Most packages and posts I found apply mean to a fixed size window or the aggregate month/week data. Is it possible to calculate rolling k month average?
For example, for 1 month rolling window, assuming the data is:
Date Value
2012-05-28 101
2012-05-25 99
2012-05-24 102
....
2012-04-30 78
2012-04-27 82
2012-04-26 77
2012-04-25 75
2012-04-24 76
The first three rolling 1 month windows should be:
1. 2012-05-28 to 2012-04-30
2. 2012-05-25 to 2012-04-26
3. 2012-05-24 to 2012-04-25
Please note that this is NOT the fixed width rolling window. The window actually changes on the daily basis.
I used this code to calculate monthly averages based on daily price data.
#function for extracting month is in the lubridate package
install.packages(c("plyr", "lubridate"))
require(plyr); require(lubridate)
#read the daily data
daily = read.csv("daily_lumber_prices.csv")
price = daily$Open
date = daily$Date
#convert date to a usable format
date = strptime(date, "%d-%b-%y")
mon = month(date)
T = length(price)
#need to know when months change
change_month = rep(0,T)
for(t in 2:T){
if(mon[t] != mon[t-1]){
change_month[t-1] = 1
}
}
month_avg = rep(0,T)
total = 0
days = 0
for(t in 1:T){
if(change_month[t] == 0){
#cumulative sums for each variable
total = total + price[t]
days = days + 1
}
else{
#need to include the current month in the calculation
month_avg[t] = (total + price[t]) / (days + 1)
#reset the variables
total = 0
days = 0
}
}
So, the variable month_avg is storing the monthly averages.
Is it something like this? This code accounts for the variable lengths of months. There's certainly a more efficient way to do it, but this works!
Assuming your data frame is df this works for me:
df$past_avg = sapply(df$Date, function(i){
i = as.POSIXct(i)
mean(subset(df, Date > (i - months(1)) & Date < i)$Value)
})
Uses just base R. You can adjust to however many months in the past you want by changing the value in months().
runner package fully supports rolling windows operations on irregulary spaced time series. To calculate 1-month moving average on x object one have to specify idx = date (to make a runner time dependent) and k = "1 months" or k = 30 (days) depending what is more important to user. User can apply any R function - in this case we execute mean.
# example data
x <- cumsum(rnorm(20))
date <- Sys.Date() + cumsum(sample(1:5, 20, replace = TRUE)) # unequaly spaced time series
# calculate rolling average
runner::runner(
x = x,
k = "1 months",
idx = date,
f = mean
)
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
I have made measurements of temperature in a high time resolution of 10 minutes on different urban Tree species, whose reactions should be compared. Therefore I am researching especially periods of heat. The Task that I fail to do on my Dataset is to choose complete days from a maximum value. E.G. Days where there is one measurement above 30 °C should be subsetted from my Dataframe completely.
Below you find a reproducible example that should illustrate my problem:
In my Measurings Dataframe I have calculated a column indicating wether the individual Measurement is above or below 30°C. I wanted to use that column to tell other functions wether they should pick a day or not to produce a New Dataframe. When anytime of the day the value is above 30 ° C i want to include it by Date from 00:00 to 23:59 in that New Dataframe for further analyses.
start <- as.POSIXct("2018-05-18 00:00", tz = "CET")
tseq <- seq(from = start, length.out = 1000, by = "hours")
Measurings <- data.frame(
Time = tseq,
Temp = sample(20:35,1000, replace = TRUE),
Variable1 = sample(1:200,1000, replace = TRUE),
Variable2 = sample(300:800,1000, replace = TRUE)
)
Measurings$heat30 <- ifelse(Measurings$Temp > 30,"heat", "normal")
Measurings$otheroption30 <- ifelse(Measurings$Temp > 30,"1", "0")
The example is yielding a Dataframe analog to the structure of my Data:
head(Measurings)
Time Temp Variable1 Variable2 heat30 otheroption30
1 2018-05-18 00:00:00 28 56 377 normal 0
2 2018-05-18 01:00:00 23 65 408 normal 0
3 2018-05-18 02:00:00 29 78 324 normal 0
4 2018-05-18 03:00:00 24 157 432 normal 0
5 2018-05-18 04:00:00 32 129 794 heat 1
6 2018-05-18 05:00:00 25 27 574 normal 0
So how do I subset to get a New Dataframe where all the days are taken where at least one entry is indicated as "heat"?
I know that for example dplyr:filter could filter the individual entries (row 5 in the head of the example). But how could I tell to take all the day 2018-05-18?
I am quite new to analyzing Data with R so I would appreciate any suggestions on a working solution to my problem. dplyris what I have been using for quite some tasks, but I am open to whatever works.
