I'm having issues removing just the right amount of information from the following data:
18,14,17,2,9,8
17,17,17,14
18,14,17,2,1,1,1,1,9,8,1,1,1
I'm applying !duplicate in order to remove the duplicates.
SplitFunction <- function(x) {
b <- unlist(strsplit(x, '[,]'))
c <- b[!duplicated(b)]
return(paste(c, collapse=","))
}
I'm having issues removing only consecutive duplicates. The result below is what I'm getting.
18,14,17,2,9,8
17,14
18,14,17,2,1,9,8
The data below is what I want to obtain.
18,14,17,2,9,8
17,14
18,14,17,2,1,9,8,1
Can you suggest a way to perform this? Ideally a vectorized approach...
Thanks,
Miguel
you can use rle function to sovle this question.
xx <- c("18,14,17,2,9,8","17,17,17,14","18,14,17,2,1,1,1,1,9,8,1,1,1")
zz <- strsplit(xx,",")
sapply(zz,function(x) rle(x)$value)
And you can refer to this link.
How to remove/collapse consecutive duplicate values in sequence in R?
We can use rle
sapply(strsplit(x, ','), function(x) paste(inverse.rle(within.list(rle(x),
lengths <- rep(1, length(lengths)))), collapse=","))
#[1] "18,14,17,2,9,8" "17,14" "18,14,17,2,1,9,8,1"
data
x <- c('18,14,17,2,9,8', '17,17,17,14', '18,14,17,2,1,1,1,1,9,8,1,1,1')
Great rle-answers. This is just to add an alternative without rle. This gives a list of numeric vectors but can of course easily expanded to return strings:
numbers <- c("18,14,17,2,9,8", "17,17,17,14", "14,17,18,2,9,8,1", "18,14,17,11,8,9,8,8,22,13,6", "14,17,2,9,8", "18,14,17,2,1,1,1,1,1,1,1,1,9,8,1,1,1,1")
result <- sapply(strsplit(numbers, ","), function(x) x[x!=c(x[-1],Inf)])
print(result)
Related
This question already has answers here:
Extracting numbers from vectors of strings
(12 answers)
Closed 1 year ago.
relatively new to R, need help with applying a regex-based substitution.
I have a data frame in one column of which I have a sequence of digits (my values of interest) followed by a string of all sorts of characters.
Example:
4623(randomcharacters)
I need to remove everything after the initial digits to continue working with the values. My idea was to use gsub to remove the non-digit characters by positive lookbehind.
The code I have is:
sub_function <- function() {
gsub("?<=[[:digit:]].", " ", fixed = T)
}
data_frame$`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$targetcol)
#[1] 4658 5847 4958
Eg : data has to follow the convention xxxx-xx-xx-xxx-xxx-xxx-xxx-xxx
the right data format is 7448-06-93-030-001 or 7448-06-93-030-001-010-030-060
but not 7448-060-030-070.Hope I made some sense
Assuming that the "correct format" means the correct number of numeric digits between dashes, here is one solution:
test_format <- function(x) {
#get number of characters of each bunch of digits
x <- paste0("-", x, "-")
dash_pos <- unlist(gregexpr("-", x))
n <- length(dash_pos)
lens <- dash_pos[2:n] - dash_pos[1:(n-1)] - 1
#check that this matches the correct convention
correct_lens <- c(4,2,2,3,3,3,3,3)
isTRUE(all.equal(lens, correct_lens[1:(n-1)]))
}
test_format("7448-06-93-030-001") #should be true
test_format("7448-06-93-030-001-010-030-060") #should be true
test_format("7448-060-030-070") #should be false
This regular expression should work, assuming you want the first pattern of x's
\d{4}-(\d{2}-){2}(\d{3}-){4}\d{3}
https://regular-expressions.mobi/rlanguage.html?wlr=1
Are you looking for a blanket gsub? These two work for those two scenarios. You could use an ifelse to determine which one to use.
df <- c("74-486993-030-001")
df <- gsub("-", "", df)
dfa <- gsub("(\\d{4})(\\d{2})(\\d{2})(\\d{3})(\\d{3})$", "\\1-\\2-\\3-\\4-\\5", df)
"7448-69-93-030-001"
df2 <- c("74480693-030-00-10-10-030-060")
df2 <- gsub("-", "", df2)
dfb <- gsub("(\\d{4})(\\d{2})(\\d{2})(\\d{3})(\\d{3})(\\d{3})(\\d{3})(\\d{3})$", "\\1-\\2-\\3-\\4-\\5-\\6-\\7-\\8", df2)
"7448-06-93-030-001-010-030-060"
I have an issue about replacing strings with the new ones conditionally.
