double datetime = clock("Nov 03, 2017 19:45:27","MDYhms")
. gen round_f = round(datetime, 60000*15)
. gen double round_d = round(datetime, 60000*15)
. format datetime round_f round_d %tc
. list, clean noobs
datetime round_f round_d
03nov2017 19:45:27 03nov2017 19:46:01 03nov2017 19:45:00

get mean of values that fit a specific criteria (pattern matching)

I asked this question before and got a reply that solved it for me. I have a dataframe that looks like this:
id weekdays halflife
241732222300860000 Friday, Aug 31, 2012, 22 0.4166666667
241689170123309000 Friday, Aug 31, 2012, 19 0.3833333333
241686878137512000 Friday, Aug 31, 2012, 19 0.4
241651117396738000 Friday, Aug 31, 2012, 16 1.5666666667
241635163505820000 Friday, Aug 31, 2012, 15 0.95
241633401382265000 Friday, Aug 31, 2012, 15 2.3666666667
And I would like to get average half life of items that were created on Monday, then on Tuesday...etc. (My date range spans over 6 months).
To get the date values I used strptime and difftime. Also, I found the maximum halflife with max(df$halflife), how can I find which id it corresponds to?
Reproducible code:
structure(list(id = c(241732222300860416, 241689170123309056,
241686878137511936, 241651117396738048, 241635163505819648, 241633401382264832
), weekdays = c("Friday, Aug 31, 2012, 22", "Friday, Aug 31, 2012, 19",
"Friday, Aug 31, 2012, 19", "Friday, Aug 31, 2012, 16", "Friday, Aug 31, 2012, 15",
"Friday, Aug 31, 2012, 15"), halflife = structure(c(0.416666666666667,
0.383333333333333, 0.4, 1.56666666666667, 0.95, 2.36666666666667
), class = "difftime", units = "mins")), .Names = c("id",
"weekdays", "halflife"), row.names = c(NA, 6L), class = "data.frame")
So now, I have an average half life value for all mondays, tuesdays...etc. How can I get the average value for all hours within those weekdays, i.e.: Average half life of all items that were created on all Mondays at 9am, then 10am, then 11am..etc. And then Tuesday at 9am, 10am, 11am..etc. The dates in the weekdays column is formatted so that the last number after the comma is the hour it was created at. I am really bad with regular expressions and pattern matching, which is why I am asking this follow-up question.

with base packages you can do following.
> mydf
id weekdays halflife
1 2.417322e+17 Friday, Aug 31, 2012, 22 0.4166667 mins
2 2.416892e+17 Friday, Aug 31, 2012, 19 0.3833333 mins
3 2.416869e+17 Friday, Aug 31, 2012, 19 0.4000000 mins
4 2.416511e+17 Friday, Aug 31, 2012, 16 1.5666667 mins
5 2.416352e+17 Friday, Aug 31, 2012, 15 0.9500000 mins
6 2.416334e+17 Friday, Aug 31, 2012, 15 2.3666667 mins
Instead of using regex, we can just use strsplit on each element of weekdays, unlist the result, and it back in 4 column format as matrix and cbind it back with mydf.
> mydf2 <- cbind(mydf, matrix(unlist(sapply(mydf$weekdays, strsplit, split=',')), byrow=TRUE, ncol=4, dimnames=list(1:nrow(mydf), c('Weekday', 'Day', 'Year', 'Hour'))))
> mydf2
id weekdays halflife Weekday Day Year Hour
1 2.417322e+17 Friday, Aug 31, 2012, 22 0.4166667 mins Friday Aug 31 2012 22
2 2.416892e+17 Friday, Aug 31, 2012, 19 0.3833333 mins Friday Aug 31 2012 19
3 2.416869e+17 Friday, Aug 31, 2012, 19 0.4000000 mins Friday Aug 31 2012 19
4 2.416511e+17 Friday, Aug 31, 2012, 16 1.5666667 mins Friday Aug 31 2012 16
5 2.416352e+17 Friday, Aug 31, 2012, 15 0.9500000 mins Friday Aug 31 2012 15
6 2.416334e+17 Friday, Aug 31, 2012, 15 2.3666667 mins Friday Aug 31 2012 15
Now we have split weekdays column appropriately, we can use aggregate function to calculate mean over desired grouping columns.
> aggregate(halflife ~ Weekday, data=mydf2, FUN = mean)
Weekday halflife
1 Friday 1.013889
If you want to group by Weekday as well as Hour then
> aggregate(halflife ~ Weekday + Hour, data=mydf2, FUN = mean)
Weekday Hour halflife
1 Friday 15 1.6583333
2 Friday 16 1.5666667
3 Friday 19 0.3916667
4 Friday 22 0.4166667
As such first parameter of aggregate function here is a forumla object which supports one ~ one, one ~ many, many ~ one, and many ~ many relationships. See ?aggregate examples to understand how to use it.
I will give brief example of how to many to many relationships.
> set.seed(12345)
> mydf2 <- cbind(mydf2, newvar = rnorm(nrow(mydf2)))
> mydf2
id weekdays halflife Weekday Day Year Hour newvar
1 2.417322e+17 Friday, Aug 31, 2012, 22 0.4166667 mins Friday Aug 31 2012 22 0.5855288
2 2.416892e+17 Friday, Aug 31, 2012, 19 0.3833333 mins Friday Aug 31 2012 19 0.7094660
3 2.416869e+17 Friday, Aug 31, 2012, 19 0.4000000 mins Friday Aug 31 2012 19 -0.1093033
4 2.416511e+17 Friday, Aug 31, 2012, 16 1.5666667 mins Friday Aug 31 2012 16 -0.4534972
5 2.416352e+17 Friday, Aug 31, 2012, 15 0.9500000 mins Friday Aug 31 2012 15 0.6058875
6 2.416334e+17 Friday, Aug 31, 2012, 15 2.3666667 mins Friday Aug 31 2012 15 -1.8179560
> aggregate(cbind(newvar,halflife) ~ Weekday + Hour, data=mydf2, FUN = mean)
Weekday Hour newvar halflife
1 Friday 15 -0.6060343 1.6583333
2 Friday 16 -0.4534972 1.5666667
3 Friday 19 0.3000814 0.3916667
4 Friday 22 0.5855288 0.4166667

