data.frame filtering function too slow - r

I am trying to filter a data frame of mine, with about 200 thousand rows, using R.
The dataframe is structured as follow:
testdf<- data.frame("CHROM"='CHR8', "POS"=c(500,510), "ID"='Some_value',
"REF"=c('A','C'), "ALT"=c('C','T,G'), "Some_more_stuff"='More_info')
I am trying to filter the rows based on how many letters are in the 'ALT' column, being equal to or lesser than a custom threshold. In the example above, if my threshold is 1, only the first row would be retained (the second row - ALT column- has 2 letters > 1).
I have written a couple of functions, which do the job. The only problem is that they take several seconds on a test dataframe with just 14 rows. On the real dataframe (200,000 rows) it takes forever. I am looking for advice on how writing better syntax and get faster results.
Here are my functions:
# Function no. 1:
allele_number_filtering<- function (snp_table, max_alleles=1, ALT_column=5) {
#here I calculate how many letters are in the ALT column
alt_allele_list_length <- function(ALT_field) {
alt_length<- length(strsplit(as.character
(ALT_field), split = ',')[[1]])
return(alt_length)}
# Create an empty dataframe with same columns as the input df
final_table<- snp_table[0,]
# Now only retain the rows that are <= max_alleles
for (i in 1:nrow(snp_table)) {
if (alt_allele_list_length(snp_table[i, ALT_column]) <= max_alleles) {
final_table<- rbind(final_table, snp_table[i,])}}
return(final_table)}
#Function no. 2:
allele_number_filtering<- function (snp_table, max_alleles=1, ALT_column=5) {
final_table<- snp_table[0,]
for (i in 1: nrow(snp_table)) {
if (length(strsplit(as.character(snp_table[i,ALT_column]),
split = ',')[[1]])<=max_alleles) {
final_table<- rbind(final_table, snp_table[i,])
}}
return(final_table)}
I would be thankful for any advice :)
Max
EDIT: I realized I also had values such as 'ALT' = 'at' (still to be counted as 1) or 'ALT' = 'aa,at' (to be counted as 2 ).

you can use lengths() for this:
testdf[lengths(strsplit(as.character(testdf$ALT), ',',fixed = TRUE))<=1,]
Thanks #docendodiscimus for strsplit( fixed=TRUE) option for speed up and to #joran for his perspicacity

I would use nchar for this (before I would remove the , via gsub):
nchar(gsub(",", "", as.character(testdf$ALT)))
# [1] 1 2
threshold <- 1
testdf[nchar(gsub(",", "", as.character(testdf$ALT))) > threshold, ]
# CHROM POS ID REF ALT Some_more_stuff
# 2 CHR8 510 Some_value C T,G More_info

Related

How to exclude a range of data points by index from a dataframe in R [duplicate]

