I have two dataframes.
I would like to make the average of sp variable for the previous 5 days defined by a specific date from a second dataframe.
For example, the mean from the day 1997.05.05 (that would be between the day 1997.05.05 and 1997.05.01) and the average between 1997.05.27 and 1997.05.31 calculate the days that have values (in this case 3).
Here are the variables:
sp < - c(7,9,9,4,2,5,2,9,NA,14,NA,NA,NA,NA,NA,14,25,NA,11,10,12,NA,9,NA,6,8,6,1,NA,7,NA)
Date <- c("1997-05-01","1997-05-02","1997-05-03","1997-05-04","1997-05-05",
"1997-05-06","1997-05-07","1997-05-08","1997-05-09","1997-05-10",
"1997-05-11","1997-05-12","1997-05-13","1997-05-14","1997-05-15",
"1997-05-16","1997-05-17","1997-05-18","1997-05-19","1997-05-20",
"1997-05-21","1997-05-22","1997-05-23","1997-05-24","1997-05-25",
"1997-05-26","1997-05-27","1997-05-28","1997-05-29","1997-05-30",
"1997-05-31")
data1 <- data.frame(sp, Date)
DateX <- c("1997-05-05","1997-05-15","1997-05-31")
data2 <- data.frame(DateX)
how to do that best? Help would be much appreciated.
Here is my expected result (in the second dataframe, data2):
1. DateX spMean
2. 1997-05-05 6.2
3. 1997-05-15 NA
4. 1997-05-31 4.6
I have made a few type changes to your initial code. Give the below a shot...I use lapply to run a quick function against the data1 object using the dates in your second object.
sp <- c(7,9,9,4,2,5,2,9,NA,14,NA,NA,NA,NA,NA,14,25,NA,11,10,12,NA,9,NA,6,8,6,1,NA,7,NA)
Date <- as.Date(c("1997-05-01","1997-05-02","1997-05-03","1997-05-04","1997-05-05",
"1997-05-06","1997-05-07","1997-05-08","1997-05-09","1997-05-10",
"1997-05-11","1997-05-12","1997-05-13","1997-05-14","1997-05-15",
"1997-05-16","1997-05-17","1997-05-18","1997-05-19","1997-05-20",
"1997-05-21","1997-05-22","1997-05-23","1997-05-24","1997-05-25",
"1997-05-26","1997-05-27","1997-05-28","1997-05-29","1997-05-30",
"1997-05-31"))
data1 <- data.frame(sp, Date)
DateX <- as.Date(c("1997-05-05","1997-05-15","1997-05-31"))
data2 <- data.frame(DateX)
#Add column for mean, NA values return NA
data2$spMean_na <- lapply(DateX,
function(m) mean(data1$sp[data1$Date >= m - 5 & data1$Date <= m]))
#Add column for mean, remove NA values
data2$spMean_na_omit <- lapply(DateX,
function(m) mean(data1$sp[data1$Date >= m - 5 & data1$Date <= m],
na.rm = TRUE))
> data2
DateX spMean_na spMean_na_omit
1 1997-05-05 6.2 6.2
2 1997-05-15 NA 14
3 1997-05-31 NA 5.5
I think you might need to change your expected result. Row 29 has an NA for the sp value and is within 5 days of 1997-05-31. So it should return an NA per your requirements as I understand them.
Related
Please see the sample data below.
I want to convert the quarterly sale data (with a start date and end date) into monthly sale data.
For example:
Data set A-Row 1 will be split into Data set B- Row 1, 2 and 3 for June, July and August separately and the sale will be pro rata based on number of days in that month, all other columns will be the same;
Data set A-Row 2 will pick up what was left in Row 1 (which ends in 5/9/2017) and formed a complete September.
Is there an efficient way to execute this, the actual data is a csv file with 100K x 15 data size, which will be split to approximately 300K x 15 new data set for monthly analysis.
Some key characteristic from sample question data includes:
The start day for the first quarterly sales data is the day that customer joins, so it could be any day;
All sales will be quarterly but in various days between 90, 91, or 92 days, but it is also possible to have imcomplete quarterly sale data as customer leave in the quarter.
