Little introduction to the question :
I am developing an ecophysiological model, and I use a reference class list called S that store every object the model need for input/output (e.g. meteo, physiological parameters etc...).
This list contains 5 objects (see example below):
- two dataframes, S$Table_Day (the outputs from the model) and S$Met_c(the meteo in input), which both have variables in columns, and observations (input or output) in row.
- a list of parameters S$Parameters.
- a matrix
- a vector
The model runs many functions with a daily time step. Each day is computed in a for loop that runs from the first day i=1 to the last day i=n. This list is passed to the functions that often take data from S$Met_c and/or S$Parameters in input and compute something that is stored in S$Table_Day, using indexes (the ith day). S is a Reference Class list because they avoid copy on modification, which is very important considering the number of computations.
The question itself :
As the model is very slow, I am trying to decrease computation time by micro-benchmarking different solutions.
Today I found something surprising when comparing two solutions to store my data. Storing data by indexing in one of the preallocated dataframes is longer than storing it into an undeclared vector. After reading this, I thought preallocating memory was always faster, but it seems that R performs more operations while modifying by index (probably comparing the length, type etc...).
My question is : is there a better way to perform such operations ? In other words, is there a way for me to use/store more efficiently the inputs/outputs (in a data.frame, a list of vector or else) to keep track of all computations of each day ? For example would it be better to use many vectors (one for each variable) and regroup them in more complex objects (e.g. list of dataframe) at then end ?
By the way, am I right to use Reference Classes to avoid copy of the big objects in S while passing it to functions and modify it from within them ?
Reproducible example for the comparison:
SimulationClass <- setRefClass("Simulation",
fields = list(Table_Day = "data.frame",
Met_c= "data.frame",
PerCohortFruitDemand_c="matrix",
Parameters= "list",
Zero_then_One="vector"))
S= SimulationClass$new()
# Initializing the table with dummy numbers :
S$Table_Day= data.frame(one= 1:10000, two= rnorm(n = 10000), three= runif(n = 10000),Bud_dd= rep(0,10000))
S$Met_c= data.frame(DegreeDays= rnorm(n=10000, mean = 10, sd = 1))
f1= function(i){
a= cumsum(S$Met_c$DegreeDays[i:(i-1000)])
}
f2= function(i){
S$Table_Day$Bud_dd[(i-1000):i]= cumsum(S$Met_c$DegreeDays[i:(i-1000)])
}
res= microbenchmark(f1(1000),f2(1000),times = 10000)
autoplot(res)
And the result :
Also if someone has any experience in programming such models, I am deeply interested in any advice for model development.
I read more about the question, and I'll just write here for prosperity some of the solutions that were proposed on other posts.
Apparently, reading and writing are both worth to consider when trying to reduce the computation time of assignation to a data.frame by index.
The sources are all found in other discussions:
How to optimize Read and Write to subsections of a matrix in R (possibly using data.table)
Faster i, j matrix cell fill
Time in getting single elements from data.table and data.frame objects
Several solutions appeared relevant :
Use a matrix instead of a data.frame if possible to leverage in place modification (Advanced R).
Use a list instead of a data.frame, because [<-.data.frame is not a primitive function (Advanced R).
Write functions in C++ and use Rcpp (from this source)
Use .subset2 to read instead of [ (third source)
Use data.table as recommanded by #JulienNavarre and #Emmanuel-Lin and the different sources, and use either set for data.frame or := if using a data.table is not a problem.
Use [[ instead of [ when possible (index by one value only). This one is not very effective, and very restrictive, so I removed it from the following comparison.
