I'm a new user of R and I'm a little bit stuck, my data looks like this:
dates temp
01/31/2011 40
01/30/2011 34
01/29/2011 30
01/28/2011 52
01/27/2011 39
01/26/2011 37
...
01/01/2011 31
i want take only temp under 40 degrees and with the dates of beginning and the end and how many days it lasts, for example:
from to days
01/29/2011 01/30/2011 2
01/26/2011 01/27/2011 2
I tried with difftime but it didn't work, maybe with a function it will.
any help would be appreciated.
I'd do something like this. I'll use data.table here.
df <- read.table(header=TRUE, text="dates temp
01/31/2011 40
01/30/2011 34
01/29/2011 30
01/28/2011 52
01/27/2011 39
01/26/2011 37", stringsAsFactors=FALSE)
require(data.table)
dt <- data.table(df)
dt <- dt[, `:=`(date.form = as.Date(dates, format="%m/%d/%Y"),
id = cumsum(as.numeric(temp >= 40)))][temp < 40]
dt[, list(from=min(date.form), to=max(date.form), count=.N), by=id]
# id from to count
# 1: 1 2011-01-29 2011-01-30 2
# 2: 2 2011-01-26 2011-01-27 2
The idea is to first create a column with the dates column converted to Date format first. Then, another column id that finds the positions where temp >= 40 and uses that to create the group of values that are within two temp>=40. That is, if you have c(40, 34, 30, 52, 39, 37), then you'd want c(1,1,1,2,2,2). That is, everything between to values >= 40, must belong to the same group (34, 30 -> 1 and 39, 37 -> 2). After doing this, I'd remove temp >= 40 entries.
then, you can split by this group and then take min and max and length(.) (which is by default stored in .N).
Not as elegant as Arun's data.table but here is base solution
DF <- read.table(text = "dates temp\n01/31/2011 40\n01/30/2011 34\n01/29/2011 30\n01/28/2011 52\n01/27/2011 39\n01/26/2011 37",
header = TRUE, stringsAsFactors = FALSE)
DF$dates <- as.POSIXct(DF$dates, format = "%m/%d/%Y")
DF <- DF[order(DF$dates), ]
DF$ID <- cumsum(DF$temp >= 40)
DF2 <- DF[DF$temp < 40, ]
# Explanation split : split DF2 by DF2$ID
# lapply : apply function on each list element given by split
# rbind : bind all the data together
do.call(rbind, lapply(split(DF2, DF2$ID), function(x)
data.frame(from = min(x$dates),
to = max(x$dates),
count = length(x$dates))))
## from to count
## 0 2011-01-26 2011-01-27 2
## 1 2011-01-29 2011-01-30 2
First read in the data. read.zoo handles many of the details all in one line including reordering the data to be ascending and converting the dates to "Date" class. If z is the resulting zoo object then coredata(z) gives the temperatures and time(z) gives the dates.
Lines <- "
dates temp
01/31/2011 40
01/30/2011 34
01/29/2011 30
01/28/2011 52
01/27/2011 39
01/26/2011 37
"
library(zoo)
z <- read.zoo(text = Lines, header = TRUE, format = "%m/%d/%Y")
The crux of all this is the use of rle which computes lengths and values from which we can derive all quantities:
tt <- time(z)
with(rle(coredata(z) < 40), {
to <- cumsum(lengths)[values]
lengths <- lengths[values]
from <- to - lengths + 1
data.frame(from = tt[from], to = tt[to], days = lengths)
})
Using the first 6 lines of the input data shown we get:
from to days
1 2011-01-26 2011-01-27 2
2 2011-01-29 2011-01-30 2
Related
I am stuck trying to combine two time series datasets that have different ranges and both are stored with item# in column1 and date as column headings. For example:
df1
#ITEM 1/1/16 1/2/16 1/3/16 ... 3/24/17
#1 350 365 370 ... 400
#2 100 95 101 ... 95
#3 5 8 9 ... 15
The other dataset range is smaller, its in the same format, and both are daily frequency.
How can I append the rows of df2 to df1 despite having different ranges, but making sure the dates are aligned when merged? Happy with NA in the new dataframe where df#2 didn't have values for dates in df1
Should I create these at xts objects so that once they are merged I can easily pull data for item1 on X date? Or is there an easy way to do that with this format as well?
Thanks in advance for you help.
