Creating an indicator variable in R - r

I'm new with R and working with the ALL dataset in R, and trying to create an indicator variable that defines =1 for all T-cell patients and =2 for all B-cell patients. Here is my code:
library(ALL)
data(ALL)
tcell<-c("T","T1","T2","T3","T4")
bcell<-c("B","B1","B2","B3","B4")
ALL.fac <- factor(ALL$BT, levels=1:2, labels=c(tcell,bcell))
ALL.fac
I'm getting an error which states:
Error in factor(ALL$BT, levels = 1:2, labels = c(tcell, bcell)) :
invalid 'labels'; length 10 should be 1 or 2
I'm not sure how to define the variable properly to get all the T-cells to print out as 1 and all B-cells to print as 2.
Suggestions? Thanks!

We can remove the numeric part from the 'BT' using sub and then change the levels
All.fac <- factor(sub('\\d+', '', ALL$BT) ,
levels=c('T', 'B'), labels=c(1,2))

Related

In R how do you factorise and add label values to specific data.table columns, using a second file of meta data?

This is part of a project to switch from SPSS to R. While there are good tools to import SPSS files into R (expss) what this question is part of is attempting to get the benefits of SPSS style labeling when data originates from CSV sources. This is to help bridge the staff training gap between SPSS and R by providing a common format for data.tables irrespective of file format origin.
Whilst CSV does a reasonable job of storing data it is hopeless for providing meaningful data. This inevitably means variable and factor levels and labels have to come from somewhere else. In most short examples of this (e.g. in documentation) it is practical to simply hard code the meta data in. But for larger projects it makes more sense to store this meta data in a second csv file.
Example data file
ID,varone,vartwo,varthree,varfour,varfive,varsix,varseven,vareight,varnine,varten
1,1,34,1,,1,,1,1,4,
2,1,21,0,1,,1,3,14,3,2
3,1,54,1,,,1,3,6,4,4
4,2,32,1,1,1,,3,7,4,
5,3,66,0,,,1,3,9,3,3
6,2,43,1,,1,,1,12,2,1
7,2,26,0,,,1,2,11,1,
8,3,,1,1,,,2,15,1,4
9,1,34,1,,1,,1,12,3,4
10,2,46,0,,,,3,13,2,
11,3,39,1,1,1,,3,7,1,2
12,1,28,0,,,1,1,6,5,1
13,2,64,0,,1,,2,11,,3
14,3,34,1,1,,,3,10,1,1
15,1,52,1,,1,1,1,8,6,
Example metadata file
Rowlabels,ID,varone,vartwo,varthree,varfour,varfive,varsix,varseven,vareight,varnine,varten
varlabel,,Question one,Question two,Question three,Question four,Question five,Question six,Question seven,Question eight,Question nine,Question ten
varrole,Unique,Attitude,Unique,Filter,Filter,Filter,Filter,Attitude,Filter,Attitude,Attitude
Missing,Error,Error,Ignored,Error,Unchecked,Unchecked,Unchecked,Error,Error,Error,Ignored
vallable,,One,,No,Checked,Checked,Checked,x,One,A,Support
vallable,,Two,,Yes,,,,y,Two,B,Neutral
vallable,,Three,,,,,,z,Three,C,Oppose
vallable,,,,,,,,,Four,D,Dont know
vallable,,,,,,,,,Five,E,
vallable,,,,,,,,,Six,F,
vallable,,,,,,,,,Seven,G,
vallable,,,,,,,,,Eight,,
vallable,,,,,,,,,Nine,,
vallable,,,,,,,,,Ten,,
vallable,,,,,,,,,Eleven,,
vallable,,,,,,,,,Twelve,,
vallable,,,,,,,,,Thirteen,,
vallable,,,,,,,,,Fourteen,,
vallable,,,,,,,,,Fifteen,,
SO the common elements are the column names which are the key to both files
The first column of the metadata file describes the role of the row for the data file
so
varlabel provides the variable label for each column
varrole describes the analytic purpose of the variable
missing describes how to treat missing data
varlabel describes the label for a factor level starting at one on up to as many labels as there are.
Right! Here's the code that works:
```#Libraries
library(expss)
library(data.table)
library(magrittr)```
readcsvdata <- function(dfile)
{
# TESTED - Working
print("OK Lets read some comma separated values")
rdata <- fread(file = dfile, sep = "," , quote = "\"" , header = TRUE, stringsAsFactors = FALSE,
na.strings = getOption("datatable.na.strings",""))
return(rdata)
}
rawdatafilename <- "testdata.csv"
rawmetadata <- "metadata.csv"
mdt <- readcsvdata(rawmetadata)
rdt <- readcsvdata(rawdatafilename)
names(rdt)[names(rdt) == "ï..ID"] <- "ID" # correct minor data error
commonnames <- intersect(names(mdt),names(rdt)) # find common variable names so metadata applies
commonnames <- commonnames[-(1)] # remove ID
qlabels <- as.list(mdt[1, commonnames, with = FALSE])
(Here I copy the rdt datatable simply so I can roll back to the original data without re-running the previous read chunks and tidying whenever I make changes that don't work out.