I put short version of my real problem so far its working however I need a better solution since there are many rows in the real data.
strings <- c("ca_A33","cb_A32","cc_A31","cd_A30")
Basicly I want to replace strings with replace_strings. First item in the strings replaced with the first item in the replace_strings.
replace_strings <- c("A1","A2","A3","A4")
So the final string should look like
final string <- c("ca_A1","cb_A2","cc_A3","cd_A4")
I write some simple function assign_new
assign_new <- function(x){
ifelse(grepl("A33",x),gsub("A33","A1",x),
ifelse(grepl("A32",x),gsub("A32","A2",x),
ifelse(grepl("A31",x),gsub("A31","A3",x),
ifelse(grepl("A30",x),gsub("A30","A4",x),x))))
}
assign_new(strings)
[1] "ca_A1" "cb_A2" "cc_A3" "cd_A4"
Ok it seems we have solution. But lets say if I have A1000 to A1 and want to replace them from A1 to A1000 I need to do 1000 of rows of ifelse statement. How can we tackle that?
If your vectors are ordered to be matched, then you can use:
> paste0(gsub("(.*_)(.*)","\\1", strings ), replace_strings)
[1] "ca_A1" "cb_A2" "cc_A3" "cd_A4"
You can use regmatches.First obtain all the characters that are followed by _ using regexpr then replace as shown below
`regmatches<-`(strings,regexpr("(?<=_).*",strings,perl = T),value=replace_strings)
[1] "ca_A1" "cb_A2" "cc_A3" "cd_A4"
Not the fastests but very tractable and easy to maintain:
for (i in 1:length(strings)) {
strings[i] <- gsub("\\d+$", 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)
I am working on a large dataset, with some rows with NAs and others with blanks:
df <- data.frame(ID = c(1:7),
home_pc = c("","CB4 2DT", "NE5 7TH", "BY5 8IB", "DH4 6PB","MP9 7GH","KN4 5GH"),
start_pc = c(NA,"Home", "FC5 7YH","Home", "CB3 5TH", "BV6 5PB",NA),
end_pc = c(NA,"CB5 4FG","Home","","Home","",NA))
How do I remove the NAs and blanks in one go (in the start_pc and end_pc columns)? I have in the past used:
df<- df[-which(is.na(df$start_pc)), ]
... to remove the NAs - is there a similar command to remove the blanks?
df[!(is.na(df$start_pc) | df$start_pc==""), ]
It is the same construct - simply test for empty strings rather than NA:
Try this:
df <- df[-which(df$start_pc == ""), ]
In fact, looking at your code, you don't need the which, but use the negation instead, so you can simplify it to:
df <- df[!(df$start_pc == ""), ]
df <- df[!is.na(df$start_pc), ]
And, of course, you can combine these two statements as follows:
df <- df[!(df$start_pc == "" | is.na(df$start_pc)), ]
And simplify it even further with with:
df <- with(df, df[!(start_pc == "" | is.na(start_pc)), ])
You can also test for non-zero string length using nzchar.
df <- with(df, df[!(nzchar(start_pc) | is.na(start_pc)), ])
Disclaimer: I didn't test any of this code. Please let me know if there are syntax errors anywhere
An elegant solution with dplyr would be:
df %>%
# recode empty strings "" by NAs
na_if("") %>%
# remove NAs
na.omit
Alternative solution can be to remove the rows with blanks in one variable:
df <- subset(df, VAR != "")
An easy approach would be making all the blank cells NA and only keeping complete cases. You might also look for na.omit examples. It is a widely discussed topic.
df[df==""]<-NA
df<-df[complete.cases(df),]