One option is to use data.table::rbindlist(df1, df2) with fill = TRUE
that fills missing columns with NAs.
Example:
library(data.table)
dt1 <- data.table(item=c(1,2,3),"d1/1/16" = c(350,100,5) ,"d1/2/16" = c(360,120,7))
dt2 <- data.table(item=c(3,4,5),"d1/2/16" = c(50,50,2) ,"d1/3/16" = c(460,150,9))
l = list(dt1,dt2)
data.table::rbindlist(l, use.names= TRUE, fill=TRUE, idcol=TRUE )
Normally in R time series are represented in columns, not rows. Assuming we have DF1 and DF2 shown reproducibly in the Note at the end here are some alternatives
1) zoo we can create zoo series from each by transposing. Then merge them:
library(zoo)
fmt <- "%m/%d/%y"
z1 <- setNames(zoo(t(DF1[-1]), as.Date(names(DF1[-1]), fmt)), DF1[[1]])
z2 <- setNames(zoo(t(DF2[-1]), as.Date(names(DF2[-1]), ftm)), DF2[[1]])
z <- merge(z1, z2)
It is probably best to leave this as the zoo series z but if you want to transform to a data frame then use: fortity.zoo(z)
2) base Alternately, without zoo using fmt from above:
d1 <- data.frame(as.Date(names(DF1[-1]), fmt), t(DF1[-1]))
names(d1) <- c("Index", DF1[[1]])
d2 <- data.frame(as.Date(names(DF2[-1]), fmt), t(DF2[-1]))
names(d2) <- c("Index", DF2[[1]])
merge(d1, d2, by = "Index", all = TRUE)
Note: The input in reproducible form is assumed to be:
Lines <- "ITEM 1/1/16 1/2/16 1/3/16 3/24/17
1 350 365 370 400
2 100 95 101 95
3 5 8 9 15"
DF <- read.table(text = Lines, header = TRUE, check.names = FALSE)
DF1 <- DF[1:2, 1:3]
DF2 <- DF[3, -3]
I have 2 dataframes, one representing daily sales figures of different stores (df1) and one representing when each store has been audited (df2). I need to create a new dataframe displaying sales information from each site taken 1 week before each audit (i.e. the information in df2). Some example data, firstly for the daily sales figures from different stores across a certain period:
Dates <- as.data.frame(seq(as.Date("2015/12/30"), as.Date("2016/4/7"),"day"))
Sales <- as.data.frame(matrix(sample(0:50, 30*10, replace=TRUE), ncol=3))
df1 <- cbind(Dates,Sales)
colnames(df1) <- c("Dates","Site.A","Site.B","Site.C")
And for the dates of each audit across different stores:
Store<- c("Store.A","Store.A","Store.B","Store.C","Store.C")
Audit_Dates <- as.data.frame(as.POSIXct(c("2016/1/4","2016/3/1","2016/2/1","2016/2/1","2016/3/1")))
df2 <- as.data.frame(cbind(Store,Audit_Dates ))
colnames(df2) <- c("Store","Audit_Dates")
Of note is that there will be an uneven amount of dates within each output (i.e. there may not be a full weeks worth of information prior to some store audits). I have previously asked a question addressing a similar problem Creating a dataframe from an lapply function with different numbers of rows. Below shows an answer from this which would work for an example if I was to consider information from only 1 store:
library(lubridate)
##Data input
Store.A_Dates <- as.data.frame(seq(as.Date("2015/12/30"), as.Date("2016/4/7"),"day"))
Store.A_Sales <- as.data.frame(matrix(sample(0:50, 10*10, replace=TRUE), ncol=1))
Store.A_df1 <- cbind(Store.A_Dates,Store.A_Sales)
colnames(Store.A_df1) <- c("Store.A_Dates","Store.A_Sales")
Store.A_df2 <- as.Date(c("2016/1/3","2016/3/1"))
##Output
Store.A_output<- lapply(Store.A_df2, function(x) {Store.A_df1[difftime(Store.A_df1[,1], x - days(7)) >= 0 & difftime(Store.A_df1[,1], x) <= 0, ]})
n1 <- max(sapply(Store.A_output, nrow))
output <- data.frame(lapply(Store.A_output, function(x) x[seq_len(n1),]))
But I don't know how I would get this for multiple sites.
