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Extracting numbers from vectors of strings
(12 answers)
Closed 1 year ago.
relatively new to R, need help with applying a regex-based substitution.
I have a data frame in one column of which I have a sequence of digits (my values of interest) followed by a string of all sorts of characters.
Example:
4623(randomcharacters)
I need to remove everything after the initial digits to continue working with the values. My idea was to use gsub to remove the non-digit characters by positive lookbehind.
The code I have is:
sub_function <- function() {
gsub("?<=[[:digit:]].", " ", fixed = T)
}
data_frame$`x` <- data_known$`x` %>%
sapply(sub_function)
But I then get the error:
Error in FUN(X[[i]], ...) : unused argument (X[[i]])
Any help would be greatly appreciated!
Here is a base R function.
It uses sub, not gsub, since there will be only one substitution. And there's no need for look behind, the meta-character ^ marks the beginning of the string, followed by an optional minus sign, followed by at least one digit. Everything else is discarded.
sub_function <- function(x){
sub("(^-*[[:digit:]]+).*", "\\1", x)
}
data <- data.frame(x = c("4623(randomcharacters)", "-4623(randomcharacters)"))
sub_function(data$x)
#[1] "4623" "-4623"
Edit
With this simple modification the function returns a numeric vector.
sub_function <- function(x){
y <- sub("(^-*[[:digit:]]+).*", "\\1", x)
as.numeric(y)
}
There are a few ways to accomplish this, but I like using functions from {tidyverse}:
library(tidyverse)
# Create some dummy data
df <- tibble(targetcol = c("4658(randomcharacters)", "5847(randomcharacters)", "4958(randomcharacters)"))
df <- mutate(df, just_digits = str_extract(targetcol, pattern = "^[[:digit:]]+"))
Output (contents of df):
targetcol just_digits
<chr> <chr>
1 4658(randomcharacters) 4658
2 5847(randomcharacters) 5847
3 4958(randomcharacters) 4958
If you always want to extract numbers from the data, you can use parse_number from readr. It will also return data in numeric form by default.
Using #Rory S' data.
sub_function <- function(x) {
readr::parse_number(x)
}
sub_function(df$targetcol)
#[1] 4658 5847 4958
Related
I'm working in R with strings like the following:
"a1_1;a1_2;a1_5;a1_6;a1_8"
"two1_1;two1_4;two1_5;two1_7"
I need to split these strings into two strings based on the last digit being less than 7 or not. For instance, the desired output for the two strings above would be:
"a1_1;a1_2;a1_5;a1_6" "a1_8"
"two1_1;two1_4;two1_5" "two1_7"
I attempted the following to no avail:
x <- "a1_1;a1_2;a1_5;a1_6;a1_8"
str_split("x", "(\\d<7);")
In an earlier version of the question I was helped by someone that provided the following function, but I don't think it's set up to handle digits both before and after the semicolon in the strings above. I'm trying to modify it but I haven't been able to get it to come out correctly.
f1 <- function(strn) {
strsplit(gsubfn("(;[A-Za-z]+\\d+)", ~ if(readr::parse_number(x) >= 7)
paste0(",", sub(";", "", x)) else x, strn), ",")[[1]]
}
Can anyone help me understand what I'd need to do to make this split as desired?
Splitting and recombining on ;, with a simple regex capture in between.
s <- c("a1_1;a1_2;a1_5;a1_6;a1_8", "two1_1;two1_4;two1_5;two1_7")
sp <- strsplit(s, ";")
lapply(sp,
function(x) {
l <- sub(".*(\\d)$", "\\1", x) < 7
c(paste(x[l], collapse=";"), paste(x[!l], collapse=";"))
}
)
# [[1]]
# [1] "a1_1;a1_2;a1_5;a1_6" "a1_8"
#
# [[2]]
# [1] "two1_1;two1_4;two1_5" "two1_7"
I have a character vector in which long names are used, which will consist of several words connected by delimiters in the form of a dot.
x <- c("Duschekia.fruticosa..Rupr...Pouzar",
"Betula.nana.L.",
"Salix.glauca.L.",
"Salix.jenisseensis..F..Schmidt..Flod.",
"Vaccinium.minus..Lodd...Worosch")
The length of the names is different. But only the first two words of the entire name are important.
