Given 1525505457, return a date object(Sat May 05 2018 15:30:57 GMT+0800 (CST)) representing it.
http://ergoemacs.org/emacs/elisp_parse_time.html this article only tell how to format a date to unix time, but not otherwise.
Use the combination of decode-time & seconds-to-time, like this:
(decode-time (seconds-to-time 1525505457))
=> (57 30 9 5 5 2018 6 t 7200)
The same module, time-date.el has other functions for conversion into time object, like string, etc.:
(decode-time (date-to-time "2018-05-05T12:33:05Z"))
=> (5 33 14 5 5 2018 6 t 7200)
Related
I have a text file of many rows containing date and time and the end goal is for me to group together the number of rows per week that their date values are in. This is so that I can plot a scatter diagram with x values being the week number and y values being the frequency. For example the text file (dates.txt):
Mon May 11 22:51:27 2013
Mon May 11 22:58:34 2013
Wed May 13 23:15:27 2013
Thu May 14 04:11:22 2013
Sat May 16 19:46:55 2013
Sat May 16 22:29:54 2013
Sun May 17 02:08:45 2013
Sun May 17 23:55:15 2013
Mon May 18 00:42:07 2013
So from here, week 1 will have a frequency of 6 and week 2 will have a frequency of 1
As I want to plot a scatter diagram for this, I want to convert them to text value first using strptime() with format %a %b
my attempt so far has been
time_stamp <- strptime(time_stamp, format='%a.%b')
However it shows the input string is too long. I'm very new to R-studio so could somebody please help me figure this out?
Thank you
Example of final output graph : https://imgur.com/a/3o3DivA
You could use readLines() to avoid the data frame, then read time using strptime, and finally strftime to format the output.
strftime(strptime(readLines('dates.txt'), '%c'), '%a.%b')
# [1] "Sat.May" "Sat.May" "Mon.May" "Tue.May" "Thu.May" "Thu.May" "Fri.May" "Fri.May" "Sat.May"
Edit
So it appears that your dates have a time zone abbreviation "Mon Apr 06 23:49:29 PDT 2009". Since it is constant during the dates we can specify it literally in the pattern.
We will use '%d_%m' for strftime to get something numeric seperated by _ with which we feed strsplit and then type.convert into numerics.
Finally we unlist, create a matrix that we fill byrow, and plot the guy.
strptime(readLines('timestamp.txt'), '%a %b %d %H:%M:%S PDT %Y') |>
strftime('%d_%m') |>
strsplit('_') |>
type.convert(as.is=TRUE) |>
unlist() |>
matrix(ncol=2, byrow=TRUE) |>
plot(pch=20, col=4, main='My Plot', xlab='day', ylab='month')
Note: Please use R>=4.1 for the |> pipes.
You need to first read (or assign) the data, parse it to a date type and then use that to e.g. get the number of the week.
Here is one example
text <- "Mon May 11 22:51:27 2013
Mon May 11 22:58:34 2013
Wed May 13 23:15:27 2013
Thu May 14 04:11:22 2013
Sat May 16 19:46:55 2013
Sat May 16 22:29:54 2013
Sun May 17 02:08:45 2013
Sun May 17 23:55:15 2013
Mon May 18 00:42:07 2013"
data <- read.table(text=text, sep='\n', col.names="dates")
data$parse <- anytime::anytime(data$dates)
data$week <- as.integer(format(data$parse, "%V"))
data
The result is a new data.frame object:
> data
dates parse week
1 Mon May 11 22:51:27 2013 2013-05-11 22:51:27 19
2 Mon May 11 22:58:34 2013 2013-05-11 22:58:34 19
3 Wed May 13 23:15:27 2013 2013-05-13 23:15:27 20
4 Thu May 14 04:11:22 2013 2013-05-14 04:11:22 20
5 Sat May 16 19:46:55 2013 2013-05-16 19:46:55 20
6 Sat May 16 22:29:54 2013 2013-05-16 22:29:54 20
7 Sun May 17 02:08:45 2013 2013-05-17 02:08:45 20
8 Sun May 17 23:55:15 2013 2013-05-17 23:55:15 20
9 Mon May 18 00:42:07 2013 2013-05-18 00:42:07 20
>
Hello I am trying to find the week number for a series of date over three years. However R is not giving the correct week number. I am generating a seq of dates from 2016-04-01 to 2019-03-30 and then I am trying to calculate week over three years such that I get the week number 54, 55 , 56 and so on.
