I've searched for several days and am still stumped.
Given a dataset defined by the following:
ids = c("a","b","c")
dates = c(as.Date("2015-01-01"), as.Date("2015-02-01"), as.Date("2015-02-15"))
test = data.frame(ids, dates)
I am trying to dynamically add new columns to the data frame whose values will be the difference between the column date (2015-03-01) and the value in the date column. I would expect the result would look like the following, but with a better column name:
d20150301 = c(59, 28, 14)
result = data.frame(ids, dates, d20150301)
Many thanks in advance.
You can subtract a vector of dates from a single date, so
test$d2015_03_01 <- as.Date('2015-03-01')-test$dates
makes test look like
> test
ids dates d2015_03_01
1 a 2015-01-01 59 days
2 b 2015-02-01 28 days
3 c 2015-02-15 14 days
Related
I have a column of dates that was stored in the format 8/7/2001, 10/21/1990, etc. Two values are just four-digit years. I converted the entire column to class Date using the following code.
lubridate::parse_date_time(eventDate, orders = c('mdy', 'Y'))
It works great, except the values that were just years are converted to yyyy-01-01 and I want them to just be yyyy. Is there a way to keep lubridate from adding on any information that wasn't already there?
Edit: Code to create data frame
id = (1:5)
eventDate = c("10/7/2001", "1989", NA, "5/5/2016", "9/18/2011")
df <- data.frame(id, eventDate)
I do not think is possible to convert your values to Dates, and keep the "yyyy" values intact. And by transforming your "yyyy" values into "yyyy-01-01" the lubridate is doing the right thing. Because dates have order, and if you have other values in your column that have days and months defined, all the other values needs to have these components too.
For example. If I produce the data.frame below. If I ask R, to order the table, according to the date column, the date in the first line ("2020"), comes before the value in the second row ("2020-02-28")? Or comes after it? The value "2020" being the year of 2020, it can actually means every possible day in this year, so how R should treate it? By adding the first day of the year, lubridate is defining these components, and avoiding that R get confused by it.
dates <- c("2020", "2020-02-28", "2020-02-20", "2020-01-10", "2020-05-12")
id <- 1:5
df <- data.frame(
id,
dates
)
id dates
1 1 2020
2 2 2020-02-28
3 3 2020-02-20
4 4 2020-01-10
5 5 2020-05-12
So if you want to mantain the "yyyy" intact, is very likely that they should not rest in your eventDate column, with other values that are in a different structure ("dd/mm/yyyy"). Now if is really necessary to mantain these values intact, I think is best, to keep the values of eventDate column as characters, and store these values as Dates in another column, like this:
df$as_dates <- lubridate::parse_date_time(df$eventDate, orders = c('mdy', 'Y'))
id eventDate as_dates
1 1 10/7/2001 2001-10-07
2 2 1989 1989-01-01
3 3 <NA> <NA>
4 4 5/5/2016 2016-05-05
5 5 9/18/2011 2011-09-18
I have a data frame with data for each month of a 26 years period (1993 - 2019), which makes 312 rows in total.
Unfortunately, I had to lag the data, so each year goes now from July t to June t+1. So I can't just extract the year from the date.
Now, I want to exclude the 12-month data for each year in a separate data frame. My first Idea is to insert in the first column the year and use the lapply function to filter afterward.
For this, I created the following loop:
n <- 1
m <- 1993
for (a in 1:26) {
for (i in n:(n+11)) {
t.monthly.ret.lag[i,1] <- m
}
n <- n+1
m <- m+1
}
Unfortunately, R isn't naming the year in steps of 12. Instead, it is counting directly in steps of 1.
Does anyone know how to solve this or maybe know a better way of doing it?
y.first <- 1993
y.last <- 2019
month.col <- rep(c(7:12, 1:6), y.last-y.first+1)
year.col <- rep(c(y.first:y.last), each=length(month.name))
df <- data.frame(year=year.col, month=month.col)
This yields a dataframe with months and year correspondingly tagged, which further allows to use dplyr::group_by() and so on.