Sample Question:
Customer.ID Country Type Sale Start..Date End.Date Days
1 1 US Commercial 91 7/06/2017 5/09/2017 91
2 1 US Commerical 92 6/09/2017 6/12/2017 92
3 2 US Casual 25 10/07/2017 3/08/2017 25
4 3 UK Commercial 64 7/06/2017 9/08/2017 64
Sample Answer:
Customer.ID Country Type Sale Start.Date End.Date Days
1 1 US Commercial 24 7/06/2017 30/06/2017 24
2 1 US Commercial 31 1/07/2017 31/07/2017 31
3 1 US Commercial 31 1/08/2017 31/08/2017 31
4 1 US Commercial 30 1/09/2017 30/09/2017 30
5 1 US Commercial 31 1/10/2017 31/10/2017 31
6 1 US Commercial 30 1/11/2017 30/11/2017 30
7 1 US Commercial 6 1/12/2017 6/12/2017 6
8 2 US Casual 22 10/07/2017 31/07/2017 22
9 2 US Casual 3 1/08/2017 3/08/2017 3
10 3 UK Commercial 24 7/06/2017 30/06/2017 24
11 3 UK Commercial 31 1/07/2017 31/07/2017 31
12 3 UK Commercial 9 1/08/2017 9/08/2017 9
I just ran CIAndrews' code. It seems to work for the most part, but it is very slow when run on a dataset with 10,000 rows. I eventually cancelled the execution after a few minutes of waiting. There's also an issue with the number of days: For example, July has 31 days, but the days variable only shows thirty. It's true that 31-1 = 30, but the first day should be counted as well.
The code below only takes about 21 seconds on my 2015 MacBook Pro (not including data generation), and takes care of the other problem, too.
library(tidyverse)
library(lubridate)
# generate data -------------------------------------------------------------
set.seed(666)
# assign variables
customer <- sample.int(n = 2000, size = 10000, replace = T)
country <- sample(c("US", "UK", "DE", "FR", "IS"), 10000, replace = T)
type <- sample(c("commercial", "casual", "other"), 10000, replace = T)
start <- sample(seq(dmy("7/06/2011"), today(), by = "day"), 10000, replace = T)
days <- sample(85:105, 10000, replace = T)
end <- start + days
sale <- sample(500:3000, 10000, replace = T)
# generate dataframe of artificial data
df_quarterly <- tibble(customer, country, type, sale, start, end, days)
# split quarters into months ----------------------------------------------
# initialize empty list with length == nrow(dataframe)
list_date_dfs <- vector(mode = "list", length = nrow(df_quarterly))
# for-loop generates new dates and adds as dataframe to list
for (i in 1:length(list_date_dfs)) {
# transfer dataframe row to variable `row`
row <- df_quarterly[i,]
# correct end date so split successful when interval doesn't cover full month
end_corr <- row$end + day(row$start) - day(row$end)
# use lubridate to compute first and last days of relevant months
m_start <- seq(row$start, end_corr, by = "month") %>%
floor_date(unit = "month")
m_end <- m_start + days_in_month(m_start) - 1
# replace first and last elements with original dates
m_start[1] <- row$start
m_end[length(m_end)] <- row$end
# compute the number of days per month as well as sales per month
# correct difference by adding 1
m_days <- as.integer(m_end - m_start) + 1
m_sale <- (row$sale / sum(m_days)) * m_days
# add tibble to list
list_date_dfs[[i]] <- tibble(customer = row$customer,
country = row$country,
type = row$type,
sale = m_sale,
start = m_start,
end = m_end,
days = m_days
)
}
# bind dataframe list elements into single dataframe
df_monthly <- bind_rows(list_date_dfs)
It's not pretty as it uses multiple functions and loops, since it consists out of multiple operations:
# Creating the dataset
library(tidyr)
customer <- c(1,1,2,3)
country <- c("US","US","US","UK")