Thanks a lot, Konrad
Create variable which specify which day (droping hours, minutes etc.). Iterate over unique dates and take only such subsets which in heat30 contains "heat" at least once:
Measurings <- Measurings %>% mutate(Time2 = format(Time, "%Y-%m-%d"))
res <- NULL
newdf <- lapply(unique(Measurings$Time2), function(x){
ss <- Measurings %>% filter(Time2 == x) %>% select(heat30) %>% pull(heat30) # take heat30 vector
rr <- Measurings %>% filter(Time2 == x) # select date x
# check if heat30 vector contains heat value at least once, if so bind that subset
if(any(ss == "heat")){
res <- rbind(res, rr)
}
return(res)
}) %>% bind_rows()
Below is one possible solution using the dataset provided in the question. Please note that this is not a great example as all days will probably include at least one observation marked as over 30 °C (i.e. there will be no days to filter out in this dataset but the code should do the job with the actual one).
# import packages
library(dplyr)
library(stringr)
# break the time stamp into Day and Hour
time_df <- as_data_frame(str_split(Measurings$Time, " ", simplify = T))
# name the columns
names(time_df) <- c("Day", "Hour")
# create a new measurement data frame with separate Day and Hour columns
new_measurings_df <- bind_cols(time_df, Measurings[-1])
# form the new data frame by filtering the days marked as heat
new_df <- new_measurings_df %>%
filter(Day %in% new_measurings_df$Day[new_measurings_df$heat30 == "heat"])
To be more precise, you are creating a random sample of 1000 observations varying between 20 to 35 for temperature across 40 days. As a result, it is very likely that every single day will have at least one observation marked as over 30 °C in your example. Additionally, it is always a good practice to set seed to ensure reproducibility.
I'm stuck on a problem calculating travel dates. I have a data frame of departure dates and return dates.
Departure Return
1 7/6/13 8/3/13
2 7/6/13 8/3/13
3 6/28/13 8/7/13
I want to create and pass a function that will take these dates and form a list of all the days away. I can do this individually by turning each column into dates.
## Turn the departure and return dates into a readable format
Dept <- as.Date(travelDates$Dept, format = "%m/%d/%y")
Retn <- as.Date(travelDates$Retn, format = "%m/%d/%y")
travel_dates <- na.omit(data.frame(dept_dates,retn_dates))
seq(from = travel_dates[1,1], to = travel_dates[1,2], by = 1)
This gives me [1] "2013-07-06" "2013-07-07"... and so on. I want to scale to cover the whole data frame, but my attempts have failed.
Here's one that I thought might work.
days_abroad <- data.frame()
get_days <- function(x,y){
all_days <- seq(from = x, to = y, by =1)
c(days_abroad, all_days)
return(days_abroad)
}
get_days(travel_dates$dept_dates, travel_dates$retn_dates)
I get this error:
Error in seq.Date(from = x, to = y, by = 1) : 'from' must be of length 1
There's probably a lot wrong with this, but what I would really like help on is how to run multiple dates through seq().
Sorry, if this is simple (I'm still learning to think in r) and sorry too for any breaches in etiquette. Thank you.
EDIT: updated as per OP comment.
How about this:
travel_dates[] <- lapply(travel_dates, as.Date, format="%m/%d/%y")
dts <- with(travel_dates, mapply(seq, Departure, Return, by="1 day"))
This produces a list with as many items as you had rows in your initial table. You can then summarize (this will be data.frame with the number of times a date showed up):
data.frame(count=sort(table(Reduce(append, dts)), decreasing=T))
# count
# 2013-07-06 3
# 2013-07-07 3
# 2013-07-08 3
# 2013-07-09 3
# ...
OLD CODE:
The following gets the #days of each trip, rather than a list with the dates.
transform(travel_dates, days_away=Return - Departure + 1)
Which produces:
# Departure Return days_away
# 1 2013-07-06 2013-08-03 29 days
# 2 2013-07-06 2013-08-03 29 days
# 3 2013-06-28 2013-08-07 41 days
If you want to put days_away in a separate list, that is trivial, though it seems more useful to have it as an additional column to your data frame.
I would like a function that counts the number of specific days per month..
i.e.. Nov '13 -> 5 fridays.. while Dec'13 would return 4 Fridays..
Is there an elegant function that would return this?
library(lubridate)
num_days <- function(date){
x <- as.Date(date)
start = floor_date(x, "month")
count = days_in_month(x)
d = wday(start)
sol = ifelse(d > 4, 5, 4) #estimate that is the first day of the month is after Thu or Fri then the week will have 5 Fridays
sol
}
num_days("2013-08-01")
num_days(today())
What would be a better way to do this?