javascript regular expression: how do I find date without year or date with year<2010

I need to find date without year, or date with year<2010.
basically,
Feb 15
Feb 20
Feb 20, 2009
Feb 20, 1995
should be accepted
Feb 20, 2010
Feb 20, 2011
should be rejected
How do I do it?
Thanks,
Cheng

Try this:
(Jan|Feb|Mar...Dec)\s\d{1,2},\s([1][0-9][0-9][0-9]|200[0-9])
Note: Expand the month list with proepr names. I was too lazy to spell it all out.

How to increment Date objects in C++

I have an assignees that I've been working on and I'm stuck on the last function.
use the function void Increment(int numDays = 1)
This function should move the date forward by the number of calendar days given in the argument. Default value on the parameter is 1 day. Examples:
Date d1(10, 31, 1998); // Oct 31, 1998
Date d2(6, 29, 1950); // June 29, 1950
d1.Increment(); // d1 is now Nov 1, 1998
d2.Increment(5); // d2 is now July 4, 1950
I don not understand how to do this.
void Date::Increment(int numDays = 1)
I'm stuck, I know how to tell the function to increment, by the ++ operator but i get confuse when I have to get the function to increment the last day of the month to the the fist, or to end at the last date of that month for example. Oct 31 to Nov 1, or June 29 to July 4. I can do July 5 to July 8 but the changing months confuse me

You will need to store a list (or array) of how many days are in each month. If you add numDays to the current date and it becomes bigger than this, you need to increment the month as well.
For example, we have a date object representing 29 March 2010. We call Increment(4) and add 4 to the day variable, ending up with 33 March 2010. We now check how many days March has and find out it's 31 (eg. daysInMonth[3] == 31). Since 33 is greater than 31, we need subtract 31 from 33 and increase the month, ending up with 2 April 2010.
You will need special handling for February in leap years (any year divisible by 4 and not divisible by 100 unless it's also divisible by 400) and for incrementing past the end of December.

30 days has September, April, June, and November. The rest have 31 days, except for February, which has 28 days except on a leap year (every 4 years, and 2008 was the last one) when it has 29 days.
This should be plenty to get you going.

First, construct a function like
int numDaysSinceBeginning( Date );
which counts number of days elapsed from a well known date (e.g. Jan 1 1900) to the specific Date.
Next, construct another function which converts that day-delta to Date
Date createDateWithDelta( int );
From your example,
Date d2(6, 29, 1950); // June 29, 1950
int d2Delta = numDaysSinceBeginnning( d2 );
Date d2Incremented = createDateWithDelta( d2Delta + 5 ); // d2Incremented is July 4, 1950