type <- c("Commercial","Commercial","Casual","Commercial")
sale <- c(91,92,25,64)
Start <- as.Date(c("7/06/2017","6/09/2017","10/07/2017","7/06/2017"),"%d/%m/%Y")
Finish <- as.Date(c("5/09/2017","6/12/2017","3/08/2017","9/08/2017"),"%d/%m/%Y")
days <- c(91,92,25,64)
df <- data.frame(customer,country, type,sale, Start,Finish,days)
# Function to split per month
library(zoo)
addrowFun <- function(y){
temp <- do.call("rbind", by(y, 1:nrow(y), function(x) with(x, {
eom <- as.Date(as.yearmon(Start), frac = 1)
if (eom < Finish)
data.frame(customer, country, type, Start = c(Start, eom+1), Finish = c(eom, Finish))
else x
})))
return(temp)
}
loop <- df
for(i in 1:10){ #not all months are split up at once
loop <- addrowFun(loop)
}
# Calculating the days per month
loop$days <- as.numeric(difftime(loop$Finish,loop$Start, units="days"))
# Creating the function to get the monthly sales pro rata
sumFun <- function(x){
tempSum <- df[x$Start >= df$Start & x$Finish <= df$Finish & df$customer == x$customer,]
totalSale <- sum(tempSum$sale)
totalDays <- sum(tempSum$days)
return(x$days / totalDays * totalSale)
}
for(i in 1:length(loop$customer)){
loop$sale[i] <- sumFun(loop[i,])
}
loop
CiAndrews,
Thanks for the help and patience. I have managed to get the answer with small change. I have replace the "rbind" with "rbind.fill" from "plyr" package and everything runs smoothly after that.
Please see the head of sample2.csv below
customer country type sale Start Finish days
1 43108181108 US Commercial 3330 17/11/2016 24/02/2017 99
2 43108181108 US Commercial 2753 24/02/2017 23/05/2017 88
3 43108181108 US Commercial 3043 13/02/2018 18/05/2018 94
4 43108181108 US Commercial 4261 23/05/2017 18/08/2017 87
5 43103703637 UK Casual 881 4/11/2016 15/02/2017 103
6 43103703637 UK Casual 1172 26/07/2018 1/11/2018 98
Please see the codes below:
library(tidyr)
#read data and change the start and finish to data type
data <- read.csv("Sample2.csv")
data$Start <- as.Date(data$Start, "%d/%m/%Y")
data$Finish <- as.Date(data$Finish, "%d/%m/%Y")
customer <- data$customer
country <- data$country
days <- data$days
Finish <- data$Finish
Start <- data$Start
sale <- data$sale
type <- data$type
df <- data.frame(customer, country, type, sale, Start, Finish, days)
# Function to split per month
library(zoo)
library(plyr)
addrowFun <- function(y){
temp <- do.call("rbind.fill", by(y, 1:nrow(y), function(x) with(x, {
eom <- as.Date(as.yearmon(Start), frac = 1)
if (eom < Finish)
data.frame(customer, country, type, Start = c(Start, eom+1), Finish = c(eom, Finish))
else x
})))
return(temp)
}
loop <- df
for(i in 1:10){ #not all months are split up at once
loop <- addrowFun(loop)
}
# Calculating the days per month
loop$days <- as.numeric(difftime(loop$Finish,loop$Start, units="days"))
# Creating the function to get the monthly sales pro rata
sumFun <- function(x){
tempSum <- df[x$Start >= df$Start & x$Finish <= df$Finish & df$customer == x$customer,]
totalSale <- sum(tempSum$sale)
totalDays <- sum(tempSum$days)
return(x$days / totalDays * totalSale)
}
for(i in 1:length(loop$customer)){
loop$sale[i] <- sumFun(loop[i,])
}
loop
Here my time period range:
start_day = as.Date('1974-01-01', format = '%Y-%m-%d')
end_day = as.Date('2014-12-21', format = '%Y-%m-%d')
df = as.data.frame(seq(from = start_day, to = end_day, by = 'day'))
colnames(df) = 'date'
I need to created 10,000 data.frames with different fake years of 365days each one. This means that each of the 10,000 data.frames needs to have different start and end of year.
