I am working with this dataset and I am trying to separate the 'Date' column into the day, month, and year but have run into a problem doing it because it has the month as a character value. Any help would be great.
Here's an image: Dataset
You can convert your Date column using as.Date(), specifying the format for the date; in this case, one option is "%d%B%y"
library(lubridate)
dataset = data.frame(Date=c("19MAY19","31MAY19"))
dataset %>% mutate(Date = as.Date(Date,"%d%B%y"),
y = year(Date),m=month(Date),d = day(Date))
Output:
Date y m d
1 2019-05-19 2019 5 19
2 2019-05-31 2019 5 31
Related
I'm trying to visualize some bird data, however after grouping by month, the resulting output is out of order from the original data. It is in order for December, January, February, and March in the original, but after manipulating it results in December, February, January, March.
Any ideas how I can fix this or sort the rows?
This is the code:
BirdDataTimeClean <- BirdDataTimes %>%
group_by(Date) %>%
summarise(Gulls=sum(Gulls), Terns=sum(Terns), Sandpipers=sum(Sandpipers),
Plovers=sum(Plovers), Pelicans=sum(Pelicans), Oystercatchers=sum(Oystercatchers),
Egrets=sum(Egrets), PeregrineFalcon=sum(Peregrine_Falcon), BlackPhoebe=sum(Black_Phoebe),
Raven=sum(Common_Raven))
BirdDataTimeClean2 <- BirdDataTimeClean %>%
pivot_longer(!Date, names_to = "Species", values_to = "Count")
You haven't shared any workable data but i face this many times when reading from csv and hence all dates and data are in character.
as suggested, please convert the date data to "date" format using lubridate package or base as.Date() and then arrange() in dplyr will work or even group_by
example :toy data created
birds <- data.table(dates = c("2020-Feb-20","2020-Jan-20","2020-Dec-20","2020-Apr-20"),
species = c('Gulls','Turns','Gulls','Sandpiper'),
Counts = c(20,30,40,50)
str(birds) will show date is character (and I have not kept order)
using lubridate convert dates
birds$dates%>%lubridate::ymd() will change to date data-type
birds$dates%>%ymd()%>%str()
Date[1:4], format: "2020-02-20" "2020-01-20" "2020-12-20" "2020-04-20"
save it with birds$dates <- ymd(birds$dates) or do it in your pipeline as follows
now simply so the dplyr analysis:
birds%>%group_by(Months= ymd(dates))%>%
summarise(N=n()
,Species_Count = sum(Counts)
)%>%arrange(Months)
will give
# A tibble: 4 x 3
Months N Species_Count
<date> <int> <dbl>
1 2020-01-20 1 30
2 2020-02-20 1 20
3 2020-04-20 1 50
However, if you want Apr , Jan instead of numbers and apply as.Date() with format etc, the dates become "character" again. I woudl suggest you keep your data that way and while representing in output for others -> format it there with as.Date or if using DT or other datatables -> check the output formatting options. That way your original data remains and users see what they want.
this will make it character
birds%>%group_by(Months= as.character.Date(dates))%>%
summarise(N=n()
,Species_Count = sum(Counts)
)%>%arrange(Months)
A tibble: 4 x 3
Months N Species_Count
<chr> <int> <dbl>
1 2020-Apr-20 1 50
2 2020-Dec-20 1 40
3 2020-Feb-20 1 20
4 2020-Jan-20 1 30
My data is in a dataframe which has a structure like this:
df2 <- data.frame(Year = c("2007"), Week = c(1:12), Measurement = c(rnorm(12, mean = 4, sd = 1)))
Unfortunately I do not have the complete date (e.g. days are missing) for each measurement, only the Year and the Weeks (these are ISO weeks).
Now I want to aggregate the Median of a Month's worth of measurements (e.g. the weekly measurements per month of the specific year) into a new column, Months. I did not find a convenient way to do this without having the exact day of the measurements available. Any inputs are much appreciated!
When it is necessary to allocate a week to a single month, the rule for first week of the year might be applied, although ISO 8601 does not consider this case. (Wikipedia)
For example, the 5th week of 2007 belongs to February, because the Thursday of the 5th week was the 1st of February.