# set var names to columns
for (each_name in commonnames) # loop through commonnames and qlabels
{
expss::var_lab(tdt[[each_name]]) <- qlabels[[each_name]]
}
OK this is where I fall down.
Failure from here
factorcols <- as.vector(commonnames) # create a vector of column names (for later use)
for (col in factorcols)
{
print( is.na(mdt[4, ..col])) # print first row of value labels (as test)
if (is.na(mdt[4, ..col])) factorcols <- factorcols[factorcols != col]
# if not a factor column, remove it from the factorcol list and dont try to factor it
else { # if it is a vector factorise
print(paste("working on",col)) # I have had a lot of problem with unrecognised ..col variables
tlabels <- as.vector(na.omit(mdt[4:18, ..col])) # get list of labels from the data column}
validrange <- seq(1,lengths(tlabels),1) # range of valid values is 1 to the length of labels list
print(as.character(tlabels)) # for testing
print(validrange) # for testing
tdt[[col]] <- factor(tdt[[col]], levels = validrange, ordered = is.ordered(validrange), labels = as.character(tlabels))
# expss::val_lab(tdt[, ..col]) <- tlabels
tlabels = c() # flush loop variable
validrange = c() # flush loop variable
}
}
So the problem is revealed here when we check the data table.
tdt
the labels have been applied as whole vectors to each column entry except where there is only one value in the vector ("checked" for varfour and varfive)
tdt
id (int) 1
varone (fctr) c("One", "Two", "Three") 1 (should be "One" 1)
vartwo (S3: labelled) 34
varthree (fctr) c("No", "Yes") 1 (should be "No" 1)
varfour (fctr) NA
varfive (fctr) Checked
And a mystery
this code works just fine on a single columns when I don't use a for loop variable
# test using column name
tlabels <- c("one","two","three")
validrange <- c(1,2,3)
factor(tdt[,varone], levels = validrange, ordered=is.ordered(validrange), labels = tlabels)
It seems the issue is in the line tlabels <- as.vector(na.omit(mdt[4:18, ..col])). It doesn't make vector as you expect. Contrary to usual data.frame data.table doesn't drop dimensions when you provide single column in the index. And as.vector do nothing with data.frames/data.tables. So tlabels remains data.table. This line need to be rewritten as tlabels <- na.omit(mdt[[col]][4:18]).
Example:
library(data.table)
mdt = as.data.table(mtcars)
col = "am"
tlabels <- as.vector(na.omit(mdt[3:6, ..col])) # ! tlabels is data.table
str(tlabels)
# Classes ‘data.table’ and 'data.frame': 4 obs. of 1 variable:
# $ am: num 1 0 0 0
# - attr(*, ".internal.selfref")=<externalptr>
as.character(tlabels) # character vector of length 1
# [1] "c(1, 0, 0, 0)"
tlabels <- na.omit(mdt[[col]][3:6]) # vector
str(tlabels)
# num [1:4] 1 0 0 0
as.character(tlabels) # character vector of length 4
# [1] "1" "0" "0" "0"

how to interpolate data within groups in R using seqtime?

I am trying to use seqtime (https://github.com/hallucigenia-sparsa/seqtime) to analyze time-serie microbiome data, as follow:
meta = data.table::data.table(day=rep(c(15:27),each=3), condition =c("a","b","c"))
meta<- meta[order(meta$day, meta$condition),]
meta.ts<-as.data.frame(t(meta))
otu=matrix(1:390, ncol = 39)
oturar<-rarefyFilter(otu, min=0)
rarotu<-oturar$rar
time<-meta.ts[1,]
interp.otu<-interpolate(rarotu, time.vector = time,
method = "stineman", groups = meta$condition)
the interpolation returns the following error:
[1] "Processing group a"
[1] "Number of members 13"
intervals
0
12
[1] "Selected interval: 1"
[1] "Length of time series: 13"
[1] "Length of time series after interpolation: 1"
Error in stinepack::stinterp(time.vector, as.numeric(x[i, ]), xout = xout, :
The values of x must strictly increasing
I tried to change method to "hyman", but it returns the error below:
Error in interpolateSub(x = x, time.vector = time.vector, method = method) :
Time points must be provided in chronological order.
I am using R version 3.6.1 and I am a bit new to R.
Please can anyone tell me what I am doing wrong/ how to go around these errors?
Many thanks!