In total df has got 14,965 days which, divided by 365 days = 41 years. In other words, df needs to be grouped 10,000 times differently by 41 years (of 365 days each one).
The start of each year has to be random, so it can be 1974-10-03, 1974-08-30, 1976-01-03, etc... and the remaining dates at the end df need to be recycled with the starting one.
The grouped fake years need to appear in a 3rd col of the data.frames.
I would put all the data.frames into a list but I don't know how to create the function which generates 10,000 different year's start dates and subsequently group each data.frame with a 365 days window 41 times.
Can anyone help me?
#gringer gave a good answer but it solved only 90% of the problem:
dates.df <- data.frame(replicate(10000, seq(sample(df$date, 1),
length.out=365, by="day"),
simplify=FALSE))
colnames(dates.df) <- 1:10000
What I need is 10,000 columns with 14,965 rows made by dates taken from df which need to be eventually recycled when reaching the end of df.
I tried to change length.out = 14965 but R does not recycle the dates.
Another option could be to change length.out = 1 and eventually add the remaining df rows for each column by maintaining the same order:
dates.df <- data.frame(replicate(10000, seq(sample(df$date, 1),
length.out=1, by="day"),
simplify=FALSE))
colnames(dates.df) <- 1:10000
How can I add the remaining df rows to each col?
The seq method also works if the to argument is unspecified, so it can be used to generate a specific number of days starting at a particular date:
> seq(from=df$date[20], length.out=10, by="day")
[1] "1974-01-20" "1974-01-21" "1974-01-22" "1974-01-23" "1974-01-24"
[6] "1974-01-25" "1974-01-26" "1974-01-27" "1974-01-28" "1974-01-29"
When used in combination with replicate and sample, I think this will give what you want in a list:
> replicate(2,seq(sample(df$date, 1), length.out=10, by="day"), simplify=FALSE)
[[1]]
[1] "1985-07-24" "1985-07-25" "1985-07-26" "1985-07-27" "1985-07-28"
[6] "1985-07-29" "1985-07-30" "1985-07-31" "1985-08-01" "1985-08-02"
[[2]]
[1] "2012-10-13" "2012-10-14" "2012-10-15" "2012-10-16" "2012-10-17"
[6] "2012-10-18" "2012-10-19" "2012-10-20" "2012-10-21" "2012-10-22"
Without the simplify=FALSE argument, it produces an array of integers (i.e. R's internal representation of dates), which is a bit trickier to convert back to dates. A slightly more convoluted way to do this is and produce Date output is to use data.frame on the unsimplified replicate result. Here's an example that will produce a 10,000-column data frame with 365 dates in each column (takes about 5s to generate on my computer):
dates.df <- data.frame(replicate(10000, seq(sample(df$date, 1),
length.out=365, by="day"),
simplify=FALSE));
colnames(dates.df) <- 1:10000;
> dates.df[1:5,1:5];
1 2 3 4 5
1 1988-09-06 1996-05-30 1987-07-09 1974-01-15 1992-03-07
2 1988-09-07 1996-05-31 1987-07-10 1974-01-16 1992-03-08
3 1988-09-08 1996-06-01 1987-07-11 1974-01-17 1992-03-09
4 1988-09-09 1996-06-02 1987-07-12 1974-01-18 1992-03-10
5 1988-09-10 1996-06-03 1987-07-13 1974-01-19 1992-03-11
To get the date wraparound working, a slight modification can be made to the original data frame, pasting a copy of itself on the end:
df <- as.data.frame(c(seq(from = start_day, to = end_day, by = 'day'),
seq(from = start_day, to = end_day, by = 'day')));
colnames(df) <- "date";
This is easier to code for downstream; the alternative being a double seq for each result column with additional calculations for the start/end and if statements to deal with boundary cases.
