I'm currently working on a project in which I have a column called 'weekday'. This column contains day numbers 0:6. I'm new to R and want to create a new column that has weekday abbreviations for each number e.g. "0 = Sun, 1 = Mon", etc. Below is an image of my dataframe I am working with. Any help and tips would be much apppreciated.
Image of dataframe
We can add your zero based weekday index to 2017-01-01, which was a Sunday.
x <- c(0:6)
days_abbrev <- substr(weekdays(as.Date("2017-01-01") + x), 1, 3)
days_abbrev
[1] "Sun" "Mon" "Tue" "Wed" "Thu" "Fri" "Sat"
Or using the excellent suggestion by #DarrenTsai we can pass TRUE as the second parameter to weekdays() to use abbreviations:
weekdays(as.Date("2017-01-01") + x, TRUE)
You can do this directly from the date column using %a format...
format(as.Date("2017-01-05"), "%a")
[1] "Thu"
Related
Trying to add a new column in my data table denoting the month (either as a numeric value or character) using an already available column of "SetDate", which is in the format mdy.
I'm new to R and having trouble. Thank you
base solution:
f = "%m/%d/%y" # note the lowercase y; it's because the year is 92, not 1992
dataset$SetDateMonth <- format(as.POSIXct(dataset$SetDate, format = f), "%m")
Basically, what it does is it converts the column from character (presumed class) to POSIXct, which allows for an easy extraction of month information.
Quick test:
format(as.POSIXct('1/1/92', format = "%m/%d/%y"), "%m")
[1] "01"
Try this (created a small example):
library(lubridate)
date_example <- "1/1/92"
lubridate::mdy(date_example)
[1] "1992-01-01"
lubridate::mdy(date_example) %>% lubridate::month()
[1] 1
If you want full month as character string, use:
lubridate::mdy(date_example) %>% lubridate::month(label = TRUE, abbr = FALSE)
I have a large dataframe of 22641 obs. and 12 variables.
The first column "year" includes extracted values from satellite images in the format below.
1_1_1_1_LT05_127024_19870517_00005ff8aac6b6bf60bc
From this format, I only want to keep the date which in this case is 19870517 and format it as date (so two different things). Usually, I use the regex to extract the words that I want, but here the date is different for each cell and I have no idea how to replace the above text with only the date. Maybe the way to do this is to search by position within the sentence but I do not know how.
Any ideas?
Thanks.
It's not clear what the "date is different in each cell" means but if it means that the value of the date is different and it is always the 7th field then either of (1) or (2) will work. If it either means that it consists of 8 consecutive digits anywhere in the text or 8 consecutive digits surrounded by _ anywhere in the text then see (3).
1) Assuming the input DF shown in reproducible form in the Note at the end use read.table to read year, pick out the 7th field and then convert it to Date class. No packages are used.
transform(read.table(text = DF$year, sep = "_")[7],
year = as.Date(as.character(V7), "%Y%m%d"), V7 = NULL)
## year
## 1 1987-05-17
2) Another alternative is separate in tidyr. 0.8.2 or later is needed.
library(dplyr)
library(tidyr)
DF %>%
separate(year, c(rep(NA, 6), "year"), extra = "drop") %>%
mutate(year = as.Date(as.character(year), "%Y%m%d"))
## year
## 1 1987-05-17
3) This assumes that the date is the only sequence of 8 digits in the year field use this or if we know it is surrounded by _ delimiters then the regular expression "_(\\d{8})_" can be used instead.
library(gsubfn)
transform(DF,
year = do.call("c", strapply(DF$year, "\\d{8}", ~ as.Date(x, "%Y%m%d"))))
## year
## 1 1987-05-17
Note
DF <- data.frame(year = "1_1_1_1_LT05_127024_19870517_00005ff8aac6b6bf60bc",
stringsAsFactors = FALSE)
Not sure if this will generalize to your whole data but maybe:
gsub(
'(^(?:.*?[^0-9])?)(\\d{8})((?:[^0-9].*)?$)',
'\\2',
'1_1_1_1_LT05_127024_19870517_00005ff8aac6b6bf60bc',
perl = TRUE
)
## [1] "19870517"
This uses group capturing and throws away anything but bounded 8 digit strings.
You can use sub to extract the data string and as.Date to convert it into R's date format:
as.Date(sub(".+?([0-9]+)_[^_]+$", "\\1", txt), "%Y%m%d")
# [1] "1987-05-17"
where txt <- "1_1_1_1_LT05_127024_19870517_00005ff8aac6b6bf60bc"
I have a dataframe with one column having all the date info. The new dataframe was created by extracting month-day format so later I could do a group_by.
df1=seq(as.Date('2011-01-01'),as.Date('2011-01-05'),by = 1)
df2=seq(as.Date('2010-12-28'),as.Date('2010-12-31'),by = 1)
df3=seq(as.Date('2011-01-16'),as.Date('2011-01-18'),by = 1)
df=c(df1,df2,df3)
s = format(df,"%m-%d")
s is the new dataframe and is character. s looks like this:
[1] "01-01" "01-02" "01-03" "01-04" "01-05" "12-28" "12-29"
[8] "12-30" "12-31" "01-16" "01-17" "01-18"
How could I detect from "12-28" to "01-05" is consecutive.
If it is in Date format I could use diff to detect, but with character, any suggestion?