My goal is to get names up to 7 symbols: 3 initial symbols from the first two words and a separator in the form of a "dot" between them.
Very close to my request are these examples, but I do not know how to apply these code variations to my case.
R How to remove characters from long column names in a data frame and
how to append names to " column names" of the output data frame in R?
What should I do to get exit names to look like this?
x <- c("Dus.fru",
"Bet.nan",
"Sal.gla",
"Sal.jen",
"Vac.min")
Any help would be appreciated.
You can do the following:
gsub("(\\w{1,3})[^\\.]*\\.(\\w{1,3}).*", "\\1.\\2", x)
# [1] "Dus.fru" "Bet.nan" "Sal.gla" "Sal.jen" "Vac.min"
First we match up to 3 characters (\\w{1,3}), then ignore anything which is not a dot [^\\.]*, match a dot \\. and then again up to 3 characters (\\w{1,3}). Finally anything, that comes after that .*. We then only use the things in the brackets and separate them with a dot \\1.\\2.
Split on dot, substring 3 characters, then paste back together:
sapply(strsplit(x, ".", fixed = TRUE), function(i){
paste(substr(i[ 1 ], 1, 3), substr(i[ 2], 1, 3), sep = ".")
})
# [1] "Dus.fru" "Bet.nan" "Sal.gla" "Sal.jen" "Vac.min"
Here a less elegant solution than kath's, but a bit more easy to read, if you are not an expert in regex.
# Your data
x <- c("Duschekia.fruticosa..Rupr...Pouzar",
"Betula.nana.L.",
"Salix.glauca.L.",
"Salix.jenisseensis..F..Schmidt..Flod.",
"Vaccinium.minus..Lodd...Worosch")
# A function that takes three characters from first two words and merges them
cleaner_fun <- function(ugly_string) {
words <- strsplit(ugly_string, "\\.")[[1]]
short_words <- substr(words, 1, 3)
new_name <- paste(short_words[1:2], collapse = ".")
return(new_name)
}
# Testing function
sapply(x, cleaner_fun)
[1]"Dus.fru" "Bet.nan" "Sal.gla" "Sal.jen" "Vac.min"
I'd like to change the variable names in my data.frame from e.g. "pmm_StartTimev4_E2_C19_1" to "pmm_StartTimev4_E2_C19". So if the name ends with an underscore followed by any number it gets removed.
But I'd like for this to happen only if the variable name has the word "Start" in it.
I've got a muddled up bit of code that doesn't work. Any help would be appreciated!
# Current data frame:
dfbefore <- data.frame(a=c("pmm_StartTimev4_E2_C19_1","pmm_StartTimev4_E2_E2_C1","delivery_C1_C12"),b=c("pmm_StartTo_v4_E2_C19_2","complete_E1_C12_1","pmm_StartTo_v4_E2_C19"))
# Desired data frame:
dfafter <- data.frame(a=c("pmm_StartTimev4_E2_C19","pmm_StartTimev4_E2_E2_C1","delivery_C1_C12"),b=c("pmm_StartTo_v4_E2_C19","complete_E1_C12_1","pmm_StartTo_v4_E2_C19"))
# Current code:
sub((.*{1,}[0-9]*).*","",grep("Start",names(df),value = TRUE)
How about something like this using gsub().
stripcol <- function(x) {
gsub("(.*Start.*)_\\d+$", "\\1", as.character(x))
}
dfnew <- dfbefore
dfnew[] <- lapply(dfbefore, stripcol)
We use the regular expression to look for "Start" and then grab everything but the underscore number at the end. We use lapply to apply the function to all columns.