You could just create a 312 element long vector giving the year (and one giving the month) using rep() and seq(). Then you can attach them as additional columns to your data.frame or just use them as reference for month and year.
month = rep(seq(1:12),27)
year = c(matrix(rep(seq(1:27),12),ncol=27,byrow=T)+1992)
month = month[7:(length(month)-6)]
year = year[7:(length(year)-6)]
The month vector counts from 1 to 12, beginning at 6, the year vector repeats the year 12 times (the first and last only 6 times).
This question is about how to replace missing days and months in a data frame using R. Considering the data frame below, 99 denotes missing day or month and NA represents dates that are completely unknown.
df<-data.frame("id"=c(1,2,3,4,5),
"date" = c("99/10/2014","99/99/2011","23/02/2016","NA",
"99/04/2009"))
I am trying to replace the missing days and months based on the following criteria:
For dates with missing day but known month and year, the replacement date would be a random selection from the middle of the interval (first day to the last day of that month). Example, for id 1, the replacement date would be sampled from the middle of 01/10/2014 to 31/10/2014. For id 5, this would be the middle of 01/04/2009 to 30/04/2009. Of note is the varying number of days for different months, e.g. 31 days for October and 30 days for April.
As in the case of id 2, where both day and month are missing, the replacement date is a random selection from the middle of the interval (first day to last day of the year), e.g 01/01/2011 to 31/12/2011.
Please note: complete dates (e.g. the case of id 3) and NAs are not to be replaced.
I have tried by making use of the seq function together with the as.POSIXct and as.Date functions to obtain the sequence of dates from which the replacement dates are to be sampled. The difficulty I am experiencing is how to automate the R code to obtain the date intervals (it varies across distinct id) and how to make a random draw from the middle of the intervals.
The expected output would have the date of id 1, 2 and 5 replaced but those of id 3 and 4 remain unchanged. Any help on this is greatly appreciated.
This isn't the prettiest, but it seems to work and adapts to differing month and year lengths:
set.seed(999)
df$dateorig <- df$date
seld <- grepl("^99/", df$date)
selm <- grepl("^../99", df$date)
md <- seld & (!selm)
mm <- seld & selm
df$date <- as.Date(gsub("99","01",as.character(df$date)), format="%d/%m/%Y")
monrng <- sapply(df$date[md], function(x) seq(x, length.out=2, by="month")[2]) - as.numeric(df$date[md])
df$date[md] <- df$date[md] + sapply(monrng, sample, 1)
yrrng <- sapply(df$date[mm], function(x) seq(x, length.out=2, by="12 months")[2]) - as.numeric(df$date[mm])
df$date[mm] <- df$date[mm] + sapply(yrrng, sample, 1)
#df
# id date dateorig
#1 1 2014-10-14 99/10/2014
#2 2 2011-02-05 99/99/2011
#3 3 2016-02-23 23/02/2016
#4 4 <NA> NA
#5 5 2009-04-19 99/04/2009
I have two data frames: rainfall data collected daily and nitrate concentrations of water samples collected irregularly, approximately once a month. I would like to create a vector of values for each nitrate concentration that is the sum of the previous 5 days' rainfall. Basically, I need to match the nitrate date with the rain date, sum the previous 5 days' rainfall, then print the sum with the nitrate data.
I think I need to either make a function, a for loop, or use tapply to do this, but I don't know how. I'm not an expert at any of those, though I've used them in simple cases. I've searched for similar posts, but none get at this exactly. This one deals with summing by factor groups. This one deals with summing each possible pair of rows. This one deals with summing by aggregate.