Slight problem where my as.Date function gives a different result when I put it in a for loop. I'm looking in a folder with subfolders (per date) that contain images. I build date_list to organize all the dates (for plotting options in a later stage). The Julian Day starts from the first of January of the year, so because I have 4 years of date, the year must be flexible.
# Set up list with 4 columns and counter Q. jan is used to set all dates to the first of january
date_list <- outer(1:52, 1:4)
q = 1
jan <- "-01-01"
for (scene in folders){
year <- as.numeric(substr(scene, start=10, stop=13))
day <- as.numeric(substr(scene, start=14, stop=16))
datum <- paste(year, day, sep='_')
date_list[q, 1] <- datum
date_list[q, 2] <- year
date_list[q, 3] <- day
date_list[q, 4] <- as.Date(day, origin = as.Date(paste(year,jan, sep="")))
q = q+1
}
Output final row:
[52,] "2016_267" "2016" "267" "17068"
What am i missing in date_list[q, 4] that doesn't transfer my integer to a date?
running the following code does work, but due to the large amount of scenes and folders I like to automate this:
as.Date(day, origin = as.Date(paste(year,jan, sep="")))
Thank you for your time!
Well, I assume this would answer your first question:
date_list[q, 4] <- as.character(as.Date(datum,format="%Y_%j"))
as.Date accept a format argument, (the %Y and %j are documented in strptime), the %jis the julian day, this is a little easier to read than using origin and multiple paste calls.
Your problem is actually linked to what a Date object is:
> dput(as.Date("2016-01-10"))
structure(16810, class = "Date")
When entered into a matrix (your date_list) it is coerced to character w
without special treatment before like this:
> d<-as.Date("2016-01-10")
> class(d)<-"character"
> d
[1] "16810"
Hence you get only the number of days since 1970-01-01. When you ask for the date as character representation with as.character, it gives the correct value because the Date class as a as.character method which first compute the date in human format before returning a character value.
Now if I understood well your problem I would go this way:
First create a function to work on one string:
name_to_list <- function(name) {
dpart <- substr(name, start=10, stop=16)
date <- as.POSIXlt(dpart, format="%Y%j")
c("datum"=paste(date$year+1900,date$yday,sep="_"), "year"=date$year+1900, "julian_day"=date$yday, "date"=as.character(date) )
}
this function just get your substring, and then convert it to POSIXlt class, which give us julian day, year and date in one pass. as the year is stored as integer since 1900 (could be negative), we have to add 1900 when storing the year in the fields.
Then if your folders variable is a vector of string:
lapply(folders,name_to_list)
wich for folders=c("LC81730382016267LGN00","LC81730382016287LGN00","LC81730382016167LGN00") gives:
[[1]]
datum year julian_day date
"2016_266" "2016" "266" "2016-09-23"
[[2]]
datum year julian_day date
"2016_286" "2016" "286" "2016-10-13"
[[3]]
datum year julian_day date
"2016_166" "2016" "166" "2016-06-15"
Do you mean to output your day as 3 numbers? Should it not be 2 numbers?
day <- as.numeric(substr(scene, start=15, stop=16))
or
day <- as.numeric(substr(scene, start=14, stop=15))
That could at least be part of the issue. Providing an example of what typical values of "scene" are would be helpful here.
I have a character string of the date in Year-week format as such:
weeks.strings <- c("2002-26", "2002-27", "2002-28", "2002-29", "2002-30", "2002-31")
However, converting this character to Date class results in a loss of week identifier:
> as.Date(weeks.strings, format="%Y-%U")
[1] "2002-08-28" "2002-08-28" "2002-08-28" "2002-08-28" "2002-08-28"
[6] "2002-08-28"
As shown above, the format is converted into year- concatenated with today's date, so any information about the original week is lost (ex - when using the format function or strptime to try and coerce back into the original format.
One solution I found in a help group is to specify the day of the week:
as.Date(weeks.strings, format="%Y-%u %U")
[1] "2002-02-12" "2002-02-19" "2002-02-26" "2002-03-05" "2002-01-02"
[6] "2002-01-09"
But it looks like this results in incorrect week numbering (doesn't match the original string).
Any guidance would be appreciated.
You just need to add a weekday to your weeks.strings in order to make the dates unambiguous (adapted from Jim Holtman's answer on R-help).
as.Date(paste(weeks.strings,1),"%Y-%U %u")
As pointed out in the comments, the Date class is not appropriate if the dates span a long horizon because--at some point--the chosen weekday will not exist in the first/last week of the year. In that case you could use a numeric vector where the whole portion is the year and the decimal portion is the fraction of weeks/year. For example:
wkstr <- sprintf("%d-%02d", rep(2000:2012,each=53), 0:52)
yrwk <- lapply(strsplit(wkstr, "-"), as.numeric)
yrwk <- sapply(yrwk, function(x) x[1]+x[2]/53)
Obviously, there's no unique solution, since each week could be represented by any of up to 7 different dates. That said, here's one idea:
weeks.strings <- c("2002-26", "2002-27", "2002-28", "2002-29",
"2002-30", "2002-31")
x <- as.Date("2002-1-1", format="%Y-%m-%d") + (0:52*7)
x[match(weeks.strings, format(x, "%Y-%U"))]
# [1] "2002-07-02" "2002-07-09" "2002-07-16" "2002-07-23"
# [5] "2002-07-30" "2002-08-06"