I am using data.table and ISOweek packages. See the example how to compute the month of the week. Then you can do any aggregation by month.
require(data.table)
require(ISOweek)
df2 <- data.table(Year = c("2007"), Week = c(1:12),
Measurement = c(rnorm(12, mean = 4, sd = 1)))
# Generate Thursday as year, week of the year, day of week according to ISO 8601
df2[, thursday_ISO := paste(Year, sprintf("W%02d", Week), 4, sep = "-")]
# Convert Thursday to date format
df2[, thursday_date := ISOweek2date(thursday_ISO)]
# Compute month
df2[, month := format(thursday_date, "%m")]
df2
Suggestion by Uwe to compute a year-month string.
# Compute year-month
df2[, yr_mon := format(ISOweek2date(sprintf("%s-W%02d-4", Year, Week)), "%Y-%m")]
df2
And finally you can do an aggregation to the new table or by adding median as a column.
df2[, median(Measurement), by = yr_mon]
df2[, median := median(Measurement), by = yr_mon]
df2
If I understand correctly, you don't know the exact day, but only the week number and year. My answer takes the first day of the year as a starting date and then compute one week intervals based on that. You can probably refine the answer.
Based on
an answer by mnel, using the lubridate package.
library(lubridate)
# Prepare week, month, year information ready for the merge
# Make sure you have all the necessary dates
wmy <- data.frame(Day = seq(ymd('2007-01-01'),ymd('2007-04-01'),
by = 'weeks'))
wmy <- transform(wmy,
Week = isoweek(Day),
Month = month(Day),
Year = isoyear(Day))
# Merge this information with your data
merge(df2, wmy, by = c("Year", "Week"))
Year Week Measurement Day Month
1 2007 1 3.704887 2007-01-01 1
2 2007 10 1.974533 2007-03-05 3
3 2007 11 4.797286 2007-03-12 3
4 2007 12 4.291169 2007-03-19 3
5 2007 2 4.305010 2007-01-08 1
6 2007 3 3.374982 2007-01-15 1
7 2007 4 3.600008 2007-01-22 1
8 2007 5 4.315184 2007-01-29 1
9 2007 6 4.887142 2007-02-05 2
10 2007 7 4.155411 2007-02-12 2
11 2007 8 4.711943 2007-02-19 2
12 2007 9 2.465862 2007-02-26 2
using dplyr you can try:
require(dplyr)
df2 %>% mutate(Date = as.Date(paste("1", Week, Year, sep = "-"), format = "%w-%W-%Y"),
Year_Mon = format(Date,"%Y-%m")) %>% group_by(Year_Mon) %>%
summarise(result = median(Measurement))
As #djhrio pointed out, Thursday is used to determine the weeks in a month. So simply switch paste("1", to paste("4", in the code above.
This can be done relatively simply in dplyr.
library(dplyr)
df2 %>%
mutate(Month = rep(1:3, each = 4)) %>%
group_by(Month) %>%
summarise(MonthlyMedian = stats::median(Measurement))
Basically, add a new column to define your months. I'm presuming since you don't have days, you are going to allocate 4 weeks per month?
Then you just group by your Month variable and calculate the median. Very simple
Hope this helps
I'm relatively new to R but I am very familiar with Excel and T-SQL.
I have a simple dataset that has a date with time and a numeric value associated it. What I'd like to do is summarize the numeric values by-hour of the day. I've found a couple resources for working with time-types in R but I was hoping to find a solution similar to is offered excel (where I can call a function and pass-in my date/time data and have it return the hour of the day).
Any suggestions would be appreciated - thanks!
library(readr)
library(dplyr)
library(lubridate)
df <- read_delim('DateTime|Value
3/14/2015 12:00:00|23
3/14/2015 13:00:00|24
3/15/2015 12:00:00|22
3/15/2015 13:00:00|40',"|")
df %>%
mutate(hour_of_day = hour(as.POSIXct(strptime(DateTime, "%m/%d/%Y %H:%M:%S")))) %>%
group_by(hour_of_day) %>%
summarise(meanValue = mean(Value))
breakdown:
Convert column of DateTime (character) into formatted time then use hour() from lubridate to pull out just that hour value and put it into new column named hour_of_day.
