I have a dataframe in R where each row corresponds to a household. One column describes a date in 2010 when that household planted crops. The remainder of the dataset contains over 1000 columns describing the temperature on every day between 2007-2010 for those households.
This is the basic form:
Date 2007-01-01 2007-01-02 2007-01-03
1 2010-05-01 70 72 61
2 2010-02-10 63 59 73
3 2010-03-06 60 59 81
I need to create columns for each household that describe the monthly mean temperatures of the two months following their planting date in each of the three years prior to 2010.
For instance: if a household planted on 2010-05-01, I would need the following columns:
mean temp of 2007-05-01 through 2007-06-01
mean temp of 2007-06-02 through 2007-07-01
mean temp of 2008-05-01 through 2008-06-01
...
mean temp of 2009-06-02 through 2009-07-01
I skipped two columns, but you get the idea. Specific code would be most helpful, but in general, I am just looking for a way to pull data from specific columns based upon a date that is described by another column.
Hi #bricevk you could use the apply function. It allows you to use a function over a data either column-wise or row-wise.
https://www.rdocumentation.org/packages/base/versions/3.6.2/topics/apply
Say your data is in a object df. It applies the mean function over the columns of df . Giving you the column-wise mean. The 2 indicates the columns. This wpuld the daily average, assuming each column, is a day.
Averages <- apply(df,2,mean)
If I didn't answer this the way you would like perhaps I have not really understood your dataset. Could you try explain it more clearly?
I suggest you to use tidyverse. However, in order to be compatible with this universe, you firstly have to make your data standard, ie tidy. In your example, the things would be easier if you transformed your data in order to have your observations ordrered by rows, and columns being variables. If I correctly understood your data, you have households planting trees (the row names are dates of plantation ?), and then controls with temperature. I'd do something like :
-----------------------------------------------------------------------------
| Household ID | planting date | Date of control | Temperature controlled |
-----------------------------------------------------------------------------
firstly, have your planting date stored as another thing than a rowname, by example :
library(dplyr)
df <- tibble::rownames_to_column(data, "PlantingDate")
You also have to get your household id var you haven't specified to us.
Then you can manage to have the tidy data with tidyr, using
library(tidyr)
df <- gather(df,"DateOfControl","Temperature",-c(PlantingDate,ID))
When you'll have that, you'll be able to use the package lubridate, something like
library(lubridate)
df %>%
group_by(ID,PlantingDate,year(ControlDate),month(ControlDate)) %>%
summarise(MeanT=mean(Temperature))
could work
Related
I have a sample data frame like this (date column format is mm-dd-YYYY):
date count grp
01-09-2009 54 1
01-09-2009 100 2
01-09-2009 546 3
01-10-2009 67 4
01-11-2009 80 5
01-11-2009 45 6
I want to convert this data frame into time series using ts(), but the problem is: the current data frame has multiple values for the same date. Can we apply time series in this case?
Can I convert data frame into time series, and build a model (ARIMA) which can forecast count value on a daily basis?
OR should I forecast count value based on grp, but in that case, I have to select only grp and count column of a data frame. So in that case, I have to skip date column, and daily forecast for count value is not possible?
Suppose if I want to aggregate count value on per day basis. I tried with aggregate function, but there we have to specify date value, but I have a very large data set? Any other option available in r?
Can somebody, please, suggest if there is a better approach to follow? My assumption is that the time series forcast works only for bivariate data? Is this assumption right?
It seems like there are two aspects of your problem:
i want to convert this data frame into time series using ts(), but the
problem is- current data frame having multiple values for the same
date. can we apply time series in this case?