doit <- function(x){
x <- as.character(x)
if(grepl("Start",x)){
x <- gsub("_([0-9])","",x)
}
return(x)
}
apply(dfbefore,c(1,2),doit)
a b
[1,] "pmm_StartTimev4_E2_C19" "pmm_StartTo_v4_E2_C19"
[2,] "pmm_StartTimev4_E2_E2_C1" "complete_E1_C12_1"
[3,] "delivery_C1_C12" "pmm_StartTo_v4_E2_C19"
We can use sub to capture groups where the 'Start' substring is also present followed by an underscore and one or more numbers. In the replacement, use the backreference of the captured group. As there are multiple columns, use lapply to loop over the columns, apply the sub and assign the output back to the original data
out <- dfbefore
out[] <- lapply(dfbefore, sub,
pattern = "^(.*_Start.*)_\\d+$", replacement ="\\1")
out
dfafter[] <- lapply(dfafter, as.character)
all.equal(out, dfafter, check.attributes = FALSE)
#[1] TRUE
I have an issue about replacing strings with the new ones conditionally.
I put short version of my real problem so far its working however I need a better solution since there are many rows in the real data.
strings <- c("ca_A33","cb_A32","cc_A31","cd_A30")
Basicly I want to replace strings with replace_strings. First item in the strings replaced with the first item in the replace_strings.
replace_strings <- c("A1","A2","A3","A4")
So the final string should look like
final string <- c("ca_A1","cb_A2","cc_A3","cd_A4")
I write some simple function assign_new
assign_new <- function(x){
ifelse(grepl("A33",x),gsub("A33","A1",x),
ifelse(grepl("A32",x),gsub("A32","A2",x),
ifelse(grepl("A31",x),gsub("A31","A3",x),
ifelse(grepl("A30",x),gsub("A30","A4",x),x))))
}
assign_new(strings)
[1] "ca_A1" "cb_A2" "cc_A3" "cd_A4"
Ok it seems we have solution. But lets say if I have A1000 to A1 and want to replace them from A1 to A1000 I need to do 1000 of rows of ifelse statement. How can we tackle that?
If your vectors are ordered to be matched, then you can use:
> paste0(gsub("(.*_)(.*)","\\1", strings ), replace_strings)
[1] "ca_A1" "cb_A2" "cc_A3" "cd_A4"
You can use regmatches.First obtain all the characters that are followed by _ using regexpr then replace as shown below
`regmatches<-`(strings,regexpr("(?<=_).*",strings,perl = T),value=replace_strings)
[1] "ca_A1" "cb_A2" "cc_A3" "cd_A4"
Not the fastests but very tractable and easy to maintain:
for (i in 1:length(strings)) {
strings[i] <- gsub("\\d+$", i, strings[i])
}
"\\d+$" just matches any number at the end of the string.
EDIT: Per #Onyambu's comment, removing map2_chr as paste is a vectorized function.
foo <- function(x, y){
x <- unlist(lapply(strsplit(x, "_"), '[', 1))
paste(x, y, sep = "_"))
}
foo(strings, replace_strings)
with x being strings and y being replace_strings. You first split the strings object at the _ character, and paste with the respective replace_strings object.
EDIT:
For objects where there is no positional relationship you could create a reference table (dataframe, list, etc.) and match your values.
reference_tbl <- data.frame(strings, replace_strings)
foo <- function(x){
y <- reference_tbl$replace_strings[match(x, reference_tbl$strings)]
x <- unlist(lapply(strsplit(x, "_"), '[', 1))
paste(x, y, sep = "_")
}
foo(strings)
Using the dplyr package:
strings <- c("ca_A33","cb_A32","cc_A31","cd_A30")
replace_strings <- c("A1","A2","A3","A4")
df <- data.frame(strings, replace_strings)
df <- mutate(rowwise(df),
strings = gsub("_.*",
paste0("_", replace_strings),
strings)
)
df <- select(df, strings)
Output:
# A tibble: 4 x 1
strings
<chr>
1 ca_A1
2 cb_A2
3 cc_A3
4 cd_A4
yet another way:
mapply(function(x,y) gsub("(\\w\\w_).*",paste0("\\1",y),x),strings,replace_strings,USE.NAMES=FALSE)
# [1] "ca_A1" "cb_A2" "cc_A3" "cd_A4"
I've been trying to remove the white space that I have in a data frame (using R). The data frame is large (>1gb) and has multiple columns that contains white space in every data entry.
