lets take an example dataframe with removal of variable columns:
frame <- data.frame("a" = 1:5, "b" = 2:6, "c" = 3:7, "d" = 4:8)
rem <- readline()
frame <- subset(frame, select = -c(rem))
How do I get the variable column to be removed? This is not my real code, just wanted to present my problem in a simple code. Thanks!
Edit: I am so sorry, I am really sleepy and don't know what I typed into my code, I edited it now.
1) Do both at once. We assume that ix contains at least one column number.
ix <- 1:2
frame[-ix]
## c d
## 1 3 4
## 2 4 5
## 3 5 6
## 4 6 7
## 5 7 8
1a) or if the case where ix is zero length, ix <- c(), is important we can do this. The output of this and all the rest are the same as for (1) so we won't repeat the output.
ix <- 1:2
frame[setdiff(seq_along(frame), ix)]
1b) or if we have names rather than column numbers. This works even if nms is a zero length vector in which case it returns the original data frame.
nms <- c("a", "b")
frame[setdiff(names(frame), nms)]
2) or if you need to do it iteratively remove the largest one first because if it were done in ascending order then after the first one is removed the second column is no longer the second but is the first. If we knew that ix is already sorted we could omit the sort. We have used frame_out to hold the result so that the input is not destroyed. This works even if ix is the empty vector.
ix <- 1:2
frame_out <- frame
for(i in rev(sort(ix))) frame_out <- frame_out[-i]
frame_out
3) One way to do it independent of order is to do it by name. In this case it would be possible to remove them in ascending order. This works even if ix the empty vector.
ix <- 1:2
nms <- names(frame)[ix]
frame_out <- frame
for(nm in nms) frame_out <- frame_out[-match(nm, names(frame_out))]
frame_out
Related
Given a 3 x 100 matrix, how could I create a vector of strings containing individual column names? Specifically, columns comprise 20 sets of 5 consecutive measures and therefore strings should match variable (i.e. varA, ... varC), sets (SET1 to SET20) and order (1 to 5). For example:
my_matrix = replicate(100, rnorm(3))
my_names <- c("varA.SET1.1", "varA.SET1.2", "varA.SET1.3", "varA.SET1.4", "varA.SET1.5",
"varA.SET2.1", "varA.SET2.2", "varA.SET2.3", "varA.SET2.4", "varA.SET2.5",
...
"varC.SET5.5")
You can use sprintf.
v <- LETTERS[1:3]
set <- 1:20
ord <- 1:5
ex <- expand.grid(v, set, ord)
my_names <- sprintf("var%s.SET%i.%i", ex[, 1],ex[, 2], ex[, 3])
head(my_names)
#[1] "varA.SET1.1" "varB.SET1.1" "varC.SET1.1" "varA.SET2.1" "varB.SET2.1"
#[6] "varC.SET2.1"
I am wondering how to subset my data based on the appearance of triplicates in a column.
t <- c(1,1,2,2,3,3,4,4,5,5,5,6,6,7,7,7,8,8)
mydf <- data.frame(t, 1:18)
I want to be able to grab only the rows that correspond to a triplicate in column t, so that I can form a new dataframe of only those rows. That would look like this where p is the vector of rows I'm looking for:
p <- c(9,10,11,14,15,16)
myidealdf[p,]
Sorry if this isn't clear, it's my first post
This should do it
keeps <- unique(t)[table(as.factor(t)) == 3]
keeps <- t %in% keeps
mydf <- mydf[keeps, ]
Using rle function.
