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$N)
Related
I am a beginner in R and while trying to make some exercises I got stuck in one of them. My data.frame is as follow:
LanguageWorkedNow LanguageNextYear
Java; PHP Java; C++; SQL
C;C++;JavaScript; JavaScript; C; SQL
And I need to know the variables which are in LanguageNextYear and are not in LanguageWorkedNow, to set a list with the different ones.
Sorry if the question is duplicated, I'm quite new here and tried to find it, but with no success.
base R
Idea: mapply setdiff on strsplitted NextYear and WorkedNow, and then paste it using collapse=";":
df$New <- with(df, {
a <- mapply(setdiff, strsplit(NextYear, ";"), strsplit(WorkedNow, ";"), SIMPLIFY = FALSE)
sapply(a, paste, collapse=";")
})
# SIMPLIFY = FALSE is needed in a general case, it doesn't
# affect the output in the example case
# Or if you use Map instead of mapply, that is the default, so
# it could also be...
df$New <- with(df,
sapply(Map(setdiff, strsplit(NextYear, ";"), strsplit(WorkedNow, ";")),
paste, collapse=";"))
data
df <- read.table(text = "WorkedNow NextYear
Java;PHP Java;C++;SQL
C;C++;JavaScript JavaScript;C;SQL
", header=TRUE, stringsAsFactors=FALSE)
Here's a solution using purrr package:
df = read.table(text = "
LanguageWorkedNow LanguageNextYear
Java;PHP Java;C++;SQL
C;C++;JavaScript JavaScript;C;SQL
", header=T, stringsAsFactors=F)
library(purrr)
df$New = map2_chr(df$LanguageWorkedNow,
df$LanguageNextYear,
~{x1 = unlist(strsplit(.x, split=";"))
x2 = unlist(strsplit(.y, split=";"))
paste0(x2[!x2%in%x1], collapse = ";")})
df
# LanguageWorkedNow LanguageNextYear New
# 1 Java;PHP Java;C++;SQL C++;SQL
# 2 C;C++;JavaScript JavaScript;C;SQL SQL
For each row you get your columns and you create vectors of values (separated by ;). Then you check which values of NextYear vector don't exist in WorkedNow vector and you create a string based on / combining those values.
The map function family will help you apply your logic / function to each row. In our case we use map2_chr as we have two inputs (your two columns) and we excpet a string / character output.
I have a very large dataset with some columns formatted as currency, some numeric, some character. When reading in the data all currency columns are identified as factor and I need to convert them to numeric. The dataset it too wide to manually identify the columns. I am trying to find a programmatic way to identify if a column contains currency data (ex. starts with '$') and then pass that list of columns to be cleaned.
name <- c('john','carl', 'hank')
salary <- c('$23,456.33','$45,677.43','$76,234.88')
emp_data <- data.frame(name,salary)
clean <- function(ttt){
as.numeric(gsub('[^a-zA-z0-9.]','', ttt))
}
sapply(emp_data, clean)
The issue in this example is that this sapply works on all columns resulting in the name column being replaced with NA. I need a way to programmatically identify just the columns that the clean function needs to be applied to.. in this example salary.
Using dplyr and stringr packages, you can use mutate_if to identify columns that have any string starting with a $ and then change the accordingly.
library(dplyr)
library(stringr)
emp_data %>%
mutate_if(~any(str_detect(., '^\\$'), na.rm = TRUE),
~as.numeric(str_replace_all(., '[$,]', '')))
Taking advantage of the powerful parsers the readr package offers out of the box:
my_parser <- function(col) {
# Try first with parse_number that handles currencies automatically quite well
res <- suppressWarnings(readr::parse_number(col))
if (is.null(attr(res, "problems", exact = TRUE))) {
res
} else {
# If parse_number fails, fall back on parse_guess
readr::parse_guess(col)
# Alternatively, we could simply return col without further parsing attempt
}
}
library(dplyr)
emp_data %>%
mutate(foo = "USD13.4",
bar = "£37") %>%
mutate_all(my_parser)
# name salary foo bar
# 1 john 23456.33 13.4 37
# 2 carl 45677.43 13.4 37
# 3 hank 76234.88 13.4 37
A base R option is to use startsWith to detect the dollar columns, and gsub to remove "$" and "," from the columns.
doll_cols <- sapply(emp_data, function(x) any(startsWith(as.character(x), '$')))
emp_data[doll_cols] <- lapply(emp_data[doll_cols],
function(x) as.numeric(gsub('\\$|,', '', x)))
I am kind of new to R and I want to apply the gsub function in columns 6 to 12 in my data frame called x. However, I am kind of stuck. I started using:
gsub("\\.", "",x[,c(6:12)])
But then it returned only a few rows.
Then, I tried to use this:
x1<-apply(x,c(6:12),function(x) gsub("\\.", "",x))
But I got the following error:
Error in if (d2 == 0L) { : missing value where TRUE/FALSE needed
Also tried this:
for (i in x[,c(6:12)])
{a<-data.frame(gsub("\\.", "",i))}
Does anybody have a tip or a solution?
It would also be great if someone showed me how to use an apply function and for.
Here is another solution. It returns all the columns of the original dataframe
library(dplyr)
mutate_at(x, 6:12, gsub("\\.", "", .))
Here is a solution in a data.table way. It worth considering when a table is big and time is a critical factor.
library(data.table) # load library
x = as.data.table(x) # convert data.frame into data.table
cols = names(x)[6:12] # define which columns to work with
x[ , (cols) := lapply(.SD, function(x) {gsub("\\.", "", x)}), .SDcols = cols] # replace
x # enjoy results
Adding the solution using apply from the comments, as this one really helped me:
x1 <- apply(x[,6:12], 2, function(x) gsub("\\.", "",x))
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),]
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),]