However when I check the week 2016-04-03 R shows the week number as 14 where as when cross checked with excel it is the week number 15 and also it simply calculates 7 days and does not reference the actual calendar days. Also the week number starts from 1 for every start of year
The code looks like this
days <- seq(as.Date("2016-04-03"),as.Date("2019-03-30"),'days')
weekdays <- data.frame('days'=days, Month = month(days), week = week(days),nweek = rep(1,length(days)))
This is how the results looks like
days week
2016-04-01 14
2016-04-02 14
2016-04-03 14
2016-04-04 14
2016-04-05 14
2016-04-06 14
2016-04-07 14
2016-04-08 15
2016-04-09 15
2016-04-10 15
2016-04-11 15
2016-04-12 15
However when checked from excel this is what I get
days week
2016-04-01 14
2016-04-02 14
2016-04-03 15
2016-04-04 15
2016-04-05 15
2016-04-06 15
2016-04-07 15
2016-04-08 15
2016-04-09 15
2016-04-10 16
2016-04-11 16
2016-04-12 16
Can someone please help me identify wherever I am going wrong.
Thanks a lot in advance!!
Not anything that you're doing wrong per se, there is just a difference in how R (I presume you're using the lubridate package) and Excel calculate week numbers.
R will calculate week numbers based on the seven day block from 1 January that year; but
Excel calculates week numbers based on a week starting from Sunday.
Taking the first few days of January 2016 for an example. On, Friday, 1 January 2016, both R and Excel will say this is week 1.
On Sunday, 3 January 2016:
this is within the first seven days of the start of the year so R will return week number 1; but
it is a Sunday, so Excel ticks over to week number 2.
Try this:
ifelse(test = weekdays.Date(days[1]) == "Sunday", yes = epiweek(days[1]), no = epiweek(days[1]) + 1) + cumsum(weekdays.Date(days) == "Sunday")
This tests whether the first day is a Sunday or not and returns an appropriate week number starting point, then adds on one more week number each Sunday. Gives the same week number if there's overlap between years.
input<-read. csv("aggregate. csv")
The csv looks like:
TimeStamp Latency Threads
7:00.06 AM 20 19
7:00.09 AM 28 18
7:00.15 AM 26 19
7:04:51 AM 45 20
7:05.07 AM 05 23
7:00.25 AM 15 24
7:10.01 AM 24 25
7:20.01 AM 35 50
8:00:10 AM 05 51
8:00:52 AM 50 10
8:05:00 AM 12 09
8:10:00 AM 100 01
But the problem I am facing is I want to filter out the input dataframe by giving user input as TimeStamp column. Means the console should askEnter your time range
Suppose If I enter bw 7:00:01 AM to 7:05:00 AM. then it should filter the dataframe according to that.
The output should be like..
TimeStamp Latency Threads
7:00.06 AM 20 19
7:00.09 AM 28 18
7:00.15 AM 26 19
7:04:51 AM 45 20
Is it possible?
I posted here because filtering it was getting hard-coded but I want that to be user input.
Please help
You can try strptime function. It will transfer a character into a POSIXlt object, which is a Date-Time Classes. After the transformation, you can compare the times by time 1 < time 2, etc. You can find more details by ?strptime.
One more thing bothering is that your TimeStamp column has multiple formats: 7:00.06 AM and 7:04:51 AM. You probably need to pre-process the format by gsub
Suppose you got your input range is from lower to upper, then
input <- read.csv("C:/Users/s0043102/Desktop/test.csv",stringsAsFactors = FALSE)
adj_TimeStamp <- gsub("\\.",":",input$TimeStamp)
adj_TimeStamp <- strptime(adj_TimeStamp ,format="%H:%M:%S %p")
lower <- strptime("7:00:01 AM",format="%H:%M:%S %p")
upper <- strptime("7:05:00 AM",format="%H:%M:%S %p")
output <- subset(input, adj_TimeStamp<=upper & adj_TimeStamp>=lower)
output
TimeStamp Latency Threads
1 7:00.06 AM 20 19
2 7:00.09 AM 28 18
3 7:00.15 AM 26 19
4 7:04:51 AM 45 20
6 7:00.25 AM 15 24
I would like to subset a timeseries dataframe based on my requirement.
I have a dataframe something similar to the one mentioned below.
> df
Date Year Month Day Time Parameter
2012-04-19 2012 04 19 7:00:00 26
2012-04-19 2012 04 19 7:00:00 20
.................................................