Now instead of doing date arithmetic, the result columns subset from the original data frame (where the arithmetic is already done). Starting with one date in the first half of the frame and choosing the next 14965 values. I'm using nrow(df)/2 instead for a more generic code:
dates.df <-
as.data.frame(lapply(sample.int(nrow(df)/2, 10000),
function(startPos){
df$date[startPos:(startPos+nrow(df)/2-1)];
}));
colnames(dates.df) <- 1:10000;
>dates.df[c(1:5,(nrow(dates.df)-5):nrow(dates.df)),1:5];
1 2 3 4 5
1 1988-10-21 1999-10-18 2009-04-06 2009-01-08 1988-12-28
2 1988-10-22 1999-10-19 2009-04-07 2009-01-09 1988-12-29
3 1988-10-23 1999-10-20 2009-04-08 2009-01-10 1988-12-30
4 1988-10-24 1999-10-21 2009-04-09 2009-01-11 1988-12-31
5 1988-10-25 1999-10-22 2009-04-10 2009-01-12 1989-01-01
14960 1988-10-15 1999-10-12 2009-03-31 2009-01-02 1988-12-22
14961 1988-10-16 1999-10-13 2009-04-01 2009-01-03 1988-12-23
14962 1988-10-17 1999-10-14 2009-04-02 2009-01-04 1988-12-24
14963 1988-10-18 1999-10-15 2009-04-03 2009-01-05 1988-12-25
14964 1988-10-19 1999-10-16 2009-04-04 2009-01-06 1988-12-26
14965 1988-10-20 1999-10-17 2009-04-05 2009-01-07 1988-12-27
This takes a bit less time now, presumably because the date values have been pre-caclulated.
Try this one, using subsetting instead:
start_day = as.Date('1974-01-01', format = '%Y-%m-%d')
end_day = as.Date('2014-12-21', format = '%Y-%m-%d')
date_vec <- seq.Date(from=start_day, to=end_day, by="day")
Now, I create a vector long enough so that I can use easy subsetting later on:
date_vec2 <- rep(date_vec,2)
Now, create the random start dates for 100 instances (replace this with 10000 for your application):
random_starts <- sample(1:14965, 100)
Now, create a list of dates by simply subsetting date_vec2 with your desired length:
dates <- lapply(random_starts, function(x) date_vec2[x:(x+14964)])
date_df <- data.frame(dates)
names(date_df) <- 1:100
date_df[1:5,1:5]
1 2 3 4 5
1 1997-05-05 2011-12-10 1978-11-11 1980-09-16 1989-07-24
2 1997-05-06 2011-12-11 1978-11-12 1980-09-17 1989-07-25
3 1997-05-07 2011-12-12 1978-11-13 1980-09-18 1989-07-26
4 1997-05-08 2011-12-13 1978-11-14 1980-09-19 1989-07-27
5 1997-05-09 2011-12-14 1978-11-15 1980-09-20 1989-07-28
I have two columns of dates. Two example dates are:
Date1= "2015-07-17"
Date2="2015-07-25"
I am trying to count the number of Saturdays and Sundays between the two dates each of which are in their own column (5 & 7 in this example code). I need to repeat this process for each row of my dataframe. The end results will be one column that represents the number of Saturdays and Sundays within the date range defined by two date columns.
I can get the code to work for one row:
sum(weekdays(seq(Date1[1,5],Date2[1,7],"days")) %in% c("Saturday",'Sunday')*1))
The answer to this will be 3. But, if I take out the "1" in the row position of date1 and date2 I get this error:
Error in seq.Date(Date1[, 5], Date2[, 7], "days") :
'from' must be of length 1
How do I go line by line and have one vector that lists the number of Saturdays and Sundays between the two dates in column 5 and 7 without using a loop? Another issue is that I have 2 million rows and am looking for something with a little more speed than a loop.
Thank you!!
map2* functions from the purrr package will be a good way to go. They take two vector inputs (eg two date columns) and apply a function in parallel. They're pretty fast too (eg previous post)!