> df %>%
mutate(hour_of_day = hour(as.POSIXct(strptime(DateTime, "%m/%d/%Y %H:%M:%S"))))
Source: local data frame [4 x 3]
DateTime Value hour_of_day
1 3/14/2015 12:00:00 23 12
2 3/14/2015 13:00:00 24 13
3 3/15/2015 12:00:00 22 12
4 3/15/2015 13:00:00 40 13
The group_by(hour_of_day) sets the groups upon which mean(Value) is computed in the via the summarise(...) call.
this gives the result:
hour_of_day meanValue
1 12 22.5
2 13 32.0
I have data for hospitalisations that records date of admission and the number of days spent in the hospital:
ID date ndays
1 2005-06-01 15
2 2005-06-15 60
3 2005-12-25 20
4 2005-01-01 400
4 2006-06-04 15
I would like to create a dataset of days spend at the hospital per year, and therefore I need to deal with cases like ID 3, whose stay at the hospital goes over the end of the year, and ID 4, whose stay at the hospital is longer than one year. There is also the problem that some people do have a record on next year, and I would like to add the `surplus' days to those when this happens.
So far I have come up with this solution:
library(lubridate)
ndays_new <- ifelse((as.Date(paste(year(data$date),"12-31",sep="-")),
format="%Y-%m-%d") - data$date) < data$ndays,
(as.Date(paste(year(data$date),"12-31",sep="-")),
format="%Y-%m-%d") - data$date) ,
data$ndays)
However, I can't think of a way to get those `surplus' days that go over the end of the year and assign them to a new record starting on the next year. Can any one point me to a good solution? I use dplyr, so solutions with that package would be specially welcome, but I'm willing to try any other tool if needed.
My solution isn't compact. But, I tried to employ dplyr and did the following. I initially changed column names for my own understanding. I calculated another date (i.e., date.2) by adding ndays to date.1. If the years of date.1 and date.2 match, that means you do not have to consider the following year. If the years do not match, you need to consider the following year. ndays.2 is basically ndays for the following year. Then, I reshaped the data using do. After filtering unnecessary rows with NAs, I changed date to year and aggregated the data by ID and year.
rename(mydf, date.1 = date, ndays.1 = ndays) %>%
mutate(date.1 = as.POSIXct(date.1, format = "%Y-%m-%d"),
date.2 = date.1 + (60 * 60 * 24) * ndays.1,
ndays.2 = ifelse(as.character(format(date.1, "%Y")) == as.character(format(date.2, "%Y")), NA,
date.2 - as.POSIXct(paste0(as.character(format(date.2, "%Y")),"-01-01"), format = "%Y-%m-%d")),
ndays.1 = ifelse(ndays.2 %in% NA, ndays.1, ndays.1 - ndays.2)) %>%
do(data.frame(ID = .$ID, date = c(.$date.1, .$date.2), ndays = c(.$ndays.1, .$ndays.2))) %>%
filter(complete.cases(ndays)) %>%
mutate(date = as.numeric(format(date, "%Y"))) %>%
rename(year = date) %>%
group_by(ID, year) %>%
summarise(ndays = sum(ndays))
# ID year ndays
#1 1 2005 15
#2 2 2005 60
#3 3 2005 7
#4 3 2006 13
#5 4 2005 365
#6 4 2006 50
I need to convert date (m/d/y format) into 3 separate columns on which I hope to run an algorithm.(I'm trying to convert my dates into Julian Day Numbers). Saw this suggestion for another user for separating data out into multiple columns using Oracle. I'm using R and am throughly stuck about how to code this appropriately. Would A1,A2...represent my new column headings, and what would the format difference be with the "update set" section?
update <tablename> set A1 = substr(ORIG, 1, 4),
A2 = substr(ORIG, 5, 6),
A3 = substr(ORIG, 11, 6),
A4 = substr(ORIG, 17, 5);
I'm trying hard to improve my skills in R but cannot figure this one...any help is much appreciated. Thanks in advance... :)
I use the format() method for Date objects to pull apart dates in R. Using Dirk's datetext, here is how I would go about breaking up a date into its constituent parts:
datetxt <- c("2010-01-02", "2010-02-03", "2010-09-10")
datetxt <- as.Date(datetxt)
df <- data.frame(date = datetxt,
year = as.numeric(format(datetxt, format = "%Y")),
month = as.numeric(format(datetxt, format = "%m")),
day = as.numeric(format(datetxt, format = "%d")))
Which gives:
> df
date year month day
1 2010-01-02 2010 1 2
2 2010-02-03 2010 2 3
3 2010-09-10 2010 9 10
Note what several others have said; you can get the Julian dates without splitting out the various date components. I added this answer to show how you could do the breaking apart if you needed it for something else.