Try this:
# Renamed vars for my convenience...
colnames(df1) <- c("t","Store.A","Store.B","Store.C")
colnames(df2) <- c("Store","t")
library(tidyr)
library(dplyr)
# Gather df1 so that df1 and df2 have the same format:
df1 = gather(df1, Store, Sales, -t)
head(df1)
t Store Sales
1 2015-12-30 Store.A 16
2 2015-12-31 Store.A 24
3 2016-01-01 Store.A 8
4 2016-01-02 Store.A 42
5 2016-01-03 Store.A 7
6 2016-01-04 Store.A 46
# This lapply call does not iterate over actual values, just indexes, which allows
# you to subset the data comfortably:
r <- lapply(1:nrow(df2), function(i) {
audit.t = df2[i, "t"] #time of audit
audit.s = df1[, "Store"] == df2[i, "Store"] #store audited
df = df1[audit.s, ] #data from audited store
df[, "audited"] = audit.t #add extra column with audit date
week_before = difftime(df[, "t"], audit.t - (7*24*3600)) >= 0
week_audit = difftime(df[, "t"], audit.t) <= 0
df[week_before & week_audit, ]
})
Does this give you the proper subsets?
Also, to summarise your results:
r = do.call("rbind", r) %>%
group_by(audited, Store) %>%
summarise(sales = sum(Sales))
r
audited Store sales
<time> <chr> <int>
1 2016-01-04 Store.A 97
2 2016-02-01 Store.B 156
3 2016-02-01 Store.C 226
4 2016-03-01 Store.A 115
5 2016-03-01 Store.C 187
I need to calculate the number of days elapsed between multiple dates in two ways and then output those results to new columns: i) number of days that has elapsed as compared to the first date (e.g., RESULTS$FIRST) and ii) between sequential dates (e.g., RESULTS$BETWEEN). Here is an example with the desired results. Thanks in advance.
library(lubridate)
DATA = data.frame(DATE = mdy(c("7/8/2013", "8/1/2013", "8/30/2013", "10/23/2013",
"12/16/2013", "12/16/2015")))
RESULTS = data.frame(DATE = mdy(c("7/8/2013", "8/1/2013", "8/30/2013", "10/23/2013",
"12/16/2013", "12/16/2015")),
FIRST = c(0, 24, 53, 107, 161, 891), BETWEEN = c(0, 24, 29, 54, 54, 730))
#Using dplyr package
library(dplyr)
df1 %>% # your dataframe
mutate(BETWEEN0=as.numeric(difftime(DATE,lag(DATE,1))),BETWEEN=ifelse(is.na(BETWEEN0),0,BETWEEN0),FIRST=cumsum(as.numeric(BETWEEN)))%>%
select(-BETWEEN0)
DATE BETWEEN FIRST
1 2013-07-08 0 0
2 2013-08-01 24 24
3 2013-08-30 29 53
4 2013-10-23 54 107
5 2013-12-16 54 161
6 2015-12-16 730 891
This will get you what you want:
d <- as.Date(DATA$DATE, format="%m/%d/%Y")
first <- c()
for (i in seq_along(d))
first[i] <- d[i] - d[1]
between <- c(0, diff(d))
This uses the as.Date() function in the base package to cast the vector of string dates to date values using the given format. Since you have dates as month/day/year, you specify format="%m/%d/%Y" to make sure it's interpreted correctly.
diff() is the lagged difference. Since it's lagged, it doesn't include the difference between element 1 and itself, so you can concatenate a 0.
Differences between Date objects are given in days by default.
Then constructing the output dataframe is simple:
RESULTS <- data.frame(DATE=DATA$DATE, FIRST=first, BETWEEN=between)
For the first part:
DATA = data.frame((c("7/8/2013", "8/1/2013", "8/30/2013", "10/23/2013","12/16/2013", "12/16/2015")))
names(DATA)[1] = "V1"
date = as.Date(DATA$V1, format="%m/%d/%Y")
print(date-date[1])
Result:
[1] 0 24 53 107 161 891
For second part - simply use a for loop
You can just add each column with the simple difftime and lagged diff calculations.