I have a data frame named "mydata" that looks like this this:
A B C D
1. 5 4 4 4
2. 5 4 4 4
3. 5 4 4 4
4. 5 4 4 4
5. 5 4 4 4
6. 5 4 4 4
7. 5 4 4 4
I'd like to delete row 2,4,6. For example, like this:
A B C D
1. 5 4 4 4
3. 5 4 4 4
5. 5 4 4 4
7. 5 4 4 4
The key idea is you form a set of the rows you want to remove, and keep the complement of that set.
In R, the complement of a set is given by the '-' operator.
So, assuming the data.frame is called myData:
myData[-c(2, 4, 6), ] # notice the -
Of course, don't forget to "reassign" myData if you wanted to drop those rows entirely---otherwise, R just prints the results.
myData <- myData[-c(2, 4, 6), ]
You can also work with a so called boolean vector, aka logical:
row_to_keep = c(TRUE, FALSE, TRUE, FALSE, TRUE, FALSE, TRUE)
myData = myData[row_to_keep,]
Note that the ! operator acts as a NOT, i.e. !TRUE == FALSE:
myData = myData[!row_to_keep,]
This seems a bit cumbersome in comparison to #mrwab's answer (+1 btw :)), but a logical vector can be generated on the fly, e.g. where a column value exceeds a certain value:
myData = myData[myData$A > 4,]
myData = myData[!myData$A > 4,] # equal to myData[myData$A <= 4,]
You can transform a boolean vector to a vector of indices:
row_to_keep = which(myData$A > 4)
Finally, a very neat trick is that you can use this kind of subsetting not only for extraction, but also for assignment:
myData$A[myData$A > 4,] <- NA
where column A is assigned NA (not a number) where A exceeds 4.
Problems with deleting by row number
For quick and dirty analyses, you can delete rows of a data.frame by number as per the top answer. I.e.,
newdata <- myData[-c(2, 4, 6), ]
However, if you are trying to write a robust data analysis script, you should generally avoid deleting rows by numeric position. This is because the order of the rows in your data may change in the future. A general principle of a data.frame or database tables is that the order of the rows should not matter. If the order does matter, this should be encoded in an actual variable in the data.frame.
For example, imagine you imported a dataset and deleted rows by numeric position after inspecting the data and identifying the row numbers of the rows that you wanted to delete. However, at some later point, you go into the raw data and have a look around and reorder the data. Your row deletion code will now delete the wrong rows, and worse, you are unlikely to get any errors warning you that this has occurred.
Better strategy
A better strategy is to delete rows based on substantive and stable properties of the row. For example, if you had an id column variable that uniquely identifies each case, you could use that.
newdata <- myData[ !(myData$id %in% c(2,4,6)), ]
Other times, you will have a formal exclusion criteria that could be specified, and you could use one of the many subsetting tools in R to exclude cases based on that rule.
Create id column in your data frame or use any column name to identify the row. Using index is not fair to delete.
Use subset function to create new frame.
updated_myData <- subset(myData, id!= 6)
print (updated_myData)
updated_myData <- subset(myData, id %in% c(1, 3, 5, 7))
print (updated_myData)
By simplified sequence :
mydata[-(1:3 * 2), ]
By sequence :
mydata[seq(1, nrow(mydata), by = 2) , ]
By negative sequence :
mydata[-seq(2, nrow(mydata), by = 2) , ]
Or if you want to subset by selecting odd numbers:
mydata[which(1:nrow(mydata) %% 2 == 1) , ]
Or if you want to subset by selecting odd numbers, version 2:
mydata[which(1:nrow(mydata) %% 2 != 0) , ]
Or if you want to subset by filtering even numbers out:
mydata[!which(1:nrow(mydata) %% 2 == 0) , ]
Or if you want to subset by filtering even numbers out, version 2:
mydata[!which(1:nrow(mydata) %% 2 != 1) , ]
For completeness, I'll add that this can be done with dplyr as well using slice. The advantage of using this is that it can be part of a piped workflow.
df <- df %>%
.
.
slice(-c(2, 4, 6)) %>%
.
.
Of course, you can also use it without pipes.
df <- slice(df, -c(2, 4, 6))
The "not vector" format, -c(2, 4, 6) means to get everything that is not at rows 2, 4 and 6. For an example using a range, let's say you wanted to remove the first 5 rows, you could do slice(df, 6:n()). For more examples, see the docs.
Delete Dan from employee.data - No need to manage a new data.frame.
employee.data <- subset(employee.data, name!="Dan")
Here's a quick and dirty function to remove a row by index.
removeRowByIndex <- function(x, row_index) {
nr <- nrow(x)
if (nr < row_index) {
print('row_index exceeds number of rows')
} else if (row_index == 1)
{
return(x[2:nr, ])
} else if (row_index == nr) {
return(x[1:(nr - 1), ])
} else {
return (x[c(1:(row_index - 1), (row_index + 1):nr), ])
}
}
It's main flaw is it the row_index argument doesn't follow the R pattern of being a vector of values. There may be other problems as I only spent a couple of minutes writing and testing it, and have only started using R in the last few weeks. Any comments and improvements on this would be very welcome!
To identify by a name:
Call out the unique ID and identify the location in your data frame (DF).
Mark to delete. If the unique ID applies to multiple rows, all these rows will be removed.
Code:
Rows<-which(grepl("unique ID", DF$Column))
DF2<-DF[-c(Rows),]
DF2
Another approach when working with Unique IDs is to subset data:
*This came from an actual report where I wanted to remove the chemical standard
Chem.Report<-subset(Chem.Report, Chem_ID!="Standard")
Chem_ID is the column name.
The ! is important for excluding