2012-05-01 2012 05 01 00:00:00 23
2012-05-01 2012 05 01 00:30:00 22
.................................................
2015-04-30 2015 04 30 23:30:00 20
.................................................
2015-05-01 2015 05 01 00:00:00 26
From the dataframe similar to this I will like to select all the data from the first of May 2012 2012-05-01 to the end of April 2015-04-30, regardless of the starting and end date of the dataframe.
However, I am familiar with the grep function to select the data from one particular column. I have been using the following code with grep and with.
# To select one particular year
> df.2012 <- df[grep("2012", df$Year),]
# To select two or more years at the same time
> df.sel.yr <- df[grep("201[2-5]", df$Year),]
# To select one particular month of a particular year.
> df.Dec.2012 <- df[with(df, Year=="2012" & Month=="12"), ]
With several Lines of commands i will be able to do it. But it would save a lot of time if I can do it with only few or one line of command.
Any help will be appreciated. Thank you in advance.
If your date column is not of class date first convert it to one by,
df$Date <- as.Date(df$Date)
and then you can subset the date by,
df[df$Date >= as.Date("2012-05-01") & df$Date <= as.Date("2015-04-30"), ]
# Date Year Month Day Time Parameter
#3 2012-05-01 2012 5 1 00:00:00 23
#4 2012-05-01 2012 5 1 00:30:00 22
#5 2015-04-30 2015 4 30 23:30:00 20
Is there a good way to get a year + week number converted a date in R? I have tried the following:
> as.POSIXct("2008 41", format="%Y %U")
[1] "2008-02-21 EST"
> as.POSIXct("2008 42", format="%Y %U")
[1] "2008-02-21 EST"
According to ?strftime:
%Y Year with century. Note that whereas there was no zero in the
original Gregorian calendar, ISO 8601:2004 defines it to be valid
(interpreted as 1BC): see http://en.wikipedia.org/wiki/0_(year). Note
that the standard also says that years before 1582 in its calendar
should only be used with agreement of the parties involved.
%U Week of the year as decimal number (00–53) using Sunday as the
first day 1 of the week (and typically with the first Sunday of the
year as day 1 of week 1). The US convention.
This is kinda like another question you may have seen before. :)
The key issue is: what day should a week number specify? Is it the first day of the week? The last? That's ambiguous. I don't know if week one is the first day of the year or the 7th day of the year, or possibly the first Sunday or Monday of the year (which is a frequent interpretation). (And it's worse than that: these generally appear to be 0-indexed, rather than 1-indexed.) So, an enumerated day of the week needs to be specified.
For instance, try this:
as.POSIXlt("2008 42 1", format = "%Y %U %u")
The %u indicator specifies the day of the week.
Additional note: See ?strptime for the various options for format conversion. It's important to be careful about the enumeration of weeks, as these can be split across the end of the year, and day 1 is ambiguous: is it specified based on a Sunday or Monday, or from the first day of the year? This should all be specified and tested on the different systems where the R code will run. I'm not certain that Windows and POSIX systems sing the same tune on some of these conversions, hence I'd test and test again.
Day-of-week == zero in the POSIXlt DateTimesClasses system is Sunday. Not exactly Biblical and not in agreement with the R indexing that starts at "1" convention either, but that's what it is. Week zero is the first (partial) week in the year. Week one (but day of week zero) starts with the first Sunday. And all the other sequence types in POSIXlt have 0 as their starting point. It kind of interesting to see what coercing the list elements of POSIXlt objects do. The only way you can actually change a POSIXlt date is to alter the $year, the $mon or the $mday elements. The others seem to be epiphenomena.
today <- as.POSIXlt(Sys.Date())
today # Tuesday
#[1] "2012-02-21 UTC"
today$wday <- 0 # attempt to make it Sunday
today
# [1] "2012-02-21 UTC" The attempt fails
today$mday <- 19
today
#[1] "2012-02-19 UTC" Success
I did not come up with this myself (it's taken from a blog post by Forester), but nevertheless I thought I'd add this to the answer list because it's the first implementation of the ISO 8601 week number convention that I've seen in R.
No doubt, week numbers are a very ambiguous topic, but I prefer an ISO standard over the current implementation of week numbers via format(..., "%U") because it seems that this is what most people agreed on, at least in Germany (calendars etc.).
I've put the actual function def at the bottom to facilitate focusing on the output first. Also, I just stumbled across package ISOweek, maybe worth a try.