Here are 2 example data frames:
# rainfall df
mm<- c(0,0,0,0,5, 0,0,2,0,0, 10,0,0,0,0)
date<- c(1:15)
rain <- data.frame(cbind(mm, date))
# b/c sums of rainfall depend on correct chronological order, make sure the data are in order by date.
rain[ do.call(order, list(rain$date)),]
# nitrate df
nconc <- c(15, 12, 14, 20, 8.5) # nitrate concentration
ndate<- c(6,8,11,13,14)
nitrate <- data.frame(cbind(nconc, ndate))
I would like to have a way of finding the matching rainfall date for each nitrate measurement, such as:
match(nitrate$date[i] %in% rain$date)
(Note: Will match work with as.Date dates?) And then sum the preceding 5 days' rainfall (not including the measurement date), such as:
sum(rain$mm[j-6:j-1]
And prints the sum in a new column in nitrate
print(nitrate$mm_sum[i])
To make sure it's clear what result I'm looking for, here's how to do the calculation 'by hand'. The first nitrate concentration was collected on day 6, so the sum of rainfall on days 1-5 is 5mm.
Many thanks in advance.
You were more or less there!
nitrate$prev_five_rainfall = NA
for (i in 1:length(nitrate$ndate)) {
day = nitrate$ndate[i]
nitrate$prev_five_rainfall[i] = sum(rain$mm[(day-6):(day-1)])
}
Step by step explanation:
Initialize empty result column:
nitrate$prev_five_rainfall = NA
For each line in the nitrate df: (i = 1,2,3,4,5)
for (i in 1:length(nitrate$ndate)) {
Grab the day we want final result for:
day = nitrate$ndate[i]
Take the rainfull sum and it put in in the results column
nitrate$prev_five_rainfall[i] = sum(rain$mm[(day-6):(day-1)])
Close the for loop :)
}
Disclaimer: This answer is basic in that:
It will break if nitrate's ndate < 6
It will be incorrect if some dates are missing in the rain dataframe
It will be slow on larger data
As you get more experience with R, you might use data manipulation packages like dplyr or data.table for these types of manipulations.
#nelsonauner's answer does all the heavy lifting. But one thing to note, in my actual data my dates are not numerical like they are in the example above, they are dates listed as MM/DD/YYYY with the appropriate as.Date(nitrate$date, "%m/%d/%Y").
I found that the for loop above gave me all zeros for nitrate$prev_five_rainfall and I suspected it was a problem with the dates.
So I changed my dates in both data sets to numerical using the difference in number of days between a common start date and the recorded date, so that the for loop would look for a matching number of days in each data frame rather than a date. First, make a column of the start date using rep_len() and format it:
nitrate$startdate <- rep_len("01/01/1980", nrow(nitrate))
nitrate$startdate <- as.Date(all$startdate, "%m/%d/%Y")
Then, calculate the difference using difftime():
nitrate$diffdays <- as.numeric(difftime(nitrate$date, nitrate$startdate, units="days"))
Do the same for the rain data frame. Finally, the for loop looks like this:
nitrate$prev_five_rainfall = NA
for (i in 1:length(nitrate$diffdays)) {
day = nitrate$diffdays[i]
nitrate$prev_five_rainfall[i] = sum(rain$mm[(day-5):(day-1)]) # 5 days
}
I need to create a time series from a data frame. The problem is variables is not well-ordered. Data frame is like below
Cases Date
15 1/2009
30 3/2010
45 12/2013
I have 60 observations like that. As you can see, data was collected randomly, which is starting from 1/2008 and ending 12/2013 ( There are many missing values(cases) in bulk of the months between these years). My assumption will be there is no cases in that months. So, how can I convert this dataset as time series? Then, I will try to make some prediction for possible number of cases in future.
Try installing the plyr library,
install.packages("plyr")
and then to sum duplicated Date2 rows:
library(plyr)
mergedData <- ddply(dat, .(Date2), .fun = function(x) {
data.frame(Cases = sum(x$Cases))
})
> head(mergedData)
Date2 Cases
1 2008-01-01 16352
2 2008-11-01 10
3 2009-01-01 23
4 2009-02-01 138
5 2009-04-01 18
6 2009-06-01 3534
you can create a separate sequence of time series and merge with data series.This will create a complete time series with missing values as NA.
if df is your data frame with Date as column of date than create new time series ts and merge as below.
ts <- data.frame(Date = seq(as.Date("2008-01-01"), as.Date("2013-12-31"), by="1 month"))
dfwithmisisng <- merge(ts, df, by="Date", all=T)