I have a data frame with several columns. One of those contains Plotids like AEG1, AEG2,..., AEG50, HEG1, HEG2,..., HEG50, SEG1, SEG2,..., SEG50. So, the data frame has 150 rows. Now I want to change only some of these Plotids, so that there is AEG01, AEG02,... instead of AEG1, AEG2, ... So, I just want to add a "0" to some of the column entries. I tried it by using lapply, a for loop, writing a function,... but nothing did the job. There was always the error message:
In if (nchar(as.character(dat_merge$EP_Plotid)) == 4)
paste(substr(dat_merge$EP_Plotid, ... :
the condition has length > 1 and only the first element will be used
So, this was my last try:
Plotid_func <- function(x) {
if(nchar(as.character(dat_merge$EP_Plotid))==4)
paste(substr(dat_merge$EP_Plotid, 1, 3), "0", substr(dat_merge$EP_Plotid, 4, 4), sep="")
}
dat_merge$Plotid <- sapply(dat_merge$EP_Plotid, Plotid_func)
Therewith, I wanted to select only those column entries with four digits. And to only those selected entries, I wanted to add a 0. Can anybody help me? dat_merge is the name of my data frame and EP_Plotid is the column I want to edit. Thanks in advance
Just extract the "string" portion and the "numeric" portion and paste them back together after using sprintf on the numeric portion.
An example:
## "x" is the "column" of plot ids. Here I go up to 12
## to demonstrate the zero padding that it sounds like
## you're looking for
x <- c(paste0("AEG", 1:12), paste0("HEG", 1:12))
## Extract the string values
Strings <- gsub("([A-Z]+)(.*)", "\\1", x)
## Extract the numeric values
Nums <- gsub("([A-Z]+)(.*)", "\\2", x)
## Put them back together
paste0(Strings, sprintf("%02d", as.numeric(Nums)))
# [1] "AEG01" "AEG02" "AEG03" "AEG04" "AEG05" "AEG06"
# [7] "AEG07" "AEG08" "AEG09" "AEG10" "AEG11" "AEG12"
# [13] "HEG01" "HEG02" "HEG03" "HEG04" "HEG05" "HEG06"
# [19] "HEG07" "HEG08" "HEG09" "HEG10" "HEG11" "HEG12"
Or you can just modify your function to actually use the input variable x (which is not happening in your original function)
dat_merge <- data.frame(EP_Plotid = c("AEG1", "AEG2", "AEG50", "HEG1", "HEG2", "HEG50", "SEG1", "SEG2", "SEG50"))
Plotid_func <- function(x) {
if(nchar(as.character(x)) == 4){
paste(substr(x, 1, 3), "0", substr(x, 4, 4), sep="")
} else as.character(x)
}
dat_merge$Plotid <- sapply(dat_merge$EP_Plotid, Plotid_func)
dat_merge
# EP_Plotid Plotid
# 1 AEG1 AEG01
# 2 AEG2 AEG02
# 3 AEG50 AEG50
# 4 HEG1 HEG01
# 5 HEG2 HEG02
# 6 HEG50 HEG50
# 7 SEG1 SEG01
# 8 SEG2 SEG02
# 9 SEG50 SEG50
A vectorized version of your function (which is much better than using sapply which is just a for loop) would be
dat_merge$Plotid <- ifelse(nchar(as.character(dat_merge$EP_Plotid))==4, paste(substr(dat_merge$EP_Plotid, 1, 3), "0", substr(dat_merge$EP_Plotid, 4, 4), sep=""), as.character(dat_merge$EP_Plotid))
Or use a combination of formatC with str_extract from library(stringr)
library(stringr)
x from Ananda's post.
Extract alphabets and numbers separately.
Flag 0's to the numbers with formatC
paste together
paste0(str_extract(x, "[[:alpha:]]+"), formatC(as.numeric(str_extract(x,"\\d+")), width=2, flag=0))
#[1] "AEG01" "AEG02" "AEG03" "AEG04" "AEG05" "AEG06" "AEG07" "AEG08" "AEG09"
#[10] "AEG10" "AEG11" "AEG12" "HEG01" "HEG02" "HEG03" "HEG04" "HEG05" "HEG06"
#[19] "HEG07" "HEG08" "HEG09" "HEG10" "HEG11" "HEG12"
Related
HEADLINE: Is there a way to get R to recognize data.frame column names contained within lists in the same way that it can recognize free-floating vectors?
