How to refer to arbitrarily names columns in R using data.table - r

I am trying to do some regression matching using the following code:
library(data.table)
colCount = 3
Test = data.table(ID=c(1,2,3),R1=c(1,1,1),R2=c(1,2,3),R3=c(4,2,1))
Compare = data.table(ID=c(1,2,3),R1=c(2,1,2),R2=c(6,2,3),R3=c(1,1,4))
# Example, in real run, I know colCount will give the arbitrary number of data columns in Test and Compare; I will not know their names
for (i in 1:nrow(Compare)){
Test[,paste0('Factor_',i):=sum(Compare[i,2:(1+colCount)]*.SD)/sum(.SD*.SD),.SDcols=2:(1+colCount),by=1:nrow(Test)]
Test[,paste0('Error_',i) :=sum((Compare[i,2:(1+colCount)]-.SD*paste0('Factor_',i))^2),.SDcols=2:(1+colCount),by=1:nrow(Test)]
}
The paste on the second to last line does not work as I was hoping; is there a better way to refer to columns with generated names?
Also more generally, is there a smarter way to do this? The sum-by-row method I'm doing seems way too complicated, but I couldn't get it working using matrix math or lapply instead

Related

R - Assign the mean of a column sub-sector to each row of that sub-sector

I am trying to create a column which has the mean of a variable according to subsectors of my data set. In this case, the mean is the crime rate of each state calculated from county observations, and then assigning this number to each county relative to the state they are located in. Here is the function wrote.
Create the new column
Data.Final$state_mean <- 0
Then calculate and assign the mean.
for (j in range[1:3136])
{
state <- Data.Final[j, "state"]
Data.Final[j, "state_mean"] <- mean(Data.Final$violent_crime_2009-2014,
which(Data.Final[, "state"] == state))
}
Here is the following error
Error in range[1:3137] : object of type 'builtin' is not subsettable
Very much appreciated if you could, take a few minutes to help a beginner out.
You've got a few problems:
range[1:3136] isn't valid syntax. range(1:3136) is valid syntax, but the range() function just returns the minimum and maximum. You don't need anything more than 1:3136, just use
for (j in 1:3136) instead.
Because of the dash, violent_crime_2009-2014 isn't a standard column name. You'll need to use it in backticks, Data.Final$\violent_crime_2009-2014`` or in quotes with [: Data.Final[["violent_crime_2009-2014"]] or Data.Final[, "violent_crime_2009-2014"]
Also, your code is very inefficient - you re-calculate the mean on every single time. Try having a look at the
Mean by Group R-FAQ. There are many faster and easier methods to get grouped means.
Without using extra packages, you could do
Data.Final$state_mean = ave(x = Data.Final[["violent_crime_2009-2014"]],
Data.Final$state,
FUN = mean)
For friendlier syntax and greater efficiency, the data.table and dplyr packages are popular. You can see examples using them at the link above.
Here is one of many ways this can be done (I'm sure someone will post a tidyverse answer soon if not before I manage to post):
# Data for my example:
data(InsectSprays)
# Note I have a response column and a column I could subset on
str(InsectSprays)
# Take the averages with the by var:
mn <- with(InsectSprays,aggregate(x=list(mean=count),by=list(spray=spray),FUN=mean))
# Map the means back to your data using the by var as the key to map on:
InsectSprays <- merge(InsectSprays,mn,by="spray",all=TRUE)
Since you mentioned you're a beginner, I'll just mention that whenever you can, avoid looping in R. Vectorize your operations when you can. The nice thing about using aggregate, and merge, is that you don't have to worry about errors in your mapping because you get an index shift while looping and something weird happens.
Cheers!