which(t %in% with(rle(t), values[lengths==3]))
[1] 9 10 11 14 15 16
I have a single column data frame - example data:
1 >PROKKA_00002 Alpha-ketoglutarate permease
2 MTESSITERGAPELADTRRRIWAIVGASSGNLVEWFDFYVYSFCSLYFAHIFFPSGNTTT
3 QLLQTAGVFAAGFLMRPIGGWLFGRIADRRGRKTSMLISVCMMCFGSLVIACLPGYAVIG
4 >PROKKA_00003 lipoprotein
5 MRTIIVIASLLLTGCSHMANDAWSGQDKAQHFLASAMLSAAGNEYAQHQGYSRDRSAAIG
Each sequence of letters is associated with the ">" line above it. I need a two-column data frame with lines starting in ">" in the first column, and the respective lines of letters concatenated as one sequence in the second column. This is what I've tried so far:
y <- matrix(0,5836,2) #empty matrix with 5836 rows and two columns
z <- 0
for(i in 1:nrow(df)){
if((grepl(pattern = "^>", x = df)) == TRUE){ #tried to set the conditional "if a line starts with ">", execute code"
z <- z + 1
y[z,1] <- paste(df[i])
} else{
y[z,2] <- paste(df[i], collapse = "")
}
}
I would eventually convert the matrix y back to a data.frame using as.data.frame, but my loop keeps getting Error: unexpected '}' in "}". I'm also not sure if my conditional is right. Can anyone help? It would be greatly appreciated!
Although I will stick with packages, here is a solution
initialize data
mydf <- data.frame(x=c(">PROKKA_00002 Alpha-ketoglutarate","MTESSITERGAPEL", "MTESSITERGAPEL",">PROKKA_00003 lipoprotein", "MTESSITERGAPEL" ,"MRTIIVIASLLLT"), stringsAsFactors = F)
process
ind <- grep(">", mydf$x)
temp<-data.frame(ind=ind, from=ind+1, to=c((ind-1)[-1], nrow(mydf)))
seqs<-rep(NA, length(ind))
for(i in 1:length(ind)) {
seqs[i]<-paste(mydf$x[temp$from[i]:temp$to[i]], collapse="")
}
fastatable<-data.frame(name=gsub(">", "", mydf[ind,1]), sequence=seqs)
> fastatable
name sequence
1 PROKKA_00002 Alpha-ketoglutarate MTESSITERGAPELMTESSITERGAPEL
2 PROKKA_00003 lipoprotein MTESSITERGAPELMRTIIVIASLLLT
Try creating an index of the rows with the target symbol with the column headers. Then split the data on that index. The call cumsum(ind1)[!ind1] first creates an id rows by coercing the logical vector into numeric, then eliminates the rows with the column headers.
ind1 <- grepl(">", mydf$x)
#split data on the index created
newdf <- data.frame(mydf$x[ind1][cumsum(ind1)], mydf$x)[!ind1,]
#Add names
names(newdf) <- c("Name", "Value")
newdf
# Name Value
# 2 >PROKKA_00002 Alpha-ketoglutarate
# 3 >PROKKA_00002 MTESSITERGAPEL
# 5 >PROKKA_00003 lipoprotein
# 6 >PROKKA_00003 MRTIIVIASLLLT
Data
mydf <- data.frame(x=c(">PROKKA_00002","Alpha-ketoglutarate","MTESSITERGAPEL", ">PROKKA_00003", "lipoprotein" ,"MRTIIVIASLLLT"))
You can use plyr to accomplish this if you are able to assigned a section number to your rows appropriately:
library(plyr)
df <- data.frame(v1=c(">PROKKA_00002 Alpha-ketoglutarate permease",
"MTESSITERGAPELADTRRRIWAIVGASSGNLVEWFDFYVYSFCSLYFAHIFFPSGNTTT",
"QLLQTAGVFAAGFLMRPIGGWLFGRIADRRGRKTSMLISVCMMCFGSLVIACLPGYAVIG",
">PROKKA_00003 lipoprotein",
"MRTIIVIASLLLTGCSHMANDAWSGQDKAQHFLASAMLSAAGNEYAQHQGYSRDRSAAIG"))
df$hasMark <- ifelse(grepl(">",df$v1,fixed=TRUE),1, 0)
df$section <- cumsum(df$hasMark)
t <- ddply(df, "section", function(x){
data.frame(v2=head(x,1),v3=paste(x$v1[2:nrow(x)], collapse=''))
})
t <- subset(t, select=-c(section,v2.hasMark,v2.section)) #drop the extra columns
if you then view 't' I believe this is what you were looking for in your original post
I have searched extensively but not found an answer to this question on Stack Overflow.