DATA$FIRST <- c(0,
with(DATA,
difftime(DATE[2:length(DATE)],DATE[1], unit="days")
)
)
DATA$BETWEEN <- c(0,
with(DATA,
diff(DATE[1:(length(DATE) - 1)], unit="days")
)
)
identical(DATA, RESULTS)
[1] TRUE
I have the following example:
Date1 <- seq(from = as.POSIXct("2010-05-01 02:00"),
to = as.POSIXct("2010-10-10 22:00"), by = 3600)
Dat <- data.frame(DateTime = Date1,
t = rnorm(length(Date1)))
I would like to find the range of values in a given day (i.e. maximum - minimum).
First, I've defined additional columns which define the unique days in terms of the date and in terms of the day of year (doy).
Dat$date <- format(Dat$DateTime, format = "%Y-%m-%d") # find the unique days
Dat$doy <- as.numeric(format(Dat$DateTime, format="%j")) # find the unique days
To then find the range I tried
by(Dat$t, Dat$doy, function(x) range(x))
but this returns the range as two values not a single value, So, my question is, how do I find the calculated range for each day and return them in a data.frame which has
new_data <- data.frame(date = unique(Dat$date),
range = ...)
Can anyone suggest a method for doing this?
I tend to use tapply for this kind of thing. ave is also useful sometimes. Here:
> dr = tapply(Dat$t,Dat$doy,function(x){diff(range(x))})
Always check tricksy stuff:
> dr[1]
121
3.084317
> diff(range(Dat$t[Dat$doy==121]))
[1] 3.084317
Use the names attribute to get the day-of-year and the values to make a data frame:
> new_data = data.frame(date=names(dr),range=dr)
> head(new_data)
date range
121 121 3.084317
122 122 4.204053
Did you want to convert the number day-of-year back to a date object?
# Use the data.table package
require(data.table)
# Set seed so data is reproducible
set.seed(42)
# Create data.table
Date1 <- seq(from = as.POSIXct("2010-05-01 02:00"), to = as.POSIXct("2010-10-10 22:00"), by = 3600)
DT <- data.table(date = as.IDate(Date1), t = rnorm(length(Date1)))
# Set key on data.table so that it is sorted by date
setkey(DT, "date")
# Make a new data.table with the required information (can be used as a data.frame)
new_data <- DT[, diff(range(t)), by = date]
# date V1
# 1: 2010-05-01 4.943101
# 2: 2010-05-02 4.309401
# 3: 2010-05-03 4.568818
# 4: 2010-05-04 2.707036
# 5: 2010-05-05 4.362990
# ---
# 159: 2010-10-06 2.659115
# 160: 2010-10-07 5.820803
# 161: 2010-10-08 4.516654
# 162: 2010-10-09 4.010017
# 163: 2010-10-10 3.311408
I don't often have to work with dates in R, but I imagine this is fairly easy. I have daily data as below for several years with some values and I want to get for each 8 days period the sum of related values.What is the best approach?
Any help you can provide will be greatly appreciated!
str(temp)
'data.frame':648 obs. of 2 variables:
$ Date : Factor w/ 648 levels "2001-03-24","2001-03-25",..: 1 2 3 4 5 6 7 8 9 10 ...
$ conv2: num -3.93 -6.44 -5.48 -6.09 -7.46 ...
head(temp)
Date amount
24/03/2001 -3.927020472
25/03/2001 -6.4427004
26/03/2001 -5.477592528
27/03/2001 -6.09462162
28/03/2001 -7.45666902
29/03/2001 -6.731540928
30/03/2001 -6.855206184
31/03/2001 -6.807210228
1/04/2001 -5.40278802
I tried to use aggregate function but for some reasons it doesn't work and it aggregates in wrong way:
z <- aggregate(amount ~ Date, timeSequence(from =as.Date("2001-03-24"),to =as.Date("2001-03-29"), by="day"),data=temp,FUN=sum)
I prefer the package xts for such manipulations.
I read your data, as zoo objects. see the flexibility of format option.
library(xts)
ts.dat <- read.zoo(text ='Date amount
24/03/2001 -3.927020472
25/03/2001 -6.4427004
26/03/2001 -5.477592528
27/03/2001 -6.09462162
28/03/2001 -7.45666902
29/03/2001 -6.731540928
30/03/2001 -6.855206184
31/03/2001 -6.807210228
1/04/2001 -5.40278802',header=TRUE,format = '%d/%m/%Y')
Then I extract the index of given period
ep <- endpoints(ts.dat,'days',k=8)
finally I apply my function to the time series at each index.
period.apply(x=ts.dat,ep,FUN=sum )
2001-03-29 2001-04-01
-36.13014 -19.06520
Use cut() in your aggregate() command.