How to append a random or arbitrary column to data frame [R]

Hear me out. Consider an arbitrary case where the new column's elements do not require any information from other columns (which I frustrates base $ and mutate assignment), and not every element in the new column is the same. Here is what I've tried:
df$rand<-rep(sample(1:100,1),nrow(df))
unique(df$rand)
[1] 58
and rest assured, nrow(df)>1. I think the correct solution might have to do with an apply function?
Your code repeats one single random number nrow(df) times. Try instead:
df$rand<-sample(1:100, nrow(df))
This samples without replacement from 1:100 nrow(df) times. Now this would give you an error if nrow(df)>100 because you would run out of numbers from 1:100 to sample. To make sure you don't get this error, you can instead sample with replacement:
df$rand<-sample(1:100, nrow(df), replace = TRUE)
If, however, you don't want any random numbers to repeat but would also like to prevent the error, you can do something like this:
df$rand<-sample(1:nrow(df), nrow(df))
if I understand this correctly ,I think this is pretty easily doable in dplyr or data.table .
for e.g dplyr soln on iris
iris%>%mutate(sample(n()))

How to quickly split values in column to create a table for plotting in R

I was wondering if anyone could offer any advice on speeding
the following up in R.
I’ve got a table in a format like this
chr1, A, G, v1,v2,v3;w1w2w3, ...
...
The header is
chr, ref, alt, sample1, sample2 ...(many samples)
In each row for each sample I’ve got 3 values for v and 3 values for w,
separated by “;"
I want to extract v1 and w1 for each sample make a table
that can be plotted using ggplot, it would look like this
chr, ref, alt, sam, v1, w1
I am doing this by strsplit and rbind one by one like the
following
varsam <- c()
for(i in 1:n.var){
chrm <- variants[i,1]
ref <- as.character(variants[i,3])
alt <- as.character(variants[i,4])
amp <- as.character(variants[i,5])
for(j in 1:n.sam){
vs <- strsplit(as.character(vcftable[i,j+6]), split=":")[[1]
vsc <- strsplit(vs[1], split=",")[[1]]
vsp <- strsplit(vs[2], split=",")[[1]]
varsam <- rbind(varsam, c(chrm, pos, ref, j, vsc[1], vsp[1]))
}
This is very slow as you would expect. Any idea how to speed this up?
As noted by others, the first thing you need is some timings, so that you can compare performance if you intend to optimize performance. This would be my first step:
Create some timings
Play around with different aspects of your code to see where the main time is being used.
Basic timing analysis can be done with system.time() method to help with performance analysis
Beyond that, there are some candidates you might like to consider to improve performance - but importantly, it is important to get the timings first so that you have something to compare against.
the dplyr library contains a mutate function which can be used to create new columns, e.g. mynewtablewithextracolumn <- mutate(table, v1 = whatever you want it to be). In the previous statement, simply insert how to calculate each column value where v1 is a new column. There are lots of examples on the internet.
In order to use dplyr, you would need to perform a call to library(dplyr) in your code.
You may need to install.packages("dplyr") if not already installed.
In order to use dplyr, you might be best converting your table into the appropriate type of table for dplyr, e.g. if your current table is data frame, then use table = tbl_df(df) to create a table
As noted, these are just some possible areas. The important thing is to get timings and explore the performance to try to get a handle on where the best place to focus is and to make sure you can measure the performance improvement.
Thanks for the comments. I think I've found way to improve this.
I used melt in "reshape" to firstly convert my input table to
chr, ref, alt, variable
I can then use apply to modify "variable", each row for which contains a concatenated string. This achieves good speed.