1) Here d is the input, a Date class object, e.g. d <- Sys.Date(). The result gives the number of Fridays in the year/month that contains d. Replace 5 with 1 to get the number of Mondays:
first <- as.Date(cut(d, "month"))
last <- as.Date(cut(first + 31, "month")) - 1
sum(format(seq(first, last, "day"), "%w") == 5)
2) Alternately replace the last line with the following line. Here, the first term is the number of Fridays from the Epoch to the next Friday on or after the first of the next month and the second term is the number of Fridays from the Epoch to the next Friday on or after the first of d's month. Again, we replace all 5's with 1's to get the count of Mondays.
ceiling(as.numeric(last + 1 - 5 + 4) / 7) - ceiling(as.numeric(first - 5 + 4) / 7)
The second solution is slightly longer (although it has the same number of lines) but it has the advantage of being vectorized, i.e. d could be a vector of dates.
UPDATE: Added second solution.
There are a number of ways to do it. Here is one:
countFridays <- function(y, m) {
fr <- as.Date(paste(y, m, "01", sep="-"))
to <- fr + 31
dt <- seq(fr, to, by="1 day")
df <- data.frame(date=dt, mon=as.POSIXlt(dt)$mon, wday=as.POSIXlt(dt)$wday)
df <- subset(df, df$wday==5 & df$mon==df[1,"mon"])
return(nrow(df))
}
It creates the first of the months, and a day in the next months.
It then creates a data frame of month index (on a 0 to 11 range, but we only use this for comparison) and weekday.
We then subset to a) be in the same month and b) on a Friday. That is your result set, and
we return the number of rows as your anwser.
Note that this only uses base R code.
Without using lubridate -
#arguments to pass to function:
whichweekday <- 5
whichmonth <- 11
whichyear <- 2013
#function code:
firstday <- as.Date(paste('01',whichmonth,whichyear,sep="-"),'%d-%m-%Y')
lastday <- if(whichmonth == 12) { '31-12-2013' } else {seq(as.Date(firstday,'%d-%m-%Y'), length=2, by="1 month")[2]-1}
sum(
strftime(
seq.Date(
from = firstday,
to = lastday,
by = "day"),
'%w'
) == whichweekday)
I have a data frame with an year of daily values of rainfall (complete dates in column 1,months in column 2, rainfall in column 3). I am trying to calculate monthly maximum rainfall and I also would like to know the date when the maximum occurred.
I tried the following code:
for (imonth in 1:12) {
month <- which(data[,2]==imonth)
monthly_max[imonth] <- max(data[month,3])
maxi[imonth] <- which.max(data[month,3])
}
tabela <- cbind(monthly_max, maxi)
write.table(tabela, col.names=TRUE, row.names=TRUE, append=FALSE, sep="\t")
The monthly maximum worked perfectly but the which.max function is not working correctly. Is giving me rows that do not correspond to the maximum values of rainfall. Can anybody tell me why or maybe suggest a better way of doing this?
Thank you for helping!
Here is a possible solution using the plyr package
library(plyr)
# create a dummy data frame
df = data.frame(date = sample(LETTERS, 100, replace = T),
month = sample(12, 100, replace = T),
rainfall = sample(1000, 100, replace = F));
# use plyr to figure out max rainfall and date for each month
df.max = ddply(df, .(month), summarize,
max.rain = max(rainfall),
date.max.rain = date[which.max(rainfall)])
Let me know if this works.
EDIT. If there are multiple dates with max rainfall, the code needs to be modified slightly
# find max rainfall for each month
df.max = ddply(df, .(month), transform, max.rain = max(rainfall))
# extract subset such that max.rain = rainfall
df.max = subset(df.max, max.rain == rainfall)
The index function works well here:
library(zoo)
data(AirPassengers)
APZ = zoo(AirPassengers)
ndx = which.max(APZ)
dmax = index(APZ[ndx])
# returns '1960.5' which is Jul 1960 once you know the series freq
frequency(APZ)
# returns 12
I have assumed that you are working with a timeseries object; for those (objects created using eg, ts, zooreg, xts) the dates are actually the value indices. If instead you have a dataframe (ie, so that date is a column in the data frame and the value is another column) then you can just access the row directly.
Edit in light of OP's comment below. For data stored as a data frame:
Suppose your data looks like this, a data frame, D0:
D0[1:10,]
# returns
Time Value
1 2011-03-12 10:48:24 -3.077784
2 2011-03-12 10:49:24 -20.145500
3 2011-03-12 10:50:24 -45.047560
4 2011-03-12 10:51:24 -69.949640
5 2011-03-12 10:52:24 -94.571920
6 2011-03-12 10:53:24 -112.199200
7 2011-03-12 10:54:24 -118.914400
8 2011-03-12 10:55:24 -114.997200
9 2011-03-12 10:56:24 -97.369900
10 2011-03-12 10:57:24 -78.063800
ndx = which.max(D0$Value)
dmax = D0[ndx,] # dmax gives the date corresponding to the max value