Here is the analysis of performance using the different solutions :
The code :
# Loading packages :
library(data.table)
library(microbenchmark)
library(ggplot2)
# Creating dummy data :
SimulationClass <- setRefClass("Simulation",
fields = list(Table_Day = "data.frame",
Met_c= "data.frame",
PerCohortFruitDemand_c="matrix",
Parameters= "list",
Zero_then_One="vector"))
S= SimulationClass$new()
S$Table_Day= data.frame(one= 1:10000, two= rnorm(n = 10000), three= runif(n = 10000),Bud_dd= rep(0,10000))
S$Met_c= data.frame(DegreeDays= rnorm(n=10000, mean = 10, sd = 1))
# Transforming data objects into simpler forms :
mat= as.matrix(S$Table_Day)
Slist= as.list(S$Table_Day)
Metlist= as.list(S$Met_c)
MetDT= as.data.table(S$Met_c)
SDT= as.data.table(S$Table_Day)
# Setting up the functions for the tests :
f1= function(i){
S$Table_Day$Bud_dd[i]= cumsum(S$Met_c$DegreeDays[i])
}
f2= function(i){
mat[i,4]= cumsum(S$Met_c$DegreeDays[i])
}
f3= function(i){
mat[i,4]= cumsum(.subset2(S$Met_c, "DegreeDays")[i])
}
f4= function(i){
Slist$Bud_dd[i]= cumsum(.subset2(S$Met_c, "DegreeDays")[i])
}
f5= function(i){
Slist$Bud_dd[i]= cumsum(Metlist$DegreeDays[i])
}
f6= function(i){
set(S$Table_Day, i=as.integer(i), j="Bud_dd", cumsum(S$Met_c$DegreeDays[i]))
}
f7= function(i){
set(S$Table_Day, i=as.integer(i), j="Bud_dd", MetDT[i,cumsum(DegreeDays)])
}
f8= function(i){
SDT[i,Bud_dd := MetDT[i,cumsum(DegreeDays)]]
}
i= 6000:6500
res= microbenchmark(f1(i),f3(i),f4(i),f5(i),f7(i),f8(i), times = 10000)
autoplot(res)
And the resulting autoplot :
With f1 the reference base assignment, f2 using a matrix instead of a data.frame, f3 using the combination of .subset2 and matrix, f4 using a list and .subset2, f5 using two lists (both reading and writing), f6 using data.table::set, f7 using data.table::set and data.table for cumulative sum, and f8using data.table :=.
As we can see the best solution is to use lists for reading and writing. This is pretty surprising to see that data.table is the worst solution. I believe I did something wrong with it, because it is supposed to be the best. If you can improve it, please tell me.
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 have a df, YearHT, 6.5M x 55 columns. There is specific information I want to extract and add but only based on an aggregate values. I am using a for loop to subset the large df, and then performing the computations.
I have heard that for loops should be avoided, and I wonder if there is a way to avoid a for loop that I have used, as when I run this query it takes ~3hrs.
Here is my code:
srt=NULL
for(i in doubletCounts$Var1){
s=subset(YearHT,YearHT$berthlet==i)
e=unlist(c(strsplit(i,'\\|'),median(s$berthtime)))
srt=rbind(srt,e)
}
srt=data.frame(srt)
s2=data.frame(srt$X2,srt$X1,srt$X3)
colnames(s2)=colnames(srt)
s=rbind(srt,s2)
doubletCounts is 700 x 3 df, and each of the values is found within the large df.
I would be glad to hear any ideas to optimize/speed up this process.
Here is a fast solution using data.table , although it is not completely clear from your question what is the output you want to get.
# load library
library(datat.table)
# convert your dataset into data.table
setDT(YearHT)
# subset YearHT keeping values that are present in doubletCounts$Var1
YearHT_df <- YearHT[ berthlet %in% doubletCounts$Var1]
# aggregate values
output <- YearHT_df[ , .( median= median(berthtime)) ]
for loops aren't necessarily something to avoid, but there are certain ways of using for loops that should be avoided. You've committed the classic for loop blunder here.
srt = NULL
for (i in index)
{
[stuff]
srt = rbind(srt, [stuff])
}
is bound to be slower than you would like because each time you hit srt = rbind(...), you're asking R to do all sorts of things to figure out what kind of object srt needs to be and how much memory to allocate to it. When you know what the length of your output needs to be up front, it's better to do
srt <- vector("list", length = doubletCounts$Var1)
for(i in doubletCounts$Var1){
s=subset(YearHT,YearHT$berthlet==i)
srt[[i]] = unlist(c(strsplit(i,'\\|'),median(s$berthtime)))
}
srt=data.frame(srt)
Or the apply alternative of
srt = lapply(doubletCounts$Var1,
function(i)
{
s=subset(YearHT,YearHT$berthlet==i)
unlist(c(strsplit(i,'\\|'),median(s$berthtime)))
}
)
Both of those should run at about the same speed
(Note: both are untested, for lack of data, so they might be a little buggy)
Something else you can try that might have a smaller effect would be dropping the subset call and use indexing. The content of your for loop could be boiled down to
unlist(c(strsplit(i, '\\|'),
median(YearHT[YearHT$berthlet == i, "berthtime"])))
But I'm not sure how much time that would save.
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 benefit from data.table fast grouping to fill a matrix (or do other stuff externally from the data.table).
