Date 1: 10/25/2020
Date 2: 01/25/2021
Difference = 3
What is the formula to find the difference between two dates in months in TOSCA?
WEEKDAY :
To find the day of the week of a given date
{CALC[WEEKDAY(DATE(2018,3,18))]}
Expected result: For the above-given date, the result will be 1 (Sunday)
Sunday= 1, Monday=2, Tuesday=3,
Wednesday=4, Thursday=5, Friday=6 Saturday=7
DATEDIF :
To find the difference between two given dates.
{CALC[DATEDIF(DATE(2018,2,10),DATE(2018,2,21), """d""")]}
Expected result: Above expression will give you the results as 11 (Difference between the given dates is 11 days.)
IF :
Given two dates we find which date is bigger amongst the two dates
{CALC[IF(DATE(2018,1,3)>DATE(2018,3,24),"""True""","""False""")]}
Expected results: This expression gives you the result as false as the first given date is smaller than the second one.
Related
I have a data written in specific expression. To simplify the data, here is the example I made:
df<-data.frame(date=c(2012034,2012044,2012051,2012063,2012074),
math=c(100,100,23,46,78))
2012034 means 4th week of march,2012. Likewise 2012044 means 4th week of April,2012. I was trying to make the values of date expressing some order. The reason why I have to do this is because when I don't change them to time expressions, x axis of the scatter plot looks really weird.
My goal is this:
Find the oldest date in date column and name it as 1. In this case, 2012034 should be 1. Next, find the second oldest date in date column and calculate how many weeks passed after that date. The second oldest date in date is 2012044.So, 5 weeks after the oldest date 2012034. So it should be changed as 1+5=6. So, likewise, I want to number the date to indicate how many weeks have passed since the oldest date
One way to do it is by also specifying the day of the week and subtract it at the end, i.e.
as.Date(paste0(df$date, '-1'), '%Y%m%U-%u') - 1
#[1] "2012-03-22" "2012-04-22" "2012-05-01" "2012-06-15" "2012-07-22"
How to create for every date hourly timestamps?
So for example from 00:00 til 23:59. The result of the function could be 10:00. I read on the internet that loop could work but we couldn't make it fit.
Data sample:
df = data.frame( id = c(1, 2, 3, 4), Date = c(2021-04-18, 2021-04-19, 2021-04-21
07:07:08.000, 2021-04-22))
A few points:
The input shown in the question is not valid R syntax so we assume what we have is the data frame shown reproducibly in the Note at the end.
the question did not describe the specific output desired so we will assume that what is wanted is a POSIXct vector of hourly values which in (1) below we assume is from the first hour of the minimum date to the last hour of the maximum date in the current time zone or in (2) below we assume that we only want hourly sequences for the dates in df also in the current time zone.
we assume that any times in the input should be dropped.
we assume that the id column of the input should be ignored.
No packages are used.
1) This calculates hour 0 of the first date and hour 0 of the day after the last date giving rng. The as.Date takes the Date part, range extracts out the smallest and largest dates into a vector of two components, adding 0:1 adds 0 to the first date leaving it as is and 1 to the second date converting it to the date after the last date. The format ensures that the Dates are converted to POSIXct in the current time zone rather than UTC. Then it creates an hourly sequence from those and uses head to drop the last value since it would be the day after the input's last date.
rng <- as.POSIXct(format(range(as.Date(df$Date)) + 0:1))
head(seq(rng[1], rng[2], "hour"), -1)
2) Another possibility is to paste together each date with each hour from 0 to 23 and then convert that to POSIXct. This will give the same result if the input dates are sequential; otherwise, it will give the hours only for those dates provided.
with(expand.grid(Date = as.Date(df$Date), hour = paste0(0:23, ":00:00")),
sort(as.POSIXct(paste(Date, hour))))
Note
df <- data.frame( id = c(1, 2, 3, 4),
Date = c("2021-04-18", "2021-04-19", "2021-04-21 07:07:08.000", "2021-04-22"))
I have a couples of weeknumbers of interest. Lets take '202124' (this week) as an example. How can I subtract x weeks from this week number?
Lets say I want to know the week number of 2 weeks prior, ideally I would like to do 202124 - 2 which would give me 202122. This is fine for most of the year however 202101 - 2 will give 202099 which is obviously not a valid week number. This would happen on a large scale so a more elegant solution is required. How could I go about this?
convert the year week values to dates subtract in days and format the output.
x <- c('202124', '202101')
format(as.Date(paste0(x, 1), '%Y%W%u') - 14, '%Y%V')
#[1] "202122" "202052"
To convert year week value to date we also need day of the week, I have used it as 1st day of the week.
For below code i am getting wrong quarters. Please help me with this issue
qy= cut.POSIXt(as.POSIXct(c("2015-09-01 IST","2016-08-1 IST")), breaks="quarter", labels=FALSE,include.lowest=T)
qy
# [1] 1 5
cut.POSIXt (with labels=FALES) gives you the quarters relative to the min(X) quarter - it starts at one with the earliest date and tells you the number of quarters between each date and that. So when you give dates in Q3 of two consecutive years, the first is 1, and the second is 4 quarters later, i.e. 5.
If you're trying to get the quarter within the year for each date use quarters or lubridate::quarter:
quarters(as.POSIXct(c("2015-09-01 IST","2016-08-1 IST")))
[1] "Q3" "Q3"
lubridate::quarter(as.POSIXct(c("2015-09-01 IST","2016-08-1 IST")))
[1] 3 3
Note that quarters comes out as a string starting with "Q", whereas lubridate::quarter comes out as an integer.
Within R, say I have a vector of some Lubridate dates:
> Date
"2012-01-01 UTC"
"2013-01-01 UTC"
Next, suppose I want to see what week number these days fall in:
> week(Date)
1
1
Lubridate is fantastic!
But wait...I'm dealing a time series with 10,000 rows of data...and the data spans 3 years.
I've been struggling with finding some way to make this happen:
> result of awesome R code here
1
54
The question: is there a succinct way to coax out a list of week numbers over multiyear periods within Lubridate? More directly, I would like the first week of the second year to be represented as the 54th week. And the first week in the third year to be represented as the 107th week, ad nauseum.
So far, I've attempted a number of hackney schemes but cannot seem to create something not fastened together with scotch tape. Any advice would be greatly appreciated. Thanks in advance.
To get the interval from a particular date to another date, you can just subtract...
If tda is your vector of dates, then
tda - min(tda)
will be the difference in seconds between them.
To get the units out in weeks:
(tda - min(tda))/eweeks(1)
To do it from a particular date:
tda - ymd(19960101)
This gives the number of days from 1996 to each value.
From there, you can divide by days per week, or seconds per week.
(tda - ymd(19960101))/eweeks(1)
To get only the integer part, and starting from January 2012:
trunc((tda - ymd(20111225))/eweeks(1))
Test data:
tda = ymd(c(20120101, 20120106, 20130101, 20130108))
Output:
1 1 53 54
Since eweeks() is now deprecated, I thought I'd add to #beroe's answer.
If tda is your date vector, you can get the week numbers with:
weeknos <- (interval(min(tda), tda) %/% weeks(1)) + 1
where %/% causes integer division. ( 5 / 3 = 1.667; 5 %/% 3 = 1)
You can do something like this :
week(dat) +53*(year(dat)-min(year(dat)))
Given you like lubridate (as do I)
year_week <- function(x,base) week(x) - week(base) + 52*(year(x) - year(base))
test <- ymd(c(20120101, 20120106, 20130101, 20130108))
year_week(test, "2012-01-01")
Giving
[1] 0 0 52 53