I am trying to put together a function that will loop thru a given data frame in blocks and return a new data frame containing stuff calculated from the original. The length of x will be different each time and the actual problem will have more loops in the function. New-ish to R and have not been able to find anything helpful (I don't think using a list will help)
func<-function(x){
tmp # need to declare this here?
for (i in 1:dim(x)[1]){
tmp[i]<-ave(x[i,]) # add things to it
}
return(tmp)
}
df<-cbind(rnorm(10),rnorm(10))
means<-func(df)
This code does not work but I hope it gets across what I want to do. thanks!
Do you mean you want to loop through each row of df and return a data frame with the calculated values?
You may want to look in to the apply function:
df <- cbind(rnorm(10),rnorm(10))
# apply(df,1,FUN) does FUN(df[i,])
# e.g. mean of each row:
apply(df,1,mean)
For more complicated looping like performing some operation on a per-factor basis, I strongly recommend package plyr, and function ddply within. Quick example:
df <- data.frame( gender=c('M','M','F','F'), height=c(183,176,157,168) )
# find mean height *per gender*
ddply(df,.(gender), function(x) c(height=mean(x$height)))
# returns:
gender height
1 F 162.5
2 M 179.5
Related
I have got a data frame with geographic position inside. The positions are strings.
This is my function to scrape the strings and get the positions by Degress.Decimal.
Example position 23º 30.0'N
latitud.decimal <- function(y) {
latregex <- str_match(y,"(\\d+)º\\s(\\d*.\\d*).(.)")
latitud <- (as.numeric(latregex[1,2])) +((as.numeric(latregex[1,3])) / 60)
if (latregex[1,4]=="S") {latitud <- -1*latitud}
return(latitud)
}
Results> 23.5
then I would like to create a new column in my original dataframe applying the function to every item in the Latitude column.
Is the same issue for the longitude. Another new column
I know how to do this using Python and Pandas buy I am newbie y R and cannot find the solution.
I am triying with
lapply(datos$Latitude, 2 , FUN= latitud.decimal(y))
but do not read the y "argument" which is every column value.
Note that the str_match is vectorized as stated in the help page of the function help("str_match").
For the sake of answering the question, I lack a reproducable example and data. This page describes how one can make questions that are more likely to be reproducable and thus obtain better answers.
As i lack data, and code, i cannot test whether i am actually hitting the spot, but i will give it a shot anyway.
Using the fact the str_match is vectorized, we can apply the entire function without using lapply, and thus create a new column simply. I'll slightly rewrite your function, to incorporate the vectorizations. Note the missing 1's in latregex[., .]
latitud.decimal <- function(y) {
latregex <- str_match(y,"(\\d+)º\\s(\\d*.\\d*).(.)")
latitud <- as.numeric(latregex[, 2]) + as.numeric(latregex[, 3]) / 60)
which_south <- which(latregex[, 4] == "S")
latitud[which_south] <- -latitud[which_south]
latitud
}
Now that the function is ready, creating a column can be done using the $ operator. If the data is very large, it can be performed more efficiently using the data.table. See this stackoverflow page for an example of how to assign via the data.table package.
In base R we would simply perform the action as
datos$new_column <- latitud.decimal(datos$Latitude)
datos$lat_decimal = sapply(datos$Latitude, latitud.decimal)
here is how I created number of data sets with names data_1,data_2,data_3 .....and so on
for initial
dim(data)<- 500(rows) 17(column) matrix
for ( i in 1:length(unique( data$cluster ))) {
assign(paste("data", i, sep = "_"),subset(data[data$cluster == i,]))
}
upto this point everything is fine
now I am trying to use these inside the other loop one by one like
for (i in 1:5) {
data<- paste(data, i, sep = "_")
}
however this is not giving me the data with required format
any help will be really appreciated.
Thank you in advance
Let me give you a tip here: Don't just assign everything in the global environment but use lists for this. That way you avoid all the things that can go wrong when meddling with the global environment. The code you have in your question, will overwrite the original dataset data, so you'll be in trouble if you want to rerun that code when something went wrong. You'll have to reconstruct the original dataframe.
Second: If you need to split a data frame based on a factor and carry out some code on each part, you should take a look at split, by and tapply, or at the plyr and dplyr packages.
Using Base R
With base R, it depends on what you want to do. In the most general case you can use a combination of split() and lapply or even a for loop:
mylist <- split( data, f = data$cluster)
for(mydata in mylist){
head(mydata)
...