Approach Comparison
x.days <- c("Mon", "Tue", "Wed", "Thu", "Fri", "Sat", "Sun")
x.names <- sapply(1:length(posix), function(x) {
x.day <- as.POSIXlt(posix[x], tz="Europe/Berlin")$wday
if (x.day == 0) {
x.day <- 7
}
out <- x.days[x.day]
})
data.frame(
posix,
name=x.names,
week.r=weeknum,
week.iso=ISOweek(as.character(posix), tzone="Europe/Berlin")$weeknum
)
# Result
posix name week.r week.iso
1 2012-01-01 Sun 1 4480458
2 2012-01-02 Mon 1 1
3 2012-01-03 Tue 1 1
4 2012-01-04 Wed 1 1
5 2012-01-05 Thu 1 1
6 2012-01-06 Fri 1 1
7 2012-01-07 Sat 1 1
8 2012-01-08 Sun 2 1
9 2012-01-09 Mon 2 2
10 2012-01-10 Tue 2 2
11 2012-01-11 Wed 2 2
12 2012-01-12 Thu 2 2
13 2012-01-13 Fri 2 2
14 2012-01-14 Sat 2 2
15 2012-01-15 Sun 3 2
16 2012-01-16 Mon 3 3
17 2012-01-17 Tue 3 3
18 2012-01-18 Wed 3 3
19 2012-01-19 Thu 3 3
20 2012-01-20 Fri 3 3
21 2012-01-21 Sat 3 3
22 2012-01-22 Sun 4 3
23 2012-01-23 Mon 4 4
24 2012-01-24 Tue 4 4
25 2012-01-25 Wed 4 4
26 2012-01-26 Thu 4 4
27 2012-01-27 Fri 4 4
28 2012-01-28 Sat 4 4
29 2012-01-29 Sun 5 4
30 2012-01-30 Mon 5 5
31 2012-01-31 Tue 5 5
Function Def
It's taken directly from the blog post, I've just changed a couple of minor things. The function is still kind of sketchy (e.g. the week number of the first date is far off), but I find it to be a nice start!
ISOweek <- function(
date,
format="%Y-%m-%d",
tzone="UTC",
return.val="weekofyear"
){
##converts dates into "dayofyear" or "weekofyear", the latter providing the ISO-8601 week
##date should be a vector of class Date or a vector of formatted character strings
##format refers to the date form used if a vector of
## character strings is supplied
##convert date to POSIXt format
if(class(date)[1]%in%c("Date","character")){
date=as.POSIXlt(date,format=format, tz=tzone)
}
# if(class(date)[1]!="POSIXt"){
if (!inherits(date, "POSIXt")) {
print("Date is of wrong format.")
break
}else if(class(date)[2]=="POSIXct"){
date=as.POSIXlt(date, tz=tzone)
}
print(date)
if(return.val=="dayofyear"){
##add 1 because POSIXt is base zero
return(date$yday+1)
}else if(return.val=="weekofyear"){
##Based on the ISO8601 weekdate system,
## Monday is the first day of the week
## W01 is the week with 4 Jan in it.
year=1900+date$year
jan4=strptime(paste(year,1,4,sep="-"),format="%Y-%m-%d")
wday=jan4$wday
wday[wday==0]=7 ##convert to base 1, where Monday == 1, Sunday==7
##calculate the date of the first week of the year
weekstart=jan4-(wday-1)*86400
weeknum=ceiling(as.numeric((difftime(date,weekstart,units="days")+0.1)/7))
#########################################################################
##calculate week for days of the year occuring in the next year's week 1.
#########################################################################
mday=date$mday
wday=date$wday
wday[wday==0]=7
year=ifelse(weeknum==53 & mday-wday>=28,year+1,year)
weeknum=ifelse(weeknum==53 & mday-wday>=28,1,weeknum)
################################################################
##calculate week for days of the year occuring prior to week 1.
################################################################
##first calculate the numbe of weeks in the previous year
year.shift=year-1
jan4.shift=strptime(paste(year.shift,1,4,sep="-"),format="%Y-%m-%d")
wday=jan4.shift$wday
wday[wday==0]=7 ##convert to base 1, where Monday == 1, Sunday==7
weekstart=jan4.shift-(wday-1)*86400
weeknum.shift=ceiling(as.numeric((difftime(date,weekstart)+0.1)/7))
##update year and week
year=ifelse(weeknum==0,year.shift,year)
weeknum=ifelse(weeknum==0,weeknum.shift,weeknum)
return(list("year"=year,"weeknum"=weeknum))
}else{
print("Unknown return.val")
break
}
}