function to subtract each column from one specific column in r

I want to subtract each column from a column called df$Means in r. I want to do this as a function but Im not sure how to iterate through each of the columns- each iteration relies on one column being subtracted from df$Means and then there is a load of downstream code that uses the output. I have simplified the code for here as this is the bit that's giving me trouble. So far I have:
CopyNumberLoop <- function (i) {df$ZScore <- (df[3:5]-df$Means)/(df$sd)
}
apply(df[3:50], 2, CopyNumberLoop)
but Im not sure how to make sure that the operation is done on one column at a time. I don't think df[3:5] is correct?
I have been asked to produce a reproducible example so all the code I want is here:
df1 <- read.delim(file.choose(),header=TRUE)
#Take the control samples and average each row for three columns excluding the first two columns- add the per row means to the data frame
df$Means <- rowMeans(df[,30:32])
RowVar <- function(x) {rowSums((x - rowMeans(x))^2)/(dim(x)[2] - 1)}
df$sd=sqrt(RowVar(df[,c(30:32)]))
#Get a Z score by dividing the test sample count at each locus by the average for the control samples and divide everything by the st dev for controls at each locus.
{
df$ZScore <- (df[,35]-df$Means)/(df$sd)
######################################### QUARTILE FILTER ###########################################################
alpha=1.5
numberofControls = 3
UL = median(df$ZScore, na.rm = TRUE) + alpha*IQR(df$ZScore, na.rm = TRUE)
LL = median(df$ZScore, na.rm = TRUE) - alpha*IQR(df$ZScore, na.rm = TRUE)
#Copy the Z score if the score is > or < a certain number, i.e. LL or UL.
Zoutliers <- which(df$ZScore > UL | df$ZScore < LL)
df$Zoutliers <- ifelse(df$ZScore > UL |df$ZScore <LL ,1,-1)
tempout = ifelse(df$ZScore[Zoutliers] > UL,1,-1)
######################################### Three neighbour Isolation filter ##############################################################################
finalSeb=c()
for(i in 2:(length(Zoutliers)-1)){
j=Zoutliers[i]
if(sum(ifelse((j-1) == Zoutliers,1,0)) > 0 & tempout[i] == tempout[i-1] & sum(ifelse((j+1) == Zoutliers,1,0)) > 0 & tempout[i] == tempout[i+1]){
finalSeb = c(finalSeb,i)
}
}
finalset_row_number = Zoutliers[finalSeb]
#View(finalset_row_number)
p_seq = rep(0,nrow(df))
for(i in 1:length(finalset_row_number)){
p_seq[(finalset_row_number[i]-1):(finalset_row_number[i]+1)] = median(df$ZScore[(finalset_row_number[i]-1):(finalset_row_number[i]+1)])
}
nrow(as.data.frame(finalset_row_number))
}
For each column between 3 and 50 I'd like to generate a nrow(as.data.frame(finalset_row_number)) and keep it in another dataframe. Admittedly my code is a mess because I dont know how to create the function that will allow me to apply this to each column
Your code isn’t using the parameter i at all. In fact, i is the current column, so that’s what you should use:
result = apply(df[, 3 : 50], 2, function (col) col - df$Means)
Or you can subtract the means directly:
result = df[, 3 : 50] - df$Means
This will return a new matrix consisting of the columns 3–50 from df, subtracting df$Means from each in turn. Or, if you want to calculate Z scores as your code seems to do:
result = (df[, 3 : 50] - df$Means) / df$sd
It appeared that you wanted the Z-scores assigned back into the original dataframe as named columns. If you want to loop over columns, it would be just as economical to use lapply or sapply. The receiving function will accept each column in turn and match it to the first parameter. Any other arguments offered after the receiving function will get matched by name or position to any other symbol/names in the parameter list. You do not do any assignment to 'df' inside the function:
CopyNumberLoop <- function (col) { col-df$Means/(df$sd)
}
df[, paste0('ZScore' , 3:50)] <- # assignment done outside the loop
lapply(df[3:50], CopyNumberLoop) # result is a list
# but the `[.data.frame<-` method will accept a list.
Usign apply coerces to a matrix which may have undesirable effects in the column is not numeric (say factor or date-time). It's better to get into he habit of using lapply when working on ranges of columns in dataframes.
If you want to assign the result of this operation to a new dataframe, then the lapply(.) result would need to be wrapped in as.data.frame and then column names could be assigned. Same effort would need to be done to a result from apply(.).

