I have a dataframe like this one:
x <- data.frame(filename = c("aa-b-c x", "c-dd-e y"), number=c(1,2))
filename number
1 aa-b-c x 1
2 c-dd-e y 2
I want to mutate the filename column so it looks like this:
filename number
1 c/aa/b 1
2 e/dd/c 2
This works on a single row: paste(str_match(x$filename[1], "(\\w+)-(\\w+)-(\\w+)")[c(4,2,3)], collapse = "/") but it fails inside the mutate. I'm sure I'm missing a simple fix.
One option is to rearrange the components after capturing as a group
library(dplyr)
library(stringr)
x %>%
mutate(filname = str_replace(filename,
"^([a-z]+)-([a-z]+)-([a-z]+)\\s.*", "\\3/\\1/\\2"))
str_match returns a matrix when you give it multiple vectors. This should work pretty well:
apply(str_match(x$filename, "(\\w+)-(\\w+)-(\\w+)")[, c(4,2,3), drop = FALSE], 1, paste, collapse = "/")
# [1] "c/aa/b" "e/c/dd"
The drop = FALSE is necessary to keep the output a matrix in case there is only one row.
Related
I have a data frame with a column (A) that contains strings such that each word is separated by a comma (still one string).
Df
A B etc.
"String1, String2, etc." ... etc.
I want to
Turn the observations in column A into a list. The list will contain elements string 1, string 2 etc.
I want to remove all strings that are not 8 characters long, start with 4 numbers and end with 4 digits (I already have the Regex for that)
I want to turn all the strings into factors
The end product should look like this
Df
A B etc.
[String1, String2, etc] ... etc.
Doing some testing, I've realised a combination of strsplit() and str_subset fulfills requirements 1 and 2
var = "ABCD1234, ABCDEFGH"
var = str_split(var, ", ")
var = str_subset(var, "^[A-Za-z]{4}\\d{4}$")
# Var = list("ABCD1234")
But I'm having trouble applying this to a dataframe column. So far, this has not worked
df = df %>% mutate(
A = strsplit(A, split = ", ")
A = case_when(
TRUE ~ str_subset(A, "^[A-Za-z]{4}\\d{4}$")
)
)
Could someone help please?
Thanks
We can combine the two steps by first splitting the string on ", " and then use str_subset to get the strings which follow a pattern.
library(tidyverse)
df %>%
mutate(new = str_split(A, ", "),
new = map(new, str_subset, pattern = "^[A-Za-z]{4}\\d{4}$"))
# A new
#1 ABCD1234, ABCDEFGH ABCD1234
#2 AQD12345, AQWE1
#3 ABCD1234, ABCD5678 ABCD1234, ABCD5678
We can do this in base R, as well
df$new <- lapply(strsplit(df$A, ", "), grep,
pattern = "^[A-Za-z]{4}\\d{4}$", value = TRUE)
data
df <- data.frame(A = c("ABCD1234, ABCDEFGH", "AQD12345, AQWE1",
"ABCD1234, ABCD5678"), stringsAsFactors = FALSE)
I know there are some answers here about splitting a string every nth character, such as this one and this one, However these are pretty question specific and mostly related to a single string and not to a data frame of multiple strings.
Example data
df <- data.frame(id = 1:2, seq = c('ABCDEFGHI', 'ZABCDJHIA'))
Looks like this:
id seq
1 1 ABCDEFGHI
2 2 ZABCDJHIA
Splitting on every third character
I want to split the string in each row every thrid character, such that the resulting data frame looks like this:
id 1 2 3
1 ABC DEF GHI
2 ZAB CDJ HIA
What I tried
I used the splitstackshape before to split a string on a single character, like so: df %>% cSplit('seq', sep = '', stripWhite = FALSE, type.convert = FALSE) I would love to have a similar function (or perhaps it is possbile with cSplit) to split on every third character.