Given a text variable x, like this:
> x
[1] "10/3/2001"
then:
> as.Date(x,"%m/%d/%Y")
[1] "2001-10-03"
converts it to a date object. Then, if you need it:
> julian(as.Date(x,"%m/%d/%Y"))
[1] 11598
attr(,"origin")
[1] "1970-01-01"
gives you a Julian date (relative to 1970-01-01).
Don't try the substring thing...
See help(as.Date) for more.
Quick ones:
Julian date converters already exist in base R, see eg help(julian).
One approach may be to parse the date as a POSIXlt and to then read off the components. Other date / time classes and packages will work too but there is something to be said for base R.
Parsing dates as string is almost always a bad approach.
Here is an example:
datetxt <- c("2010-01-02", "2010-02-03", "2010-09-10")
dates <- as.Date(datetxt) ## you could examine these as well
plt <- as.POSIXlt(dates) ## now as POSIXlt types
plt[["year"]] + 1900 ## years are with offset 1900
#[1] 2010 2010 2010
plt[["mon"]] + 1 ## and months are on the 0 .. 11 intervasl
#[1] 1 2 9
plt[["mday"]]
#[1] 2 3 10
df <- data.frame(year=plt[["year"]] + 1900,
month=plt[["mon"]] + 1, day=plt[["mday"]])
df
# year month day
#1 2010 1 2
#2 2010 2 3
#3 2010 9 10
And of course
julian(dates)
#[1] 14611 14643 14862
#attr(,"origin")
#[1] "1970-01-01"
To convert date (m/d/y format) into 3 separate columns,consider the df,
df <- data.frame(date = c("01-02-18", "02-20-18", "03-23-18"))
df
date
1 01-02-18
2 02-20-18
3 03-23-18
Convert to date format
df$date <- as.Date(df$date, format="%m-%d-%y")
df
date
1 2018-01-02
2 2018-02-20
3 2018-03-23
To get three seperate columns with year, month and date,
library(lubridate)
df$year <- year(ymd(df$date))
df$month <- month(ymd(df$date))
df$day <- day(ymd(df$date))
df
date year month day
1 2018-01-02 2018 1 2
2 2018-02-20 2018 2 20
3 2018-03-23 2018 3 23
Hope this helps.
Hi Gavin: another way [using your idea] is:
The data-frame we will use is oilstocks which contains a variety of variables related to the changes over time of the oil and gas stocks.
The variables are:
colnames(stocks)
"bpV" "bpO" "bpC" "bpMN" "bpMX" "emdate" "emV" "emO" "emC"
"emMN" "emMN.1" "chdate" "chV" "cbO" "chC" "chMN" "chMX"
One of the first things to do is change the emdate field, which is an integer vector, into a date vector.
realdate<-as.Date(emdate,format="%m/%d/%Y")
Next we want to split emdate column into three separate columns representing month, day and year using the idea supplied by you.
> dfdate <- data.frame(date=realdate)
year=as.numeric (format(realdate,"%Y"))
month=as.numeric (format(realdate,"%m"))
day=as.numeric (format(realdate,"%d"))
ls() will include the individual vectors, day, month, year and dfdate.
Now merge the dfdate, day, month, year into the original data-frame [stocks].
ostocks<-cbind(dfdate,day,month,year,stocks)
colnames(ostocks)
"date" "day" "month" "year" "bpV" "bpO" "bpC" "bpMN" "bpMX" "emdate" "emV" "emO" "emC" "emMN" "emMX" "chdate" "chV"
"cbO" "chC" "chMN" "chMX"
Similar results and I also have date, day, month, year as separate vectors outside of the df.