Deleting many, specific rows in R [duplicate]

I have a data frame named "mydata" that looks like this this:
A B C D
1. 5 4 4 4
2. 5 4 4 4
3. 5 4 4 4
4. 5 4 4 4
5. 5 4 4 4
6. 5 4 4 4
7. 5 4 4 4
I'd like to delete row 2,4,6. For example, like this:
A B C D
1. 5 4 4 4
3. 5 4 4 4
5. 5 4 4 4
7. 5 4 4 4
The key idea is you form a set of the rows you want to remove, and keep the complement of that set.
In R, the complement of a set is given by the '-' operator.
So, assuming the data.frame is called myData:
myData[-c(2, 4, 6), ] # notice the -
Of course, don't forget to "reassign" myData if you wanted to drop those rows entirely---otherwise, R just prints the results.
myData <- myData[-c(2, 4, 6), ]
You can also work with a so called boolean vector, aka logical:
row_to_keep = c(TRUE, FALSE, TRUE, FALSE, TRUE, FALSE, TRUE)
myData = myData[row_to_keep,]
Note that the ! operator acts as a NOT, i.e. !TRUE == FALSE:
myData = myData[!row_to_keep,]
This seems a bit cumbersome in comparison to #mrwab's answer (+1 btw :)), but a logical vector can be generated on the fly, e.g. where a column value exceeds a certain value:
myData = myData[myData$A > 4,]
myData = myData[!myData$A > 4,] # equal to myData[myData$A <= 4,]
You can transform a boolean vector to a vector of indices:
row_to_keep = which(myData$A > 4)
Finally, a very neat trick is that you can use this kind of subsetting not only for extraction, but also for assignment:
myData$A[myData$A > 4,] <- NA
where column A is assigned NA (not a number) where A exceeds 4.
Problems with deleting by row number
For quick and dirty analyses, you can delete rows of a data.frame by number as per the top answer. I.e.,
newdata <- myData[-c(2, 4, 6), ]
However, if you are trying to write a robust data analysis script, you should generally avoid deleting rows by numeric position. This is because the order of the rows in your data may change in the future. A general principle of a data.frame or database tables is that the order of the rows should not matter. If the order does matter, this should be encoded in an actual variable in the data.frame.
For example, imagine you imported a dataset and deleted rows by numeric position after inspecting the data and identifying the row numbers of the rows that you wanted to delete. However, at some later point, you go into the raw data and have a look around and reorder the data. Your row deletion code will now delete the wrong rows, and worse, you are unlikely to get any errors warning you that this has occurred.
Better strategy
A better strategy is to delete rows based on substantive and stable properties of the row. For example, if you had an id column variable that uniquely identifies each case, you could use that.
newdata <- myData[ !(myData$id %in% c(2,4,6)), ]