SETUP: Say I have a vector named varA:
(varA <- 1:6)
# [1] 1 2 3 4 5 6
To get the length of varA, I could do:
length(varA)
#[1] 6
and if the variable was contained within a larger list, the variable and its length could still be found by doing:
list <- list(vars = "varA")
length(get(list$vars[1]))
#[1] 6
PROBLEM:
This is not the case when I substitute the vector for a dataframe column and I don't know how to work around this:
rows <- 1:6
cols <- c("colA")
(df <- data.frame(matrix(NA,
nrow = length(rows),
ncol = length(cols),
dimnames = list(rows, cols))))
# colA
# 1 NA
# 2 NA
# 3 NA
# 4 NA
# 5 NA
# 6 NA
list <- list(vars = "varA",
cols = "df$colA")
length(get(list$vars[1]))
#[1] 6
length(get(list$cols[1]))
#Error in get(list$cols[1]) : object 'df$colA' not found
Though this contrived example seems inane, because I could always use the simple length(variable) approach, I'm actually interested in writing data from hundreds of variables varying in lengths onto respective dataframe columns, and so keeping them in a list that I could iterate through would be very helpful. I've tried everything I could think of, but it may be the case that it's just not possible in R, especially given that I cannot find any posts with solutions to the issue.
You could try:
> length(eval(parse(text = list$cols[1])))
[1] 6
Or:
list <- list(vars = "varA",
cols = "colA")
length(df[, list$cols[1]])
[1] 6
Or with regex:
list <- list(vars = "varA",
cols = "df$colA")
length(df[, sub(".*\\$", "", list$cols[1])])
[1] 6
If you are truly working with a data frame d, then nrow(d) is the length of all of the variables in d. There should be no reason to use length in this case.
If you are actually working with a list x containing variables of potentially different lengths, then you should use the [[ operator to extract those variables by name (see ?Extract):
x <- list(a = 1:10, b = rnorm(20L))
l <- list(vars = "a")
length(d[[l$vars[1L]]]) # 10
If you insist on using get (you shouldn't), then you need to supply a second argument telling it where to look for the variable (see ?get):
length(get(l$vars[1L], x)) # 10
I am trying to use ifelse to populate a new column in a data frame.
I want to extract the last digits of a character string in a column if this is longer than 3. if the charachter string is shorter I just want it to give -1...
I already figured out how to extract the last characters of the string if the string is longer than 3 characters.
x<- c("ABCD1", "ABCD2", "ABCD3", "ABCD4", "BC5", "BC6", "BC7")
y<-NULL
dat<-cbind(x,y)
ifelse (nchar(x>3), y=substr(x, 5,5), y=-1)
dat<-cbind(x,y)
view(dat)
when I run this, I get the next error
Error in ifelse(nchar(x > 3), y = substr(x, 4, 5), y = substr(x, 3)) :
formal argument "yes" matched by multiple actual arguments`
What I want is that vector "y" gets the numbers 1,2,3,4,-1,-1,-1
so I can bind both columns latter. If you have a better way of doing this I would appreciate it.
You're almost there! This will work as long as the strings with length > 3 are 4 characters long.
ifelse(nchar(x) > 3, substr(x, 5, 5), -1)
If your strings might be longer than 4 characters:
ifelse(nchar(x) > 3, sub(".*([0-9]).*", "\\1", x), -1)
I am guessing you need a dataframe. Here's what you probably need -
x <- c("ABCD1", "ABCD2", "ABCD3", "ABCD4", "BC5", "BC6", "BC7")
dat <- data.frame(x, stringsAsFactors = F)
dat$y <- ifelse(nchar(dat$x) > 3, as.numeric(substr(dat$x, 5,5)), -1)
x y
1 ABCD1 1
2 ABCD2 2
3 ABCD3 3
4 ABCD4 4
5 BC5 -1
6 BC6 -1
7 BC7 -1
I have a vector as below
data <- c("6X75ML","24X37.5ML (KKK)", "6X2X75ML", "168X5CL (UUU)")
here i want to extract the first number before the "X" for each of the elements.
In case of situations with 2 "X" i.e. "6X2X75CL" the number 12 (6 multiplied by 2) should be calculated.
expected output
6, 24, 12, 168
Thank you for the help...