Some sample data:
set.seed(1)
mydf <- data.frame(
DATE = seq(as.Date("2000/1/1"), by="day", length.out = 365),
VALS = runif(365, -5, 5))
Now, the aggregation. See ?cut.Date for details. You can specify the number of days you want in each group using cut:
output <- aggregate(VALS ~ cut(DATE, "8 days"), mydf, sum)
list(head(output), tail(output))
# [[1]]
# cut(DATE, "8 days") VALS
# 1 2000-01-01 8.242384
# 2 2000-01-09 -5.879011
# 3 2000-01-17 7.910816
# 4 2000-01-25 -6.592012
# 5 2000-02-02 2.127678
# 6 2000-02-10 6.236126
#
# [[2]]
# cut(DATE, "8 days") VALS
# 41 2000-11-16 17.8199285
# 42 2000-11-24 -0.3772209
# 43 2000-12-02 2.4406024
# 44 2000-12-10 -7.6894484
# 45 2000-12-18 7.5528077
# 46 2000-12-26 -3.5631950
rollapply. The zoo package has a rolling apply function which can also do non-rolling aggregations. First convert the temp data frame into zoo using read.zoo like this:
library(zoo)
zz <- read.zoo(temp)
and then its just:
rollapply(zz, 8, sum, by = 8)
Drop the by = 8 if you want a rolling total instead.
(Note that the two versions of temp in your question are not the same. They have different column headings and the Date columns are in different formats. I have assumed the str(temp) output version here. For the head(temp) version one would have to add a format = "%d/%m/%Y" argument to read.zoo.)
aggregate. Here is a solution that does not use any external packages. It uses aggregate based on the original data frame.
ix <- 8 * ((1:nrow(temp) - 1) %/% 8 + 1)
aggregate(temp[2], list(period = temp[ix, 1]), sum)
Note that ix looks like this:
> ix
[1] 8 8 8 8 8 8 8 8 16
so it groups the indices of the first 8 rows, the second 8 and so on.
Those are NOT Date classed variables. (No self-respecting program would display a date like that, not to mention the fact that these are labeled as factors.) [I later noticed these were not the same objects.] Furthermore, the timeSequence function (at least the one in the timeDate package) does not return a Date class vector either. So your expectation that there would be a "right way" for two disparate non-Date objects to be aligned in a sensible manner is ill-conceived. The irony is that just using the temp$Date column would have worked since :
> z <- aggregate(amount ~ Date, data=temp , FUN=sum)
> z
Date amount
1 1/04/2001 -5.402788
2 24/03/2001 -3.927020
3 25/03/2001 -6.442700
4 26/03/2001 -5.477593
5 27/03/2001 -6.094622
6 28/03/2001 -7.456669
7 29/03/2001 -6.731541
8 30/03/2001 -6.855206
9 31/03/2001 -6.807210
But to get it in 8 day intervals use cut.Date:
> z <- aggregate(temp$amount ,
list(Dts = cut(as.Date(temp$Date, format="%d/%m/%Y"),
breaks="8 day")), FUN=sum)
> z
Dts x
1 2001-03-24 -49.792561
2 2001-04-01 -5.402788
A more cleaner approach extended to #G. Grothendieck appraoch. Note: It does not take into account if the dates are continuous or discontinuous, sum is calculated based on the fixed width.
code
interval = 8 # your desired date interval. 2 days, 3 days or whatevea
enddate = interval-1 # this sets the enddate
nrows = nrow(z)
z <- aggregate(.~V1,data = df,sum) # aggregate sum of all duplicate dates
z$V1 <- as.Date(z$V1)
data.frame ( Start.date = (z[seq(1, nrows, interval),1]),
End.date = z[seq(1, nrows, interval)+enddate,1],
Total.sum = rollapply(z$V2, interval, sum, by = interval, partial = TRUE))
output
Start.date End.date Total.sum
1 2000-01-01 2000-01-08 9.1395926
2 2000-01-09 2000-01-16 15.0343960
3 2000-01-17 2000-01-24 4.0974712
4 2000-01-25 2000-02-01 4.1102645
5 2000-02-02 2000-02-09 -11.5816277
data
df <- data.frame(
V1 = seq(as.Date("2000/1/1"), by="day", length.out = 365),
V2 = runif(365, -5, 5))