Other times, you will have a formal exclusion criteria that could be specified, and you could use one of the many subsetting tools in R to exclude cases based on that rule.
Create id column in your data frame or use any column name to identify the row. Using index is not fair to delete.
Use subset function to create new frame.
updated_myData <- subset(myData, id!= 6)
print (updated_myData)
updated_myData <- subset(myData, id %in% c(1, 3, 5, 7))
print (updated_myData)
By simplified sequence :
mydata[-(1:3 * 2), ]
By sequence :
mydata[seq(1, nrow(mydata), by = 2) , ]
By negative sequence :
mydata[-seq(2, nrow(mydata), by = 2) , ]
Or if you want to subset by selecting odd numbers:
mydata[which(1:nrow(mydata) %% 2 == 1) , ]
Or if you want to subset by selecting odd numbers, version 2:
mydata[which(1:nrow(mydata) %% 2 != 0) , ]
Or if you want to subset by filtering even numbers out:
mydata[!which(1:nrow(mydata) %% 2 == 0) , ]
Or if you want to subset by filtering even numbers out, version 2:
mydata[!which(1:nrow(mydata) %% 2 != 1) , ]
For completeness, I'll add that this can be done with dplyr as well using slice. The advantage of using this is that it can be part of a piped workflow.
df <- df %>%
.
.
slice(-c(2, 4, 6)) %>%
.
.
Of course, you can also use it without pipes.
df <- slice(df, -c(2, 4, 6))
The "not vector" format, -c(2, 4, 6) means to get everything that is not at rows 2, 4 and 6. For an example using a range, let's say you wanted to remove the first 5 rows, you could do slice(df, 6:n()). For more examples, see the docs.
Delete Dan from employee.data - No need to manage a new data.frame.
employee.data <- subset(employee.data, name!="Dan")
Here's a quick and dirty function to remove a row by index.
removeRowByIndex <- function(x, row_index) {
nr <- nrow(x)
if (nr < row_index) {
print('row_index exceeds number of rows')
} else if (row_index == 1)
{
return(x[2:nr, ])
} else if (row_index == nr) {
return(x[1:(nr - 1), ])
} else {
return (x[c(1:(row_index - 1), (row_index + 1):nr), ])
}
}
It's main flaw is it the row_index argument doesn't follow the R pattern of being a vector of values. There may be other problems as I only spent a couple of minutes writing and testing it, and have only started using R in the last few weeks. Any comments and improvements on this would be very welcome!
To identify by a name:
Call out the unique ID and identify the location in your data frame (DF).
Mark to delete. If the unique ID applies to multiple rows, all these rows will be removed.
Code:
Rows<-which(grepl("unique ID", DF$Column))
DF2<-DF[-c(Rows),]
DF2
Another approach when working with Unique IDs is to subset data:
*This came from an actual report where I wanted to remove the chemical standard
Chem.Report<-subset(Chem.Report, Chem_ID!="Standard")
Chem_ID is the column name.
The ! is important for excluding