An option would be separate
library(tidyverse)
df %>%
separate(seq, into = paste0("x", 1:3), sep = c(3, 6))
# id x1 x2 x3
#1 1 ABC DEF GHI
#2 2 ZAB CDJ HIA
If we want to create it more generic
n1 <- nchar(as.character(df$seq[1])) - 3
s1 <- seq(3, n1, by = 3)
nm1 <- paste0("x", seq_len(length(s1) +1))
df %>%
separate(seq, into = nm1, sep = s1)
Or using base R, using strsplit, split the 'seq' column for each instance of 3 characters by passing a regex lookaround into a list and then rbind the list elements
df[paste0("x", 1:3)] <- do.call(rbind,
strsplit(as.character(df$seq), "(?<=.{3})", perl = TRUE))
NOTE: It is better to avoid column names that start with non-standard labels such as numbers. For that reason, appended 'x' at the beginning of the names
You can split a string each x characters in base also with read.fwf (Read Fixed Width Format Files), which needs either a file or a connection.
read.fwf(file=textConnection(as.character(df$seq)), widths=c(3,3,3))
V1 V2 V3
1 ABC DEF GHI
2 ZAB CDJ HIA
I have variables with names such as r1a r3c r5e r7g r9i r11k r13g r15i etc. I am trying to select variables which starts with r5 - r12 and create a dataframe in R.
The best code that I could write to get this done is,
data %>% select(grep("r[5-9][^0-9]" , names(data), value = TRUE ),
grep("r1[0-2]", names(data), value = TRUE))
Given my experience with regular expressions span a day, I was wondering if anyone could help me write a better and compact code for this!
Here's a regex that gets all the columns at once:
data %>% select(grep("r([5-9]|1[0-2])", names(data), value = TRUE))
The vertical bar represents an 'or'.
As the comments have pointed out, this will fail for items such as r51, and can also be shortened. Instead, you will need a slightly longer regex:
data %>% select(matches("r([5-9]|1[0-2])([^0-9]|$)"))
Suppose that in the code below x represents your names(data). Then the following will do what you want.
# The names of 'data'
x <- scan(what = character(), text = "r1a r3c r5e r7g r9i r11k r13g r15i")
y <- unlist(strsplit(x, "[[:alpha:]]"))
y <- as.numeric(y[sapply(y, `!=`, "")])
x[y > 4]
#[1] "r5e" "r7g" "r9i" "r11k" "r13g" "r15i"
EDIT.
You can make a function with a generalization of the above code. This function has three arguments, the first is the vector of variables names, the second and the third are the limits of the numbers you want to keep.
var_names <- function(x, from = 1, to = Inf){
y <- unlist(strsplit(x, "[[:alpha:]]"))
y <- as.integer(y[sapply(y, `!=`, "")])
x[from <= y & y <= to]
}
var_names(x, 5)
#[1] "r5e" "r7g" "r9i" "r11k" "r13g" "r15i"
Remove the non-digits, scan the remainder in and check whether each is in 5:12 :
DF <- data.frame(r1a=1, r3c=2, r5e=3, r7g=4, r9i=5, r11k=6, r13g=7, r15i=8) # test data
DF[scan(text = gsub("\\D", "", names(DF)), quiet = TRUE) %in% 5:12]
## r5e r7g r9i r11k
## 1 3 4 5 6
Using magrittr it could also be written like this:
library(magrittr)
DF %>% .[scan(text = gsub("\\D", "", names(.)), quiet = TRUE) %in% 5:12]
## r5e r7g r9i r11k
## 1 3 4 5 6
I have a data frame like the following:
sampleid <- c("patient_sdlkfjd_2354_CSF_CD19+", "control_sdlkfjd_2632_CSF_CD8+", "control_sdlkfjd_2632_CSF")
values = rnorm(3, 8, 3)
df <- data.frame(sampleid, values)
I also have a vector like the following:
matches <- c("632_CSF_CD8+", "632_CSF").
I want to extract rows in this data frame which contain the matches at the end of the value in the sampleid column. From this example, you can see why the end of string is important,as I have two samples which contain "632_CSF," but they are distinct samples. If I chose to change matches to only:
matches <- c("632_CSF").
Then I want only the third row of the data frame to be outputted, because this is the only one where this matches at the end of the sampleid.
How can this be achieved?
Thanks!