I used quite some time stumbling around trying to figure this out. It all comes down to the data structure of meta and the resulting time variable used as input for the time.vector parameter.
When meta.ts is being converted to a data frame, all strings are automatically converted to factors - this includes day.
To adjust, you can edit your code to the following:
library(seqtime)
meta <- data.table::data.table(day=rep(c(15:27),each=3), condition =c("a","b","c"))
meta <- meta[order(meta$day, meta$condition),]
meta.ts <- as.data.frame(t(meta), stringsAsFactors = FALSE) # Set stringsAsFactors = FALSE
otu <- matrix(1:390, ncol = 39)
oturar <- rarefyFilter(otu, min=0)
rarotu <- oturar$rar
time <- as.integer(meta.ts[1,]) # Now 'day' is character, so convert to integer
interp.otu <- interpolate(rarotu, time.vector = time,
method = "stineman", groups = meta$condition)
As a bonus, read this blogpost for information on the stringsAsFactors parameter. Strings automatically being converted to Factors is a common bewilderment.

Selecting features from a feature set using mRMRe package

I am a new user of R and trying to use mRMRe R package (mRMR is one of the good and well known feature selection approaches) to obtain feature subset from a feature set. Please excuse if my question is simple as I really want to know how I can fix an error. Below is the detail.
Suppose, I have a csv file (gene.csv) having feature set of 6 attributes ([G1.1.1.1], [G1.1.1.2], [G1.1.1.3], [G1.1.1.4], [G1.1.1.5], [G1.1.1.6]) and a target class variable [Output] ('1' indicates positive class and '-1' stands for negative class). Here's a sample gene.csv file:
[G1.1.1.1] [G1.1.1.2] [G1.1.1.3] [G1.1.1.4] [G1.1.1.5] [G1.1.1.6] [Output]
11.688312 0.974026 4.87013 7.142857 3.571429 10.064935 -1
12.538226 1.223242 3.669725 6.116208 3.363914 9.174312 1
10.791367 0.719424 6.115108 6.47482 3.597122 10.791367 -1
13.533835 0.37594 6.766917 7.142857 2.631579 10.902256 1
9.737828 2.247191 5.992509 5.992509 2.996255 8.614232 -1
11.864407 0.564972 7.344633 4.519774 3.389831 7.909605 -1
11.931818 0 7.386364 5.113636 3.409091 6.818182 1
16.666667 0.333333 7.333333 4.333333 2 8.333333 -1
I am trying to get best feature subset of 2 attributes (out of above 6 attributes) and wrote following R code.
library(mRMRe)
file_n<-paste0("E:\\gene", ".csv")
df <- read.csv(file_n, header = TRUE)
f_data <- mRMR.data(data = data.frame(df))
featureData(f_data)
mRMR.ensemble(data = f_data, target_indices = 7,
feature_count = 2, solution_count = 1)
When I run this code, I am getting following error for the statement f_data <- mRMR.data(data = data.frame(df)):
Error in .local(.Object, ...) :
data columns must be either of numeric, ordered factor or Surv type
However, data in each column of the csv file are real number.So, how can I change the R code to fix this problem? Also, I am not sure what should be the value of target_indices in the statement mRMR.ensemble(data = f_data, target_indices = 7,feature_count = 2, solution_count = 1) as my target class variable name is "[Output]" in the gene.csv file.
I will appreciate much if anyone can help me to obtain the best feature subset based on the gene.csv file using mRMRe R package.
I solved the problem by modifying my code as follows.
library(mRMRe)
file_n<-paste0("E:\\gene", ".csv")
df <- read.csv(file_n, header = TRUE)
df[[7]] <- as.numeric(df[[7]])
f_data <- mRMR.data(data = data.frame(df))
results <- mRMR.classic("mRMRe.Filter", data = f_data, target_indices = 7,
feature_count = 2)
solutions(results)
It worked fine. The output of the code gives the indices of the selected 2 features.
I think it has to do with your Output column which is probably of class integer. You can check that using class(df[[7]]).
To convert it to numeric as required by the warning, just type:
df[[7]] <- as.numeric(df[[7]])
That worked for me.
As for the other question, after reading the documentation, setting target_indices = 7 seems the right choice.

How to choose a specific line in a dataset using R

dataset
And I want to pick the row which the Date is 17/12/2006 and 18/12/2006, the type of Date is character, I use the code:
a<-c('17/12/2006','18/12/2006')
NewTable<-WholeTable[which($Date %in% a)]
The error is "Error in which$Date : object of type 'closure' is not subsettable"
Then I try another code:
WholeTable$Date <- as.character(WholeTable$Date)
NewTable<-subset(WholeTable, Date == "17/12/2006"|Date == "18/12/2006")
It can create a new subset but with 0 rows.