If you are happy making use of the xts package you could attempt:
dta2$date <- as.Date(dta2$date, "%d-%m-%Y")
dtaXTS <- xts::as.xts(dta2[,2:3], dta2$date)
which would result in:
>> head(dtaXTS)
count grp
2009-09-01 54 1
2009-09-01 100 2
2009-09-01 546 3
2009-10-01 67 4
2009-11-01 80 5
2009-11-01 45 6
of the following classes:
>> class(dtaXTS)
[1] "xts" "zoo"
You could then use your time series object as univariate time series and refer to the selected variable or as a multivariate time series, example using PerformanceAnalytics packages:
PerformanceAnalytics::chart.TimeSeries(dtaXTS)
Side points
Concerning your second question:
can somebody plz suggest me what is the better approach to follow, my
assumption is time series forcast is works only for bivariate data? is
this assumption also right?
IMHO, this is rather broad. I would suggest that you use created xts object and elaborate on the model you want to utilise and why, if it's a conceptual question about nature of time series analysis you may prefer to post your follow-up question on CrossValidated.
Data sourced via: dta2 <- read.delim(pipe("pbpaste"), sep = "") using the provided example.
Since daily forecasts are wanted we need to aggregate to daily. Using DF from the Note at the end, read the first two columns of data into a zoo series z using read.zoo and argument aggregate=sum. We could optionally convert that to a "ts" series (tser <- as.ts(z)) although this is unnecessary for many forecasting functions. In particular, checking out the source code of auto.arima we see that it runs x <- as.ts(x) on its input before further processing. Finally run auto.arima, forecast or other forecasting function.
library(forecast)
library(zoo)
z <- read.zoo(DF[1:2], format = "%m-%d-%Y", aggregate = sum)
auto.arima(z)
forecast(z)
Note: DF is given reproducibly here:
Lines <- "date count grp
01-09-2009 54 1
01-09-2009 100 2
01-09-2009 546 3
01-10-2009 67 4
01-11-2009 80 5
01-11-2009 45 6"
DF <- read.table(text = Lines, header = TRUE)
Updated: Revised after re-reading question.
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$mm[(day-5):(day-1)]) # 5 days
}
I have two dataframes that I am trying to merge. One is daily data with days missing (but at least one observation for each month). The other is monthly data (with no months missing). They both span the same time frame.
I would like to merge the data by month (i.e. the month-year of the daily data corresponding with the month-year of the monthly data), keeping the higher frequency.
df1 = daily data (unequal frequency ... i.e. missing days)
df2 = monthly data (equal frequency)
merge(df1, df2) ???
df1.date df1.x df2.y
1/1/2005 5.5 10
1/2/2005 5.9 10
1/5/2005 6.5 10
...
11/2/2005 2.5 12
11/4/2005 3.9 12
11/6/2005 1.3 12
...
Is there anyway to do this in R? (I have been struggling with zoo and ts and haven't found anything even close ... hence this post).
Thank you #user1945827 & #Gregor
I was making a mountain out of a molehill.
As you suggested, all that I needed to do was to create a common index for both datasets to merge on:
lo$monthyear <- format(lo$ListingCreationDate, format='%B-%Y')
ue$monthyear <- format(ue$Month, format='%B-%Y')
lonew <- data.frame(merge(lo, ue, by="monthyear"))
I am posting this as an answer because I literally spent hours using different packages trying to accomplish something that a one sentence answer resolved. Hopefully it will be useful to someone else.
I am wondering how to create a subset of data in R based on a list of dates, rather than by a date range.
For example, I have the following data set data which contains 3 years of 6-minute data.
date zone month day year hour minute temp speed gust dir
1 09/06/2009 00:00 PDT 9 6 2009 0 0 62 2 15 156
2 09/06/2009 00:06 PDT 9 6 2009 0 6 62 13 16 157
I have used breeze<-subset(data, ws>=15 & wd>=247.5 & wd<=315, select=date:dir) to select the rows which meet my criteria for a sea breeze, which is fine, but what I want to do is create a subset of the days which contain those times that meet my criteria.