Loop and clear the basic function in R

I've got this dataset
install.packages("combinat")
install.packages("quantmod")
library(quantmod)
library(combinat)
library(utils)
getSymbols("AAPL",from="2012-01-01")
data<-AAPL
p1<-4
dO<-data[,1]
dC<-data[,4]
emaO<-EMA(dO,n=p1)
emaC<-EMA(dC,n=p1)
Pos_emaO_dO_UP<-emaO>dO
Pos_emaO_dO_D<-emaO<dO
Pos_emaC_dC_UP<-emaC>dC
Pos_emaC_dC_D<-emaC<dC
Pos_emaC_dO_D<-emaC<dO
Pos_emaC_dO_UP<-emaC>dO
Pos_emaO_dC_UP<-emaO>dC
Pos_emaO_dC_D<-emaO<dC
Profit_L_1<-((lag(dC,-1)-lag(dO,-1))/(lag(dO,-1)))*100
Profit_L_2<-(((lag(dC,-2)-lag(dO,-1))/(lag(dO,-1)))*100)/2
Profit_L_3<-(((lag(dC,-3)-lag(dO,-1))/(lag(dO,-1)))*100)/3
Profit_L_4<-(((lag(dC,-4)-lag(dO,-1))/(lag(dO,-1)))*100)/4
Profit_L_5<-(((lag(dC,-5)-lag(dO,-1))/(lag(dO,-1)))*100)/5
Profit_L_6<-(((lag(dC,-6)-lag(dO,-1))/(lag(dO,-1)))*100)/6
Profit_L_7<-(((lag(dC,-7)-lag(dO,-1))/(lag(dO,-1)))*100)/7
Profit_L_8<-(((lag(dC,-8)-lag(dO,-1))/(lag(dO,-1)))*100)/8
Profit_L_9<-(((lag(dC,-9)-lag(dO,-1))/(lag(dO,-1)))*100)/9
Profit_L_10<-(((lag(dC,-10)-lag(dO,-1))/(lag(dO,-1)))*100)/10
which are given to this frame
frame<-data.frame(Pos_emaO_dO_UP,Pos_emaO_dO_D,Pos_emaC_dC_UP,Pos_emaC_dC_D,Pos_emaC_dO_D,Pos_emaC_dO_UP,Pos_emaO_dC_UP,Pos_emaO_dC_D,Profit_L_1,Profit_L_2,Profit_L_3,Profit_L_4,Profit_L_5,Profit_L_6,Profit_L_7,Profit_L_8,Profit_L_9,Profit_L_10)
colnames(frame)<-c("Pos_emaO_dO_UP","Pos_emaO_dO_D","Pos_emaC_dC_UP","Pos_emaC_dC_D","Pos_emaC_dO_D","Pos_emaC_dO_UP","Pos_emaO_dC_UP","Pos_emaO_dC_D","Profit_L_1","Profit_L_2","Profit_L_3","Profit_L_4","Profit_L_5","Profit_L_6","Profit_L_7","Profit_L_8","Profit_L_9","Profit_L_10")
There is vector with variables for later usage
vector<-c("Pos_emaO_dO_UP","Pos_emaO_dO_D","Pos_emaC_dC_UP","Pos_emaC_dC_D","Pos_emaC_dO_D","Pos_emaC_dO_UP","Pos_emaO_dC_UP","Pos_emaO_dC_D")
I made all possible combination with 4 variables of the vector (there are no depended variables)
comb<-as.data.frame(combn(vector,4))
comb
and get out the ,,nonsense" combination (where are both possible values of variable)
rc<-comb[!sapply(comb, function(x) any(duplicated(sub('_D|_UP', '', x))))]
rc
Then I prepare the first combination to later subseting
var<-paste(rc[,1],collapse=" & ")
var
and subset the frame (with all DVs)
kr<-eval(parse(text=paste0('subset(frame,' , var,')' )))
kr
Now I have the subseted df by the first combination of 4 variables.
Then I used the evaluation function on it
evaluation<-function(x){
s_1<-nrow(x[x$Profit_L_1>0,])/nrow(x)
s_2<-nrow(x[x$Profit_L_2>0,])/nrow(x)
s_3<-nrow(x[x$Profit_L_3>0,])/nrow(x)
s_4<-nrow(x[x$Profit_L_4>0,])/nrow(x)
s_5<-nrow(x[x$Profit_L_5>0,])/nrow(x)
s_6<-nrow(x[x$Profit_L_6>0,])/nrow(x)
s_7<-nrow(x[x$Profit_L_7>0,])/nrow(x)
s_8<-nrow(x[x$Profit_L_8>0,])/nrow(x)
s_9<-nrow(x[x$Profit_L_9>0,])/nrow(x)
s_10<-nrow(x[x$Profit_L_10>0,])/nrow(x)
n_1<-nrow(x[x$Profit_L_1>0,])/nrow(frame)
n_2<-nrow(x[x$Profit_L_2>0,])/nrow(frame)
n_3<-nrow(x[x$Profit_L_3>0,])/nrow(frame)
n_4<-nrow(x[x$Profit_L_4>0,])/nrow(frame)
n_5<-nrow(x[x$Profit_L_5>0,])/nrow(frame)
n_6<-nrow(x[x$Profit_L_6>0,])/nrow(frame)
n_7<-nrow(x[x$Profit_L_7>0,])/nrow(frame)
n_8<-nrow(x[x$Profit_L_8>0,])/nrow(frame)
n_9<-nrow(x[x$Profit_L_9>0,])/nrow(frame)
n_10<-nrow(x[x$Profit_L_10>0,])/nrow(frame)
pr_1<-sum(kr[,"Profit_L_1"])/nrow(kr[,kr=="Profit_L_1"])
pr_2<-sum(kr[,"Profit_L_2"])/nrow(kr[,kr=="Profit_L_2"])
pr_3<-sum(kr[,"Profit_L_3"])/nrow(kr[,kr=="Profit_L_3"])
pr_4<-sum(kr[,"Profit_L_4"])/nrow(kr[,kr=="Profit_L_4"])
pr_5<-sum(kr[,"Profit_L_5"])/nrow(kr[,kr=="Profit_L_5"])
pr_6<-sum(kr[,"Profit_L_6"])/nrow(kr[,kr=="Profit_L_6"])
pr_7<-sum(kr[,"Profit_L_7"])/nrow(kr[,kr=="Profit_L_7"])
pr_8<-sum(kr[,"Profit_L_8"])/nrow(kr[,kr=="Profit_L_8"])
pr_9<-sum(kr[,"Profit_L_9"])/nrow(kr[,kr=="Profit_L_9"])
pr_10<-sum(kr[,"Profit_L_10"])/nrow(kr[,kr=="Profit_L_10"])
mat<-matrix(c(s_1,n_1,pr_1,s_2,n_2,pr_2,s_3,n_3,pr_3,s_4,n_4,pr_4,s_5,n_5,pr_5,s_6,n_6,pr_6,s_7,n_7,pr_7,s_8,n_8,pr_8,s_9,n_9,pr_9,s_10,n_10,pr_10),ncol=3,nrow=10,dimnames=list(c(1:10),c("s","n","pr")))
df<-as.data.frame(mat)
return(df)
}
result<-evaluation(kr)
result
And I need to help in several cases.
1, in evaluation function the way the matrix is made is wrong (s_1,n_1,pr_1 are starting in first column but I need to start the order by rows)
2, I need to use some loop/lapply function to go trough all possible combinations (not only the first one like in this case (var<-paste(rc[,1],collapse=" & ")) and have the understandable output where is evaluation function used on every combination and I will be able to see for which combination of variables is the evaluation done (understand I need to recognize for what is this evaluation made) and compare evaluation results for each combination.
3, This is not main point, BUT I generally want to evaluate all possible combinations (it means for 2:n number of variables and also all combinations in each of them) and then get the best possible combination according to specific DV (Profit_L_1 or Profit_L_2 and so on). And I am so weak in looping now, so, if it this possible, keep in mind what am I going to do with it later.
Thanks, feel free to update, repair or improve the question (if there is something which could be done way more easily, effectively - do it - I am open for every senseful advice.