Really confused
May be easier if you provide a minimum dataset, if I understand correctly though, this should work:
# In this example date is a factor variable with 4 levels
Wholetable <- data.frame(date = c("16/12/2006", "17/12/2006", "18/12/2006", "19/12/2006"), a = c(1:4))
Newtable <- subset(Wholetable, date == "17/12/2006" | date == "18/12/2006")

replacing a value in column X based on columns Y with R

i've gone through several answers and tried the following but each either yields an error or an un-wanted result:
here's the data:
Network Campaign
Moburst_Chartboost Test Campaign
Moburst_Chartboost Test Campaign
Moburst_Appnext unknown
Moburst_Appnext 1065
i'd like to replace "Test Campaign" with "1055" whenever "Network" == "Moburst_Chartboost". i realize this should be very simple but trying out these:
dataset = read.csv('C:/Users/User/Downloads/example.csv')
for( i in 1:nrow(dataset)){
if(dataset$Network == 'Moburst_Chartboost') dataset$Campaign <- '1055'
}
this yields an error: Warning messages:
1: In if (dataset$Network == "Moburst_Chartboost") dataset$Campaign <- "1055" :
the condition has length > 1 and only the first element will be used
2: In if (dataset$Network == "Moburst_Chartboost") dataset$Campaign <- "1055" :
the condition has length > 1 and only the first element will be used
etc.
then i tried:
within(dataset, {
dataset$Campaign <- ifelse(dataset$Network == 'Moburst_Chartboost', '1055', dataset$Campaign)
})
this turned ALL 4 values in row "Campaign" into "1055" over running what was there even when condition isn't met
also this:
dataset$Campaign[which(dataset$Network == 'Moburst_Chartboost')] <- 1055
yields this error, and replaced the values in the two first rows of "Campaign" with NA:
Warning message:
In `[<-.factor`(`*tmp*`, which(dataset$Network == "Moburst_Chartboost"), :
invalid factor level, NA generated
scratching my head here. new to R but this shouldn't be so hard :(
In your first attempt, you're trying to iterate over all the columns, when you only want to change the 2nd column.
In your second, you're trying to assign the value "1055" to all of the 2nd column.
The way to think about it is as an if else, where if the condition in col 1 is met, col 2 is changed, otherwise it remains the same.
dataset <- data.frame(Network = c("Moburst_Chartboost", "Moburst_Chartboost",
"Moburst_Appnext", "Moburst_Appnext"),
Campaign = c("Test Campaign", "Test Campaign",
"unknown", "1065"))
dataset$Campaign <- ifelse(dataset$Network == "Moburst_Chartboost",
"1055",
dataset$Campaign)
head(dataset)
Network Campaign
1 Moburst_Chartboost 1055
2 Moburst_Chartboost 1055
3 Moburst_Appnext unknown
4 Moburst_Appnext 1065
You may also try dataset$Campaign[dataset$Campaign=="Test Campaign"]<-1055 to avoid the use of loops and ifelse statements.
Where dataset
dataset <- data.frame(Network = c("Moburst_Chartboost", "Moburst_Chartboost",
"Moburst_Appnext", "Moburst_Appnext"),
Campaign = c("Test Campaign", "Test Campaign",
"unknown", 1065))
Try the following
dataset = read.csv('C:/Users/User/Downloads/example.csv', stringsAsFactors = F)
for( i in 1:nrow(dataset)){
if(dataset$Network[i] == 'Moburst_Chartboost') dataset$Campaign[i] <- '1055'
}
It seems your forgot the index variable. Without [i] you work on the whole vector of the data frame, resulting in the error/warning you mentioned.
Note that I added stringsAsFactors = F to the read.csv() function to make sure the strings are indeed interpreted as strings and not factors. Using factors this would result in an error like this
In `[<-.factor`(`*tmp*`, i, value = c(NA, 2L, 3L, 1L)) :
invalid factor level, NA generated
Alternatively you can do the following without using a for loop:
idx <- which(dataset$Network == 'Moburst_Chartboost')
dataset$Campaign[idx] <- '1055'
Here, idx is a vector containing the positions where Network has the value 'Moburst_Chartboost'
thank you for the help! not elegant, but since this lingered with me when going to sleep last night i decided to try to bludgeon this with some ugly code but it worked too - just as a workaround...separated to two data frames, replaced all values and then binded back...
# subsetting only chartboost
chartboost <- subset(dataset, dataset$Network=='Moburst_Chartboost')
# replace all values in Campaign
chartboost$Campaign <-sub("^.*", "1055",chartboost$Campaign)
#subsetting only "not chartboost"
notChartboost <-subset(dataset, dataset$Network!='Moburst_Chartboost')
# binding back to single dataframe
newSet <- rbind(chartboost, notChartboost)
Ugly as a duckling but worked :)

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