}
Or
mylist <- split( data, f = data$cluster)
result <- lapply(mylist, function(mydata){
doSomething(mydata)
})
Which one you use, depends largely on what the result should be. If you need some kind of a summary for every subset, using lapply will give you a list with the results per subset. If you need this for a simulation or plotting or so, you better use the for loop.
If you want to add some variables based on other variables, then the plyr or dplyr packages come in handy
Using plyr and dplyr
These packages come especially handy if the result of your code is going to be an array or data frame of some kind. This would be similar to using split and lapply but then in a way Hadley approves of :-)
For example:
library(plyr)
result <- ddply(data, .(cluster),
function(mydata){
doSomething(mydata)
})
Use dlply if the result should be a list.
I am writing a script that ultimately returns a data frame. My question is around if there are any good practices on how to use a unit test package to make sure that the data frame that is returned is correct. (I'm a beginning R programmer, plus new to the concept of unit testing)
My script effectively looks like the following:
# initialize data frame
df.out <- data.frame(...)
# function set
function1 <- function(x) {...}
function2 <- function(x) {...}
# do something to this data frame
df.out$new.column <- function1(df.out)
# do something else
df.out$other.new.column <- function2(df.out)
# etc ....
... and I ultimately end up with a data frame with many new columns. However, what is the best approach to test that the data frame that is produced is what is anticipated, using unit tests?
So far I have created unit tests that check the results of each function, but I want to make sure that running all of these together produces what is intended. I've looked at Hadley Wickham's page on testing but can't see anything obvious regarding what to do when returning data frames.
My thoughts to date are:
Create an expected data frame by hand
Check that the output equals this data frame, using expect_that or similar
Any thoughts / pointers on where to look for guidance? My Google-fu has let me down considerably on this one to date.
Your intuition seems correct. Construct a data.frame manually based on the expected output of the function and then compare that against the function's output.
# manually created data
dat <- iris[1:5, c("Species", "Sepal.Length")]
# function
myfun <- function(row, col, data) {
data[row, col]
}
# result of applying function
outdat <- myfun(1:5, c("Species", "Sepal.Length"), iris)
# two versions of the same test
expect_true(identical(dat, outdat))
expect_identical(dat, outdat)
If your data.frame may not be identical, you could also run tests in parts of the data.frame, including:
dim(outdat), to check if the size is correct
attributes(outdat) or attributes of columns
sapply(outdat, class), to check variable classes
summary statistics for variables, if applicable
and so forth
If you would like to test this at runtime, you should check out the excellent ensurer package, see here. At the bottom of the page you can see how to construct a template that you can test your dataframe against, you can make it as detailed and specific as you like.
I'm just using something like this
d1 <- iris
d2 <- iris
expect_that(d1, equals(d2)) # passes
d3 <- iris
d3[141,3] <- 5
expect_that(d1, equals(d3)) # fails
I need to do a quality control in a dataset with more than 3000 variables (columns). However, I only want to apply some conditions in a couple of them. A first step would be to replace outliers by NA. I want to replace the observations that are greater or smaller than 3 standard deviations from the mean by NA. I got it, doing column by column:
height = ifelse(abs(height-mean(height,na.rm=TRUE)) <
3*sd(height,na.rm=TRUE),height,NA)
And I also want to create other variables based on different columns. For example:
data$CGmark = ifelse(!is.na(data$mark) & !is.na(data$height) ,
paste(data$age, data$mark,sep=""),NA)
An example of my dataset would be:
name = factor(c("A","B","C","D","E","F","G","H","H"))
height = c(120,NA,150,170,NA,146,132,210,NA)
age = c(10,20,0,30,40,50,60,NA,130)
mark = c(100,0.5,100,50,90,100,NA,50,210)
data = data.frame(name=name,mark=mark,age=age,height=height)
data
I have tried this (for one condition):
d1=names(data)
list = c("age","height","mark")
ntraits=length(list)
nrows=dim(data)[1]
for(i in 1:ntraits){
a=list[i]
b=which(d1==a)
d2=data[,b]
for (j in 1:nrows){
d2[j] = ifelse(abs(d2[j]-mean(d2,na.rm=TRUE)) < 3*sd(d2,na.rm=TRUE),d2[j],NA)
}
}
Someone told me that I am not storing d2. How can I create for loops to apply the conditions I want? I know that there are similar questions but i didnt get it yet. Thanks in advance.
You pretty much wrote the answer in your first line. You're overthinking this one.
First, it's good practice to encapsulate this kind of operation in a function. Yes, function dispatch is a tiny bit slower than otherwise, but the code is often easier to read and debug. Same goes for assigning "helper" variables like mean_x: the cost of assigning the variable is very, very small and absolutely not worth worrying about.