Lets say I have a data frame a.
I define:
a <- NULL
a <- as.data.frame(a)
If I wanted to add a column to this data frame as so:
a$col1 <- c(1,2,3)
I get the following error:
Error in `$<-.data.frame`(`*tmp*`, "a", value = c(1, 2, 3)) :
replacement has 3 rows, data has 0
Why is the row dimension fixed but the column is not?
How do I change the number of rows in a data frame?
If I do this (inputting the data into a list first and then converting to a df), it works fine:
a <- NULL
a$col1 <- c(1,2,3)
a <- as.data.frame(a)
The row dimension is not fixed, but data.frames are stored as list of vectors that are constrained to have the same length. You cannot add col1 to a because col1 has three values (rows) and a has zero, thereby breaking the constraint. R does not by default auto-vivify values when you attempt to extend the dimension of a data.frame by adding a column that is longer than the data.frame. The reason that the second example works is that col1 is the only vector in the data.frame so the data.frame is initialized with three rows.
If you want to automatically have the data.frame expand, you can use the following function:
cbind.all <- function (...)
{
nm <- list(...)
nm <- lapply(nm, as.matrix)
n <- max(sapply(nm, nrow))
do.call(cbind, lapply(nm, function(x) rbind(x, matrix(, n -
nrow(x), ncol(x)))))
}
This will fill missing values with NA. And you would use it like: cbind.all( df, a )
You could also do something like this where I read in data from multiple files, grab the column I want, and store it in the dataframe. I check whether the dataframe has anything in it, and if it doesn't, create a new one rather than getting the error about mismatched number of rows:
readCounts = data.frame()
for(f in names(files)){
d = read.table(files[f], header=T, as.is=T)
d2 = round(data.frame(d$NumReads))
colnames(d2) = f
if(ncol(readCounts) == 0){
readCounts = d2
rownames(readCounts) = d$Name
} else{
readCounts = cbind(readCounts, d2)
}
}
if you have an empty dataframe, called for example df, in my opinion another quite simple solution is the following:
df[1,]=NA # ad a temporary new row of NA values
df[,'new_column'] = NA # adding new column, called for example 'new_column'
df = df[0,] # delete row with NAs
I hope this may help.
I have two vectors in R, e.g.
a <- c(2,6,4,9,8)
b <- c(8,9,4,2,1)
Vectors a and b are ordered in a way that I wish to conserve (I will be plotting them against each other). I want to remove certain values from vector a and remove the values at the same indices in b. e.g. if I wanted to remove values ≥ 8 from a:
a <- a[a<8]
... which gives a new vector without those values.
Is there now an easy way of removing values from the same indices in b (in this example indices 4 and 5)? Perhaps by using a data frame?
Something like this:
keep <- a < 8
a <- a[keep]
b <- b[keep]
You could also use:
keep <- which( a < 8 )
If the vectors are logically part of the same data, use a data frame. It is better programming practice.
df <- data.frame(a = a, b = b)
df <- df[df$a < 8, ]
Otherwise, set another vector to be the indices removed:
keep <- a < 8
a <- a[keep]
b <- b[keep]
Why not:
d <- data.frame(a=a, b=b)
d <- d[d$a < 8, ]
Or even:
d <- subset(d, a < 8)
First remove the indices from b then from a
b <- b[a<8]
a <- a[a<8]
a<8 returns a vector which defines which indices are smaller than 8.
If this is purely for plotting, you can avoid messing with b and the x-axis by using NA .
a[a>8]<-NA
plot(b,a) #works for point or line graphs