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)
Related
I have a tibble called 'Volume' in which I store some data (10 columns - the first 2 columns are characters, 30 rows).
Now I want to calculate the relative Volume of every column that corresponds to Column 3 of my tibble.
My current solution looks like this:
rel.Volume_unmod = tibble(
"Volume_OD" = Volume[[3]] / Volume[[3]],
"Volume_Imp" = Volume[[4]] / Volume[[3]],
"Volume_OD_1" = Volume[[5]] / Volume[[3]],
"Volume_WS_1" = Volume[[6]] / Volume[[3]],
"Volume_OD_2" = Volume[[7]] / Volume[[3]],
"Volume_WS_2" = Volume[[8]] / Volume[[3]],
"Volume_OD_3" = Volume[[9]] / Volume[[3]],
"Volume_WS_3" = Volume[[10]] / Volume[[3]])
rel.Volume_unmod
I would like to keep the tibble structure and the labels. I am sure there is a better solution for this, but I am relative new to R so I it's not obvious to me. What I tried is something like this, but I can't actually run this:
rel.Volume = NULL
for(i in Volume[,3:10]){
rel.Volume[i] = tibble(Volume = Volume[[i]] / Volume[[3]])
}
Mockup Data
Since you did not provide some data, I've followed the description you provided to create some mockup data. Here:
set.seed(1)
Volume <- data.frame(ID = sample(letters, 30, TRUE),
GR = sample(LETTERS, 30, TRUE))
Volume[3:10] <- rnorm(30*8)
Solution with Dplyr
library(dplyr)
# rename columns [brute force]
cols <- c("Volume_OD","Volume_Imp","Volume_OD_1","Volume_WS_1","Volume_OD_2","Volume_WS_2","Volume_OD_3","Volume_WS_3")
colnames(Volume)[3:10] <- cols
# divide by Volumn_OD
rel.Volume_unmod <- Volume %>%
mutate(across(all_of(cols), ~ . / Volume_OD))
# result
rel.Volume_unmod
Explanation
I don't know the names of your columns. Probably, the names correspond to the names of the columns you intended to create in rel.Volume_unmod. Anyhow, to avoid any problem I renamed the columns (kinda brutally). You can do it with dplyr::rename if you wan to.
There are many ways to select the columns you want to mutate. mutate is a verb from dplyr that allows you to create new columns or perform operations or functions on columns.
across is an adverb from dplyr. Let's simplify by saying that it's a function that allows you to perform a function over multiple columns. In this case I want to perform a division by Volum_OD.
~ is a tidyverse way to create anonymous functions. ~ . / Volum_OD is equivalent to function(x) x / Volumn_OD
all_of is necessary because in this specific case I'm providing across with a vector of characters. Without it, it will work anyway, but you will receive a warning because it's ambiguous and it may work incorrectly in same cases.
More info
Check out this book to learn more about data manipulation with tidyverse (which dplyr is part of).
Solution with Base-R
rel.Volume_unmod <- Volume
# rename columns
cols <- c("Volume_OD","Volume_Imp","Volume_OD_1","Volume_WS_1","Volume_OD_2","Volume_WS_2","Volume_OD_3","Volume_WS_3")
colnames(rel.Volume_unmod)[3:10] <- cols
# divide by columns 3
rel.Volume_unmod[3:10] <- lapply(rel.Volume_unmod[3:10], `/`, rel.Volume_unmod[3])
rel.Volume_unmod
Explanation
lapply is a base R function that allows you to apply a function to every item of a list or a "listable" object.
in this case rel.Volume_unmod is a listable object: a dataframe is just a list of vectors with the same length. Therefore, lapply takes one column [= one item] a time and applies a function.
the function is /. You usually see / used like this: A / B, but actually / is a Primitive function. You could write the same thing in this way:
`/`(A, B) # same as A / B
lapply can be provided with additional parameters that are passed directly to the function that is being applied over the list (in this case /). Therefore, we are writing rel.Volume_unmod[3] as additional parameter.
lapply always returns a list. But, since we are assigning the result of lapply to a "fraction of a dataframe", we will just edit the columns of the dataframe and, as a result, we will have a dataframe instead of a list. Let me rephrase in a more technical way. When you are assigning rel.Volume_unmod[3:10] <- lapply(...), you are not simply assigning a list to rel.Volume_unmod[3:10]. You are technically using this assigning function: [<-. This is a function that allows to edit the items in a list/vector/dataframe. Specifically, [<- allows you to assign new items without modifying the attributes of the list/vector/dataframe. As I said before, a dataframe is just a list with specific attributes. Then when you use [<- you modify the columns, but you leave the attributes (the class data.frame in this case) untouched. That's why the magic works.