NA_outside_3s <- function(x) {
mean_x <- mean(x)
sd_x <- sd(x,na.rm=TRUE)
x_outside_3s <- abs(x - mean(x)) < 3 * sd_x
x[x_outside_3s] <- NA # no need for ifelse here
x
}
of course, you can choose any function name you want. More descriptive is better.
Then if you want to apply the function to very column, just loop over the columns. That function NA_outside_3s is already vectorized, i.e. it takes a logical vector as an argument and returns a vector of the same length.
cols_to_loop_over <- 1:ncol(my_data) # or, some subset of columns.
for (j in cols_to_loop_over) {
my_data[, j] <- NA_if_3_sd(my_data[, j])
}
I'm not sure why you wrote your code the way you did (and it took me a minute to even understand what you were trying to do), but looping over columns is usually straightforward.
In my comment I said not to worry about efficiency, but once you understand how the loop works, you should rewrite it using lapply:
my_data[cols_to_loop_over] <- lapply(my_data[cols_to_loop_over], NA_outside_3s)
Once you know how the apply family of functions works, they are very easy to read if written properly. And yes, they are somewhat faster than looping, but not as much as they used to be. It's more a matter of style and readability.
Also: do NOT name a variable list! This masks the function list, which is an R built-in function and a fairly important one at that. You also shouldn't generally name variables data because there is also a data function for loading built-in data sets.
I am trying to run some Monte Carlo simulations on animal position data. So far, I have sampled 100 X and Y coordinates, 100 times. This results in a list of 200. I then convert this list into a dataframe that is more condusive to eventual functions I want to run for each sample (kernel.area).
Now I have a data frame with 200 columns, and I would like to perform the kernel.area function using each successive pair of columns.
I can't reproduce my own data here very well, so I've tried to give a basic example just to show the structure of the data frame I'm working with. I've included the for loop I've tried so far, but I am still an R novice and would appreciate any suggestions.
# generate dataframe representing X and Y positions
df <- data.frame(x=seq(1:200),y=seq(1:200))
# 100 replications of sampling 100 "positions"
resamp <- replicate(100,df[sample(nrow(df),100),])
# convert to data frame (kernel.area needs an xy dataframe)
df2 <- do.call("rbind", resamp[1:2,])
# xy positions need to be in columns for kernel.area
df3 <- t(df2)
#edit: kernel.area requires you have an id field, but I am only dealing with one individual, so I'll construct a fake one of the same length as the positions
id=replicate(100,c("id"))
id=data.frame(id)
Here is the structure of the for loop I've tried (edited since first post):
for (j in seq(1,ncol(df3)-1,2)) {
kud <- kernel.area(df3[,j:(j+1)],id=id,kern="bivnorm",unin=c("m"),unout=c("km2"))
print(kud)
}
My end goal is to calculate kernel.area for each resampling event (ie rows 1:100 for every pair of columns up to 200), and be able to combine the results in a dataframe. However, after running the loop, I get this error message:
Error in df[, 1] : incorrect number of dimensions
Edit: I realised my id format was not the same as my data frame, so I change it and now have the error:
Error in kernelUD(xy, id, h, grid, same4all, hlim, kern, extent) :
id should have the same length as xy
First, a disclaimer: I have never worked with the package adehabitat, which has a function kernel.area, which I assume you are using. Perhaps you could confirm which package contains the function in question.
I think there are a couple suggestions I can make that are independent of knowledge of the specific package, though.
The first lies in the creation of df3. This should probably be
df3 <- t(df2), but this is most likely correct in your actual code
and just a typo in your post.
The second suggestion has to do with the way you subset df3 in the
loop. j:j+1 is just a single number, since the : has a higher
precedence than + (see ?Syntax for the order in which
mathematical operations are conducted in R). To get the desired two
columns, use j:(j+1) instead.
EDIT:
When loading adehabitat, I was warned to "Be careful" and use the related new packages, among which is adehabitatHR, which also contains a function kernel.area. This function has slightly different syntax and behavior, but perhaps it would be worthwhile examining. Using adehabitatHR (I had to install from source since the package is not available for R 2.15.0), I was able to do the following.
library(adehabitatHR)
for (j in seq(1,ncol(df3)-1,2)) {
kud <-kernelUD(SpatialPoints(df3[,j:(j+1)]),kern="bivnorm")
kernAr<-kernel.area(kud,unin=c("m"),unout=c("km2"))
print(kernAr)
}
detach(package:adehabitatHR, unload=TRUE)
This prints something, and as is mentioned in a comment below, kernelUD() is called before kernel.area().