Just use $ in your pattern to indicate that it occurs at the end of the string.
grep("632_CSF$", sampleid, value=TRUE)
[1] "control_sdlkfjd_2632_CSF"
You can make this with stringr and some manipulations.
You need to encode regex, it's done with quotemeta function.
Next step would be to append $ to ensure the match is in the end of the string and then concatenate all matches into one with regex OR - |.
And then it should be used with str_detect to get boolean indices.
library(stringr)
# taken from here
# https://stackoverflow.com/a/14838753/1030110
quotemeta <- function(string) {
str_replace_all(string, "(\\W)", "\\\\\\1")
}
matches_with_end <- sapply(matches, function(x) { paste0(quotemeta(x), '$') })
joined_matches <- paste(matches_with_end, collapse = '|')
ind <- str_detect(df$sampleid, joined_matches)
# [1] FALSE TRUE TRUE
df[ind, ]
# sampleid values
# 2 control_sdlkfjd_2632_CSF_CD8+ 10.712634
# 3 control_sdlkfjd_2632_CSF 7.001628
Suggest making your dataset more regular.
library(tidyverse)
df_regular <- df %>%
separate(
sampleid,
into = c("patient_type",
"test_number",
"patient_group",
"patient_id"),
extra = "merge") %>%
mutate(patient_id = str_pad(patient_id, 9, side = c("left"), pad = "0"))
df_regular
df_regular %>%
filter(patient_group %in% "2632" & patient_id %in% "000000CSF")
My dataframe which I read from a csv file has column names like this
abc.def, ewf.asd.fkl, qqit.vsf.addw.coil
I want to remove the '.' from all the names and convert them to
abcdef, eqfasdfkl, qqitvsfaddwcoil.
I tried using the sub command sub(".","",colnames(dataframe)) but this command took out the first letter of each column name and the column names changed to
bc.def, wf.asd.fkl, qit.vsf.addw.coil
Anyone know another command to do this. I can change the column name one by one, but I have a lot of files with 30 or more columns in each file.
Again, I want to remove the "." from all the colnames. I am trying to do this so I can use "sqldf" commands, which don't deal well with "."
Thank you for your help
1) sqldf can deal with names having dots in them if you quote the names:
library(sqldf)
d0 <- read.csv(text = "A.B,C.D\n1,2")
sqldf('select "A.B", "C.D" from d0')
giving:
A.B C.D
1 1 2
2) When reading the data using read.table or read.csv use the check.names=FALSE argument.
Compare:
Lines <- "A B,C D
1,2
3,4"
read.csv(text = Lines)
## A.B C.D
## 1 1 2
## 2 3 4
read.csv(text = Lines, check.names = FALSE)
## A B C D
## 1 1 2
## 2 3 4
however, in this example it still leaves a name that would have to be quoted in sqldf since the names have embedded spaces.
3) To simply remove the periods, if DF is a data frame:
names(DF) <- gsub(".", "", names(DF), fixed = TRUE)
or it might be nicer to convert the periods to underscores so that it is reversible:
names(DF) <- gsub(".", "_", names(DF), fixed = TRUE)
This last line could be alternatively done like this:
names(DF) <- chartr(".", "_", names(DF))
UPDATE dplyr 0.8.0
As of dplyr 0.8 funs() is soft deprecated, use formula notation.
a dplyr way to do this using stringr.
library(dplyr)
library(stringr)
data <- data.frame(abc.def = 1, ewf.asd.fkl = 2, qqit.vsf.addw.coil = 3)
renamed_data <- data %>%
rename_all(~str_replace_all(.,"\\.","_")) # note we have to escape the '.' character with \\
Make sure you install the packages with install.packages().
Remember you have to escape the . character with \\. in regex, which functions like str_replace_all use, . is a wildcard.
To replace all the dots in the names you'll need to use gsub, rather than sub, which will only replace the first occurrence.
This should work.
test <- data.frame(abc.def = NA, ewf.asd.fkl = NA, qqit.vsf.addw.coil = NA)
names(test) <- gsub( ".", "", names(test), fixed = TRUE)
test
abcdef ewfasdfkl qqitvsfaddwcoil
1 NA NA NA
You can also try:
names(df) = gsub(pattern = ".", replacement = "", x = names(df))