I have used...
as.character(breeze$date)
trimdate<-strtrim(breeze$date, 10)
breezedate<-as.Date(trimdate, "%m/%d/%Y")
breezedate<-format(breezedate, format="%m/%d/%Y")
...to extract the dates from each row that meets my criteria so I have a variable called breezedate that contains a list of the dates that I want (not the most eloquent coding to do this, I'm sure). There are about two-hundred dates in the list. What I am trying to do with the next command is in my original dataset data to create a subset which contains only those days which meet the seabreeze criteria, not just the specific times.
breezedays<-(data$date==breezedate)
I think one of my issues here is that I am comparing one value to a list of values, but I am not sure how to make it work.
Lets assume your breezedate list looks like this and data$date is simple string:
breezedate <- as.Date(c("2009-09-06", "2009-10-01"))
This is probably want you want:
breezedays <- data[as.Date(data$date, '%m/%d/%Y') %in% breezedate]
The intersect() function (docs) will allow you to compare one data frame to another and return those records that are the same.
To use, run the following:
breezedays <- intersect(data$date,breezedate) # returns into breezedays all records that are shared between data$date and breezedate
I am facing a problem concerning aggregating my data to daily data.
I have a data frame where NAs have been removed (Link of picture of data is given below). Data has been collected 3 times a day, but sometimes due to NAs, there is just 1 or 2 entries per day; some days data is missing completely.
I am now interested in calculating the daily mean of "dist": this means summing up the data of "dist" of one day and dividing it by number of entries per day (so 3 if there is no data missing that day). I would like to do this via a loop.
How can I do this with a loop? The problem is that sometimes I have 3 entries per day and sometimes just 2 or even 1. I would like to tell R that for every day, it should sum up "dist" and divide it by the number of entries that are available for every day.
I just have no idea how to formulate a for loop for this purpose. I would really appreciate if you could give me any advice on that problem. Thanks for your efforts and kind regards,
Jan
Data frame: http://www.pic-upload.de/view-11435581/Data_loop.jpg.html
Edit: I used aggregate and tapply as suggested, however, the mean value of the data was not really calculated:
Group.1 x
1 2006-10-06 12:00:00 636.5395
2 2006-10-06 20:00:00 859.0109
3 2006-10-07 04:00:00 301.8548
4 2006-10-07 12:00:00 649.3357
5 2006-10-07 20:00:00 944.8272
6 2006-10-08 04:00:00 136.7393
7 2006-10-08 12:00:00 360.9560
8 2006-10-08 20:00:00 NaN
The code used was:
dates<-Dis_sub$date
distance<-Dis_sub$dist
aggregate(distance,list(dates),mean,na.rm=TRUE)
tapply(distance,dates,mean,na.rm=TRUE)
Don't use a loop. Use R. Some example data :
dates <- rep(seq(as.Date("2001-01-05"),
as.Date("2001-01-20"),
by="day"),
each=3)
values <- rep(1:16,each=3)
values[c(4,5,6,10,14,15,30)] <- NA
and any of :
aggregate(values,list(dates),mean,na.rm=TRUE)
tapply(values,dates,mean,na.rm=TRUE)
gives you what you want. See also ?aggregate and ?tapply.
If you want a dataframe back, you can look at the package plyr :
Data <- as.data.frame(dates,values)
require(plyr)
ddply(data,"dates",mean,na.rm=TRUE)
Keep in mind that ddply is not fully supporting the date format (yet).
Look at the data.table package especially if your data is huge. Here is some code that calculates the mean of dist by day.
library(data.table)
dt = data.table(Data)
Data[,list(avg_dist = mean(dist, na.rm = T)),'date']
It looks like your main problem is that your date field has times attached. The first thing you need to do is create a column that has just the date using something like
Dis_sub$date_only <- as.Date(Dis_sub$date)
Then using Joris Meys' solution (which is the right way to do it) should work.
However if for some reason you really want to use a loop you could try something like
newFrame <- data.frame()
for d in unique(Dis_sub$date){
meanDist <- mean(Dis_sub$dist[Dis_sub$date==d],na.rm=TRUE)
newFrame <- rbind(newFrame,c(d,meanDist))
}
But keep in mind that this will be slow and memory-inefficient.