Is there a quick way to remove the white space from the whole data frame? I've been trying to do this on a subset of the first 10 rows of data using:
gsub( " ", "", mydata)
This didn't seem to work, although R returned an output which I have been unable to interpret.
str_replace( " ", "", mydata)
R returned 47 warnings and did not remove the white space.
erase_all(mydata, " ")
R returned an error saying 'Error: could not find function "erase_all"'
I would really appreciate some help with this as I've spent the last 24hrs trying to tackle this problem.
Thanks!
A lot of the answers are older, so here in 2019 is a simple dplyr solution that will operate only on the character columns to remove trailing and leading whitespace.
library(dplyr)
library(stringr)
data %>%
mutate_if(is.character, str_trim)
## ===== 2020 edit for dplyr (>= 1.0.0) =====
df %>%
mutate(across(where(is.character), str_trim))
You can switch out the str_trim() function for other ones if you want a different flavor of whitespace removal.
# for example, remove all spaces
df %>%
mutate(across(where(is.character), str_remove_all, pattern = fixed(" ")))
If i understood you correctly then you want to remove all the white spaces from entire data frame, i guess the code which you are using is good for removing spaces in the column names.I think you should try this:
apply(myData, 2, function(x)gsub('\\s+', '',x))
Hope this works.
This will return a matrix however, if you want to change it to data frame then do:
as.data.frame(apply(myData, 2, function(x) gsub('\\s+', '', x)))
EDIT In 2020:
Using lapply and trimws function with both=TRUE can remove leading and trailing spaces but not inside it.Since there was no input data provided by OP, I am adding a dummy example to produce the results.
DATA:
df <- data.frame(val = c(" abc", " kl m", "dfsd "),
val1 = c("klm ", "gdfs", "123"),
num = 1:3,
num1 = 2:4,
stringsAsFactors = FALSE)
#situation: 1 (Using Base R), when we want to remove spaces only at the leading and trailing ends NOT inside the string values, we can use trimws
cols_to_be_rectified <- names(df)[vapply(df, is.character, logical(1))]
df[, cols_to_be_rectified] <- lapply(df[, cols_to_be_rectified], trimws)
# situation: 2 (Using Base R) , when we want to remove spaces at every place in the dataframe in character columns (inside of a string as well as at the leading and trailing ends).
(This was the initial solution proposed using apply, please note a solution using apply seems to work but would be very slow, also the with the question its apparently not very clear if OP really wanted to remove leading/trailing blank or every blank in the data)
cols_to_be_rectified <- names(df)[vapply(df, is.character, logical(1))]
df[, cols_to_be_rectified] <- lapply(df[, cols_to_be_rectified],
function(x) gsub('\\s+', '', x))
## situation: 1 (Using data.table, removing only leading and trailing blanks)
library(data.table)
setDT(df)
cols_to_be_rectified <- names(df)[vapply(df, is.character, logical(1))]
df[, c(cols_to_be_rectified) := lapply(.SD, trimws), .SDcols = cols_to_be_rectified]
Output from situation1:
val val1 num num1
1: abc klm 1 2
2: kl m gdfs 2 3
3: dfsd 123 3 4
## situation: 2 (Using data.table, removing every blank inside as well as leading/trailing blanks)
cols_to_be_rectified <- names(df)[vapply(df, is.character, logical(1))]
df[, c(cols_to_be_rectified) := lapply(.SD, function(x) gsub('\\s+', '', x)), .SDcols = cols_to_be_rectified]
Output from situation2:
val val1 num num1
1: abc klm 1 2
2: klm gdfs 2 3
3: dfsd 123 3 4
Note the difference between the outputs of both situation, In row number 2: you can see that, with trimws we can remove leading and trailing blanks, but with regex solution we are able to remove every blank(s).