For example, I have a data.table like this:
DT = data.table(x_id=rep(c(1,2),c(100,100)),x_value = rnorm(200))
setkey(DT,x_id)
(representing two different time-series)
I want to put the same information a matrix of 100 rows and 2 columns.
I tried
A = matrix(NA,100,2)
DT[,{A[,.GRP] = x_value},by=x_id]
But it doesn't work. This raises two questions for me: (I was unable to find help in the doc)
1) Is there a nice way (without loops) to transform the data.table into the matrix.
2) Generally speaking, can we assign value to outside variables in the j environment.
Many thanks for your help.
Try:
DT[,A[,.GRP] <<- x_value,by=x_id]
<<- assigns through to the global environment, which is what you need to do since the data.table expressions are evaluated in a child environment that doesn't contain A.
I would add this is a fairly odd way to use data.table. If you are guaranteed that each group has the same number of rows, then all you need to do is (assuming you have already sorted by x_id:
A <- matrix(DT[, x_value], 100)
Which takes advantage of the underlying vector-like nature of matrices.
[Update 1: As Matthew Dowle noted, I'm using data.table version 1.6.7 on R-Forge, not CRAN. You won't see the same behavior with an earlier version of data.table.]
As background: I am porting some little utility functions to do set operations on rows of a data frame or pairs of data frames (i.e. each row is an element in a set), e.g. unique - to create a set from a list, union, intersection, set difference, etc. These mimic Matlab's intersect(...,'rows'), setdiff(...,'rows'), etc., which don't appear to have counterparts in R (R's set operations are limited to vectors and lists, but not rows of matrices or data frames). Examples of these little functions are below. If this functionality for data frames already exists in some package or base R, I'm open to suggestions.
I have been migrating these to data tables and one necessary step in the current approach is to find duplicated rows. When duplicated() is executed an error is returned stating that data tables must have keys. This is an unfortunate roadblock - other than setting keys, which isn't a universal solution and adds to computational costs, is there some other way to find duplicated objects?
Here is a reproducible example:
library(data.table)
set.seed(0)
x <- as.data.table(matrix(sample(2, 100, replace = TRUE), ncol = 4))
y <- as.data.table(matrix(sample(2, 100, replace = TRUE), ncol = 4))
res3 <- dt_intersect(x,y)
Yielding this error message:
Error in duplicated.data.table(z_rbind) : data table must have keys
The code works as-is for data frames, though I've named each function with the pattern dt_operation.
Is there some way to get around this issue? Setting keys only works for integers, which is a constraint I can't assume for the input data. So, perhaps I'm missing a clever way to use data tables?
Example set operation functions, where the elements of the sets are rows of data:
dt_unique <- function(x){
return(unique(x))
}
dt_union <- function(x,y){
z_rbind <- rbind(x,y)
z_unique <- dt_unique(z_rbind)
return(z_unique)
}
dt_intersect <- function(x,y){
zx <- dt_unique(x)
zy <- dt_unique(y)
z_rbind <- rbind(zy,zx)
ixDupe <- which(duplicated(z_rbind))
z <- z_rbind[ixDupe,]
return(z)
}
dt_setdiff <- function(x,y){
zx <- dt_unique(x)
zy <- dt_unique(y)
z_rbind <- rbind(zy,zx)
ixRangeX <- (nrow(zy) + 1):nrow(z_rbind)
ixNotDupe <- which(!duplicated(z_rbind))
ixDiff <- intersect(ixNotDupe, ixRangeX)
diffX <- z_rbind[ixDiff,]
return(diffX)
}
Note 1: One intended use for these helper functions is to find rows where key values in x are not among the key values in y. This way, I can find where NAs may appear when calculating x[y] or y[x]. Although this usage allows for setting of keys for the z_rbind object, I'd prefer not to constrain myself to just this use case.
Note 2: For related posts, here is a post on running unique on data frames, with excellent results for running it with the updated data.table package.
And this is an earlier post on running unique on data tables.
duplicated.data.table needs the same fix unique.data.table got [EDIT: Now done in v1.7.2]. Please raise another bug report: bug.report(package="data.table"). For the benefit of others watching, you're already using v1.6.7 from R-Forge, not 1.6.6 on CRAN.
But, on Note 1, there's a 'not join' idiom :
x[-x[y,which=TRUE]]
See also FR#1384 (New 'not' and 'whichna' arguments?) to make that easier for users, and that links to the keys that don't match thread which goes into more detail.
Update. Now in v1.8.3, not-join has been implemented.
DT[-DT["a",which=TRUE,nomatch=0],...] # old idiom
DT[!"a",...] # same result, now preferred.