Recursive function in R to find unique rows of a list of data tables

I am working on a function that takes a list of data tables with the same column names as an input and returns a single data table that has the unique rows from each data frame combined using successive rbind as shown below.
The function would be applied on a "very" large data.table (10s of millions of rows) which is why I had to split it up into several smaller data tables and assign them into a list to use recursion. At each step depending upon the length of the list of data tables (odd or even), I find the unique of data.table at that list index and the data table at the list index x - 1 and then successively rbind the 2 and assign to list index x - 1, and more list index x.
I must be missing something obvious, because although I can produce the final unique-d data.table when I print it (eg., print (listelement[[1]]), when I return (listelement[[1]]) I get NULL. Would help if someone can spot what I am missing ... or suggest if there is perhaps any other more efficient way to perform this.
Also, instead of having to add each data.table to a list, can I add them as "references" in the list ? I believe doing something like list(datatable1, datatable2 ...) would actually copy them ?
## CODE
returnUnique2 <- function (alist) {
if (length(alist) == 1) {
z <- (alist[[1]])
print (class(z))
print (z) ### This is the issue, if I change to return (z), I get NULL (?)
}
if (length(alist) %% 2 == 0) {
alist[[length(alist) - 1]] <- unique(rbind(unique(alist[[length(alist)]]), unique(alist[[length(alist) - 1]])))
alist[[length(alist)]] <- NULL
returnUnique2(alist)
}
if (length(alist) %% 2 == 1 && length(alist) > 2) {
alist[[length(alist) - 1]] <- unique(rbind(unique(alist[[length(alist)]]), unique(alist[[length(alist) - 1]])))
alist[[length(alist)]] <- NULL
returnUnique2(alist)
}
}
## OUTPUT with print statement
t1 <- data.table(col1=rep("a",10), col2=round(runif(10,1,10)))
t2 <- data.table(col1=rep("a",10), col2=round(runif(10,1,10)))
t3 <- data.table(col1=rep("a",10), col2=round(runif(10,1,10)))
tempList <- list(t1, t2, t3)
returnUnique2(tempList)
[1] "list"
[[1]]
col1 col2
1: a 3
2: a 2
3: a 5
4: a 9
5: a 10
6: a 7
7: a 1
8: a 8
9: a 4
10: a 6
Changing the following,
print (z) ### This is the issue, if I change to return (z), I get NULL (?)
to read
return(z)
returns NULL
Thanks in advance.
Please correct me if I misunderstand what you're doing, but it sounds like you have one big data.table and are trying to split it up to run some function on it and would then combine everything back and run a unique on that. The data.table way of doing that would be to use by, e.g.
fn = function(d) {
# do whatever to the subset and return the resulting data.table
# in this case, do nothing
d
}
N = 10 # number of pieces you like
dt[, fn(.SD), by = (seq_len(nrow(dt)) - 1) %/% (nrow(dt)/N)][, seq_len := NULL]
dt = dt[!duplicated(dt)]
Seems like this could be a good use case for a for loop. With many rows the overhead of using a for loop should be relatively small compared to the computation time. I would try combining my data.table's into a list (called ll in my example), then for each one remove duplicated rows, then rbind to the previous data.table with unique rows and then subset by unique rows again.
If you have many duplicated rows in each chunk then this might save some time, overall I'm not sure how effective it will be, but worth a shot?
# Create empty data.table for results (I have columns x and y in this case)
res <- data.table( x= numeric(0),y=numeric(0))
# loop over all data.tables in a list called 'll'
for( i in 1:length(ll) ){
# rbind the unique rows from the current list element to the results from all previous iterations
res <- rbind( res , ll[[i]][ ! duplicated(ll[[i]]) , ] )
# Keep only unique records at each iteration
res <- res[ ! duplicated(res) , ]
}
On another note, have you looked at the documentation for data.table? It explicitly states,
Because data.tables are usually sorted by key, tests for duplication
are especially quick.
So you might just be better off running on the entire data.table?
DT[ ! duplicated(DT) , ]
Add an id column to each data.table
t1$id=1
t2$id=2
t3$id=3
then combine them all at once and do a unique using by=.
If the data.tables are huge you could use setkey(...) to create an index on id before calling unique.
tall=rbind(t1,t2,t3)
tall[,unique(col1,col2),by=id]

How do I delete rows in a data frame?