Here's an example. Note that the _int requests an integer vector back.
library(purrr)
# Example data
d <- data.frame(
Date1 = as.Date(c("2015-07-17", "2015-07-28", "2015-08-15")),
Date2 = as.Date(c("2015-07-25", "2015-08-14", "2015-08-20"))
)
# Wrapper function to compute number of weekend days between dates
n_weekend_days <- function(date_1, date_2) {
sum(weekdays(seq(date_1, date_2, "days")) %in% c("Saturday",'Sunday'))
}
# Iterate row wise
map2_int(d$Date1, d$Date2, n_weekend_days)
#> [1] 3 4 2
If you want to add the results back to your original data frame, mutate() from the dplyr package can help:
library(dplyr)
d <- mutate(d, end_days = map2_int(Date1, Date2, n_weekend_days))
d
#> Date1 Date2 end_days
#> 1 2015-07-17 2015-07-25 3
#> 2 2015-07-28 2015-08-14 4
#> 3 2015-08-15 2015-08-20 2
Here is a solution that uses dplyr to clean things up. It's not too difficult to use with to assign the columns in the dataframe directly.
Essentially, use a reference date, calculate the number of full weeks (by floor or ceiling). Then take the difference between the two. The code does not include cases in which the start date or end data fall on Saturday or Sunday.
# weekdays(as.Date(0,"1970-01-01")) -> "Friday"
require(dplyr)
startDate = as.Date(0,"1970-01-01") # this is a friday
df <- data.frame(start = "2015-07-17", end = "2015-07-25")
df$start <- as.Date(df$start,"", format = "%Y-%m-%d", origin="1970-01-01")
df$end <- as.Date(df$end, format = "%Y-%m-%d","1970-01-01")
# you can use with to define the columns directly instead of %>%
df <- df %>%
mutate(originDate = startDate) %>%
mutate(startDayDiff = as.numeric(start-originDate), endDayDiff = as.numeric(end-originDate)) %>%
mutate(startWeekDiff = floor(startDayDiff/7),endWeekDiff = floor(endDayDiff/7)) %>%
mutate(NumSatsStart = startWeekDiff + ifelse(startDayDiff %% 7>=1,1,0),
NumSunsStart = startWeekDiff + ifelse(startDayDiff %% 7>=2,1,0),
NumSatsEnd = endWeekDiff + ifelse(endDayDiff %% 7 >= 1,1,0),
NumSunsEnd = endWeekDiff + ifelse(endDayDiff %% 7 >= 2,1,0)
) %>%
mutate(NumSats = NumSatsEnd - NumSatsStart, NumSuns = NumSunsEnd - NumSunsStart)
Dates are number of days since 1970-01-01, a Thursday.
So the following is the number of Saturdays or Sundays since that date
f <- function(d) {d <- as.numeric(d); r <- d %% 7; 2*(d %/% 7) + (r>=2) + (r>=3)}
For the number of Saturdays or Sundays between two dates, just subtract, after decrementing the start date to have an inclusive count.
g <- function(d1, d2) f(d2) - f(d1-1)
These are all vectorized functions so you can just call directly on the columns.
# Example data, as in Simon Jackson's answer
d <- data.frame(
Date1 = as.Date(c("2015-07-17", "2015-07-28", "2015-08-15")),
Date2 = as.Date(c("2015-07-25", "2015-08-14", "2015-08-20"))
)
As follows
within(d, end_days<-g(Date1,Date2))
# Date1 Date2 end_days
# 1 2015-07-17 2015-07-25 3
# 2 2015-07-28 2015-08-14 4
# 3 2015-08-15 2015-08-20 2
I have a dataset with dates occurring randomly. For example:
10/21/15, 11/21/15, 11/22/15, 11/28/15,11/30/15, 12/12/15...etc
I am looking to create a rolling average by time-period NOT by at the observation level. For instance if I wanted to do a moving average of the last 7 days. I would not want to look up at the last 7 rows, but rather the last 7 days
For a tiny example:
dates = c('2015-08-07', '2015-08-08','2015-08-09','2015-09-09','2015-10-10')
value = c(5,10,5,3,2)
df=data.frame(dates, value)
df$desired = c(NA,5,7.5, NA,NA)
I am obviously looking to do this for much larger dataset, but I hope you get the idea. If I was to use 7 days for example this is the result I would expect.