Whithout a minimal working example it's hard to guess what the Variable Volume actually refers to. Apart from that there seems to be a problem with your for-loop:
for(i in Volume[,3:10]){
Assuming Volume refers to a data.frame or tibble, this causes the actual column-vectors with indices between 3 and 10 to be assigned to i successively. You can verify this by putting print(i) inside the loop. But inside the loop it seems like you actually want to use i as a variable containing just the index of the current column as a number (not the column itself):
rel.Volume[i] = tibble(Volume = Volume[[i]] / Volume[[3]])
Also, two brackets are usually used with lists, not data.frames or tibbles. (You can, however, do so, because data.frames are special cases of lists.)
Last but not least, initialising the variable rel.Volume with NULL will result in an error, when trying to reassign to that variable, since you haven't told R, what rel.Volume should be.
Try this, if you like (thanks #Edo for example data):
set.seed(1)
Volume <- data.frame(ID = sample(letters, 30, TRUE),
GR = sample(LETTERS, 30, TRUE),
Vol1 = rnorm(30),
Vol2 = rnorm(30),
Vol3 = rnorm(30))
rel.Volume <- Volume[1:2] # Assuming you want to keep the IDs.
# Your data.frame will need to have the correct number of rows here already.
for (i in 3:ncol(Volume)){ # ncol gives the total number of columns in data.frame
rel.Volume[i] = Volume[i]/Volume[3]
}
A more R-like approach would be to avoid using a for-loop altogether, since R's strength is implicit vectorization. These expressions will produce the same result without a loop:
# OK, this one messes up variable names...
rel.V.2 <- data.frame(sapply(X = Volume[3:5], FUN = function(x) x/Volume[3]))
rel.V.3 <- data.frame(Map(`/`, Volume[3:5], Volume[3]))
Since you said you were new to R, frankly I would recommend avoiding the Tidyverse-packages while you are still learing the basics. From my experience, in the long run you're better off learning base-R first and adding the "sugar" when you're more familiar with the core language. You can still learn to use Tidyverse-functions later (but then, why would anybody? ;-) ).
I'm importing some raster files from a PostgreSQL connection into R in a loop. I want to assign my newly gained rasters automatically to a variable whose name is derived from the input variable like this: substring(crop, 12)
crop <- "efsa_capri_barley"
ras <- readGDAL(sprintf("PG:dbname='' host='' port='' user='' schema='' table='%s' mode=2", crop))
paste0(substring(crop, 12)) <- raster(ras, 1)
What function do I have to use that R recognizes the result of substring() as a character string and not as the function itself? I was thinking about paste() but it doesn't work.
Probably this question has already been asked but I couldn't find a proper answer.
Based on your description, assign is technically correct, but recommending it is bad advice.
If you are pulling in multiple rasters in a loop, best practice in R is to initialize a list to hold all the resulting rasters and name each list element accordingly. You can do this one at a time:
# n is number of rasters
raster_list <- vector("list",n)
for (i in seq_len(n)){
...
#crop[i] is the ith crop name
raster_list[[substring(crop[i],12)]] <- raster(...)
}
You can also set the names of each element of the list all at once via setNames. But you should try to avoid using assign pretty much at all costs.
If I understand your question correctly, you are looking for something like assign. For example you can try this:
assign(substring(crop, 12), raster(ras, 1))
To understand how assign works, you can check this code:
x <- 2
# x is now 2
var_to_assign <- "x"
assign(var_to_assign, 3)
# x is now set to 3
x
# 3
Does that give you what you want?
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().
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