I have a data frame named "mydata" that looks like this this:
A B C D
1. 5 4 4 4
2. 5 4 4 4
3. 5 4 4 4
4. 5 4 4 4
5. 5 4 4 4
6. 5 4 4 4
7. 5 4 4 4
I'd like to delete row 2,4,6. For example, like this:
A B C D
1. 5 4 4 4
3. 5 4 4 4
5. 5 4 4 4
7. 5 4 4 4
The key idea is you form a set of the rows you want to remove, and keep the complement of that set.
In R, the complement of a set is given by the '-' operator.
So, assuming the data.frame is called myData:
myData[-c(2, 4, 6), ] # notice the -
Of course, don't forget to "reassign" myData if you wanted to drop those rows entirely---otherwise, R just prints the results.
myData <- myData[-c(2, 4, 6), ]
You can also work with a so called boolean vector, aka logical:
row_to_keep = c(TRUE, FALSE, TRUE, FALSE, TRUE, FALSE, TRUE)
myData = myData[row_to_keep,]
Note that the ! operator acts as a NOT, i.e. !TRUE == FALSE:
myData = myData[!row_to_keep,]
This seems a bit cumbersome in comparison to #mrwab's answer (+1 btw :)), but a logical vector can be generated on the fly, e.g. where a column value exceeds a certain value:
myData = myData[myData$A > 4,]
myData = myData[!myData$A > 4,] # equal to myData[myData$A <= 4,]
You can transform a boolean vector to a vector of indices:
row_to_keep = which(myData$A > 4)
Finally, a very neat trick is that you can use this kind of subsetting not only for extraction, but also for assignment:
myData$A[myData$A > 4,] <- NA
where column A is assigned NA (not a number) where A exceeds 4.
Problems with deleting by row number
For quick and dirty analyses, you can delete rows of a data.frame by number as per the top answer. I.e.,
newdata <- myData[-c(2, 4, 6), ]
However, if you are trying to write a robust data analysis script, you should generally avoid deleting rows by numeric position. This is because the order of the rows in your data may change in the future. A general principle of a data.frame or database tables is that the order of the rows should not matter. If the order does matter, this should be encoded in an actual variable in the data.frame.
For example, imagine you imported a dataset and deleted rows by numeric position after inspecting the data and identifying the row numbers of the rows that you wanted to delete. However, at some later point, you go into the raw data and have a look around and reorder the data. Your row deletion code will now delete the wrong rows, and worse, you are unlikely to get any errors warning you that this has occurred.
Better strategy
A better strategy is to delete rows based on substantive and stable properties of the row. For example, if you had an id column variable that uniquely identifies each case, you could use that.
newdata <- myData[ !(myData$id %in% c(2,4,6)), ]
Other times, you will have a formal exclusion criteria that could be specified, and you could use one of the many subsetting tools in R to exclude cases based on that rule.
Create id column in your data frame or use any column name to identify the row. Using index is not fair to delete.
Use subset function to create new frame.
updated_myData <- subset(myData, id!= 6)
print (updated_myData)
updated_myData <- subset(myData, id %in% c(1, 3, 5, 7))
print (updated_myData)
By simplified sequence :
mydata[-(1:3 * 2), ]
By sequence :
mydata[seq(1, nrow(mydata), by = 2) , ]
By negative sequence :
mydata[-seq(2, nrow(mydata), by = 2) , ]
Or if you want to subset by selecting odd numbers:
mydata[which(1:nrow(mydata) %% 2 == 1) , ]
Or if you want to subset by selecting odd numbers, version 2:
mydata[which(1:nrow(mydata) %% 2 != 0) , ]
Or if you want to subset by filtering even numbers out:
mydata[!which(1:nrow(mydata) %% 2 == 0) , ]
Or if you want to subset by filtering even numbers out, version 2:
mydata[!which(1:nrow(mydata) %% 2 != 1) , ]
For completeness, I'll add that this can be done with dplyr as well using slice. The advantage of using this is that it can be part of a piped workflow.
df <- df %>%
.
.
slice(-c(2, 4, 6)) %>%
.
.
Of course, you can also use it without pipes.
df <- slice(df, -c(2, 4, 6))
The "not vector" format, -c(2, 4, 6) means to get everything that is not at rows 2, 4 and 6. For an example using a range, let's say you wanted to remove the first 5 rows, you could do slice(df, 6:n()). For more examples, see the docs.
Delete Dan from employee.data - No need to manage a new data.frame.
employee.data <- subset(employee.data, name!="Dan")
Here's a quick and dirty function to remove a row by index.
removeRowByIndex <- function(x, row_index) {
nr <- nrow(x)
if (nr < row_index) {
print('row_index exceeds number of rows')
} else if (row_index == 1)
{
return(x[2:nr, ])
} else if (row_index == nr) {
return(x[1:(nr - 1), ])
} else {
return (x[c(1:(row_index - 1), (row_index + 1):nr), ])
}
}
It's main flaw is it the row_index argument doesn't follow the R pattern of being a vector of values. There may be other problems as I only spent a couple of minutes writing and testing it, and have only started using R in the last few weeks. Any comments and improvements on this would be very welcome!
To identify by a name:
Call out the unique ID and identify the location in your data frame (DF).
Mark to delete. If the unique ID applies to multiple rows, all these rows will be removed.
Code:
Rows<-which(grepl("unique ID", DF$Column))
DF2<-DF[-c(Rows),]
DF2
Another approach when working with Unique IDs is to subset data:
*This came from an actual report where I wanted to remove the chemical standard
Chem.Report<-subset(Chem.Report, Chem_ID!="Standard")
Chem_ID is the column name.
The ! is important for excluding

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