Notice that I don't include the current observations value into the rolling average, only the previous. I want rolling average by time period, not observation row number.
I tried looking at rollmean and dplyr but I couldnt figure it out. I don't really care how it happens though.
Thanks!
try this:
rollavgbyperiod <- function(i,window){
startdate <- dates[i]-window
enddate <- dates[i]-1
interval <- seq(startdate,enddate,1)
tmp <- value[dates %in% interval]
return(mean(tmp))
}
dates <- as.Date(dates)
window <- 7
res <- sapply(1:length(dates),function(m) rollavgbyperiod(m,window))
res[is.nan(res)] <- NA
> data.frame(dates,value,res)
dates value res
1 2015-08-07 5 NA
2 2015-08-08 10 5.0
3 2015-08-09 5 7.5
4 2015-09-09 3 NA
5 2015-10-10 2 NA
I suggest using runner package in this case. What is needed here is mean_run with k = 7 window, lagged by 1 period. Simple one-liner:
library(runner)
dates = c('2015-08-07', '2015-08-08','2015-08-09','2015-09-09','2015-10-10')
value = c(5, 10, 5, 3, 2)
mean_run(x = value, k = 7, lag = 1, idx = as.Date(dates))
#[1] NA 5.0 7.5 NA NA
Check package and function documentation
Please help me on this..
so I have daily observations (data frame) for 32-year period. (thus total around 11659 rows: there's some missing rows)
I want to calculate average of each column at every 365th interval (i.e. every jan-01 for 32 year period, every Jan-02 for 32 year period, etc.
so the output would have total 365 rows and each row is average of 32 rows at 365 interval.
any suggestions? I found similar case and tried their solution and modified a bit but the output is not correct. especially I don't understand sapply part below..
df <-data.frame(x=c(1:10000),y=c(1:10000))
byapply <- function(x, by, fun, ...)
{
# Create index list
if (length(by) == 1)
{
nr <- nrow(x)
split.index <- rep(1:ceiling(nr / by), each = by, length.out = nr)
} else
{
nr <- length(by)
split.index <- by
}
index.list <- split(seq(from = 1, to = nr), split.index)
# Pass index list to fun using sapply() and return object #this is where I am lost
sapply(index.list, function(i)
{
do.call(fun, list(x[, i], ...))
})
}
thank you for your time..
How about using the plyr package:
require(plyr) # for aggregating data
require(plyr) # for aggregating data
series<-data.frame(date=as.Date("1964-01-01")+(1:100000),
obs=runif(10000),
obs2=runif(10000),
obs3=runif(10000))
ddply(series, # run on series df
.(DOY=format(date,"%j")), # group by string of day and month (call col DOY)
summarise, # tell the function to summarise by group (day of year)
daymean=mean(obs), # calculate the mean
daymean2=mean(obs2), # calculate the mean
daymean3=mean(obs3) # calculate the mean
)
# DOY daymean daymean2 daymean3
#1 001 0.4957763 0.4882559 0.4944281
#2 002 0.5184197 0.4970996 0.4720893
#3 003 0.5192313 0.5185357 0.4878891
#4 004 0.4787227 0.5150596 0.5317068
#5 005 0.4972933 0.5065012 0.4956527
#6 006 0.5112484 0.5276013 0.4785681
#...
Although there's possibly a special function, which does exactly what you need, here is a solution using ave:
set.seed(1)
dates = seq(from=as.Date("1970-01-01"), as.Date("2000-01-01"), by="day")
df <- data.frame(val1=runif(length(dates)),
val2=rchisq(length(dates), 10))
day <- format(dates, "%j") # day of year (1:366)
df <- cbind(df, setNames(as.data.frame(sapply(df, function(x) {
ave(x, day) # calculate mean by day for df$val1 and df$val2
})), paste0(names(df), "_mean")))
head(df[1:365, 3:4], 3)
# val1_mean val2_mean
# 1 0.5317151 10.485001
# 2 0.5555664 10.490968
# 3 0.6428217 10.763027
That is, if I understood your task correctly.