Bandwidth selection using NP package

New to R and having problem with a very simple task! I have read a few columns of .csv data into R, the contents of which contains of variables that are in the natural numbers plus zero, and have missing values. After trying to use the non-parametric package, I have two problems: first, if I use the simple command bw=npregbw(ydat=y, xdat=x, na.omit), where x and y are column vectors, I get the error that "number of regression data and response data do not match". Why do I get this, as I have the same number of elements in each vector?
Second, I would like to call the data ordered and tell npregbw this, using the command bw=npregbw(ydat=y, xdat=ordered(x)). When I do that, I get the error that x must be atomic for sort.list. But how is x not atomic, it is just a vector with natural numbers and NA's?
Any clarifications would be greatly appreciated!
1) You probably have a different number of NA's in y and x.
2) Can't be sure about this, since there is no example. If it is of following type:
x <- c(3,4,NA,2)
Then ordered(x) should work fine. Please provide an example of your case.
EDIT: You of course tried bw=npregbw(ydat=y, xdat=x)? ordered() makes your vector an ordered factor (see ?ordered), which is not an atomic vector (see 2.1.1 link and ?factor)
EDIT2: So the problem was the way of subsetting data. Note the difference in various ways of subsetting. data$x and data[,i] (where i = column number of column x) give you vectors, while data[c("x")] and data[i] give a data frame. Functions expect vectors, unless they call for data = (your data). In that case they work with column names

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