I hope this helps , Thanks
One possibility involving just dplyr could be:
data %>%
mutate_if(is.character, trimws)
Or considering that all variables are of class character:
data %>%
mutate_all(trimws)
Since dplyr 1.0.0 (only strings):
data %>%
mutate(across(where(is.character), trimws))
Or if all columns are strings:
data %>%
mutate(across(everything(), trimws))
Picking up on Fremzy and the comment from Stamper, this is now my handy routine for cleaning up whitespace in data:
df <- data.frame(lapply(df, trimws), stringsAsFactors = FALSE)
As others have noted this changes all types to character. In my work, I first determine the types available in the original and conversions required. After trimming, I re-apply the types needed.
If your original types are OK, apply the solution from MarkusN below https://stackoverflow.com/a/37815274/2200542
Those working with Excel files may wish to explore the readxl package which defaults to trim_ws = TRUE when reading.
Picking up on Fremzy and Mielniczuk, I came to the following solution:
data.frame(lapply(df, function(x) if(class(x)=="character") trimws(x) else(x)), stringsAsFactors=F)
It works for mixed numeric/charactert dataframes manipulates only character-columns.
You could use trimws function in R 3.2 on all the columns.
myData[,c(1)]=trimws(myData[,c(1)])
You can loop this for all the columns in your dataset. It has good performance with large datasets as well.
If you're dealing with large data sets like this, you could really benefit form the speed of data.table.
library(data.table)
setDT(df)
for (j in names(df)) set(df, j = j, value = df[[trimws(j)]])
I would expect this to be the fastest solution. This line of code uses the set operator of data.table, which loops over columns really fast. There is a nice explanation here: Fast looping with set.
R is simply not the right tool for such file size. However have 2 options :
Use ffdply and ff base
Use ff and ffbase packages:
library(ff)
library(ffabse)
x <- read.csv.ffdf(file=your_file,header=TRUE, VERBOSE=TRUE,
first.rows=1e4, next.rows=5e4)
x$split = as.ff(rep(seq(splits),each=nrow(x)/splits))
ffdfdply( x, x$split , BATCHBYTES=0,function(myData)
apply(myData,2,function(x)gsub('\\s+', '',x))
Use sed (my preference)
sed -ir "s/(\S)\s+(/S)/\1\2/g;s/^\s+//;s/\s+$//" your_file
If you want to maintain the variable classes in your data.frame - you should know that using apply will clobber them because it outputs a matrix where all variables are converted to either character or numeric. Building upon the code of Fremzy and Anthony Simon Mielniczuk you can loop through the columns of your data.frame and trim the white space off only columns of class factor or character (and maintain your data classes):
for (i in names(mydata)) {
if(class(mydata[, i]) %in% c("factor", "character")){
mydata[, i] <- trimws(mydata[, i])
}
}
I think that a simple approach with sapply, also works, given a df like:
dat<-data.frame(S=LETTERS[1:10],
M=LETTERS[11:20],
X=c(rep("A:A",3),"?","A:A ",rep("G:G",5)),
Y=c(rep("T:T",4),"T:T ",rep("C:C",5)),
Z=c(rep("T:T",4),"T:T ",rep("C:C",5)),
N=c(1:3,'4 ','5 ',6:10),
stringsAsFactors = FALSE)
You will notice that dat$N is going to become class character due to '4 ' & '5 ' (you can check with class(dat$N))
To get rid of the spaces on the numeic column simply convert to numeric with as.numeric or as.integer.
dat$N<-as.numeric(dat$N)
If you want to remove all the spaces, do:
dat.b<-as.data.frame(sapply(dat,trimws),stringsAsFactors = FALSE)
And again use as.numeric on col N (ause sapply will convert it to character)
dat.b$N<-as.numeric(dat.b$N)