Here's a possible solution using regular expressions :
data <- c("6X75ML","24X37.5ML (KKK)", "6X2X75ML", "168X5CL (UUU)")
# this regular expression finds any group of digits followed
# by a upper-case 'X' in each string and returns a list of the matches
tokens <- regmatches(data,gregexpr('[[:digit:]]+(?=X)',data,perl=TRUE))
res <- sapply(tokens,function(x)prod(as.numeric(x)))
> res
[1] 6 24 12 168
Here is a method using base R:
dataList <- strsplit(data, split="X")
sapply(dataList, function(x) Reduce("*", as.numeric(head(x, -1))))
[1] 6 24 12 168
strplit breaks up the vector along "X". The resulting list is fed to sapply which the performs an operation on all but the final element of each vector in the list. The operation is to transform the elements into numerics and the multiply them. The final element is dropped using head(x, -1).
As #zheyuan-li comments, prod can fill in for Reduce and will probably be a bit faster:
sapply(dataList, function(x) prod(as.numeric(head(x, -1))))
[1] 6 24 12 168
We can also use str_extract_all
library(stringr)
sapply(str_extract_all(data, "\\d+(?=X)"), function(x) prod(as.numeric(x)))
#[1] 6 24 12 168
ind=regexpr("X",data)
val=as.integer(substr(data, 1, ind-1))
data2=substring(data,ind+1)
ind2=regexpr("[0-9]+X", data2)
if (!all(ind2!=1)) {
val2 = as.integer(substr(data2[ind2==1], 1, attr(ind2,"match.length")[ind2==1]-1))
val[ind2==1] = val[ind2==1] * val2
}
I have character vector of the following form (this is just a sample):
R1Ng(10)
test(0)
n.Ex1T(34)
where as can be seen above, the first part is always some combination of alphanumeric and punctuation marks, then there are parentheses with a number inside. I want to create a numeric vector which will store the values inside the parentheses, and each number should have name attribute, and the name attribute should be the string before the number. So, for example, I want to store 10, 0, 34, inside a numeric vector and their name attributes should be, R1Ng, test, n.Ex1T, respectively.
I can always do something like this to get the numbers and create a numeric vector:
counts <- regmatches(data, gregexpr("[[:digit:]]+", data))
as.numeric(unlist(counts))
But, how can I extract the first string part, and store it as the name attribute of that numberic array?
How about this:
x <- c("R1Ng(10)", "test(0)", "n.Ex1T(34)")
data.frame(Name = gsub( "\\(.*", "", x),
Count = as.numeric(gsub(".*?\\((.*?)\\).*", "\\1", x)))
# Name Count
# 1 R1Ng 10
# 2 test 0
# 3 n.Ex1T 34
Or alternatively as a vector
setNames(as.numeric(gsub(".*?\\((.*?)\\).*", "\\1", x)),
gsub( "\\(.*", "", x ))
# R1Ng test n.Ex1T
# 10 0 34
Here is another variation using the same expression and capturing parentheses:
temp <- c("R1Ng(10)", "test(0)", "n.Ex1T(34)")
data.frame(Name=gsub("^(.*)\\((\\d+)\\)$", "\\1", temp),
count=gsub("^(.*)\\((\\d+)\\)$", "\\2", temp))
We can use str_extract_all
library(stringr)
lst <- str_extract_all(x, "[^()]+")
Or with strsplit from base R
lst <- strsplit(x, "[()]")
If we need to store as a named vector
sapply(lst, function(x) setNames(as.numeric(x[2]), x[1]))
# R1Ng test n.Ex1T
# 10 0 34
data
x <- c("R1Ng(10)", "test(0)", "n.Ex1T(34)")
Suppose that I have a vector x whose elements I want to use to extract columns from a matrix or data frame M.
If x[1] = "A", I cannot use M$x[1] to extract the column with header name A, because M$A is recognized while M$"A" is not. How can I remove the quotes so that M$x[1] is M$A rather than M$"A" in this instance?
Don't use $ in this case; use [ instead. Here's a minimal example (if I understand what you're trying to do).
mydf <- data.frame(A = 1:2, B = 3:4)
mydf
# A B
# 1 1 3
# 2 2 4
x <- c("A", "B")
x
# [1] "A" "B"
mydf[, x[1]] ## As a vector
# [1] 1 2
mydf[, x[1], drop = FALSE] ## As a single column `data.frame`
# A
# 1 1
# 2 2
I think you would find your answer in the R Inferno. Start around Circle 8: "Believing it does as intended", one of the "string not the name" sub-sections.... You might also find some explanation in the line The main difference is that $ does not allow computed indices, whereas [[ does. from the help page at ?Extract.
Note that this approach is taken because the question specified using the approach to extract columns from a matrix or data frame, in which case, the [row, column] mode of extraction is really the way to go anyway (and the $ approach would not work with a matrix).