Function to Change Blanks to Na

Change the Blank Cells to NA

I'm assuming you are talking about row 5 column "sex." It could be the case that in the data2.csv file, the cell contains a space and hence is not considered empty by R.

Also, I noticed that in row 5 columns "axles" and "door", the original values read from data2.csv are string "NA". You probably want to treat those as na.strings as well. To do this,

dat2 <- read.csv("data2.csv", header=T, na.strings=c("","NA"))

EDIT:

I downloaded your data2.csv. Yes, there is a space in row 5 column "sex". So you want

na.strings=c(""," ","NA")

Function to change blanks to NA

You can directly index fields that match a logical criterion. So you can just write:

df[is_empty(df)] = NA

Where is_empty is your comparison, e.g. df == "":

df[df == ""] = NA

But note that is.null(df) won’t work, and would be weird anyway¹. I would advise against merging the logic for columns of different types, though! Instead, handle them separately.

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¹ You’ll almost never encounter NULL inside a table since that only works if the underlying vector is a list. You can create matrices and data.frames with this constraint, but then is.null(df) will never be TRUE because the NULL values are wrapped inside the list).

How to replace empty string with NA in R dataframe?

I'm not sure why df[df==""]<-NA would not have worked for OP. Let's take a sample data.frame and investigate options.

Option#1: Base-R

df[df==""]<-NA

df
#    One  Two Three Four
# 1    A    A  <NA>  AAA
# 2 <NA>    B    BA <NA>
# 3    C <NA>    CC  CCC

Option#2: dplyr::mutate_all and na_if. Or mutate_if if the data frame has multiple types of columns

library(dplyr)

mutate_all(df, list(~na_if(.,"")))

OR

#if data frame other types of character Then
df %>% mutate_if(is.character, list(~na_if(.,""))) 

#    One  Two Three Four
# 1    A    A  <NA>  AAA
# 2 <NA>    B    BA <NA>
# 3    C <NA>    CC  CCC

Toy Data:

df <- data.frame(One=c("A","","C"), 
                 Two=c("A","B",""), 
                 Three=c("","BA","CC"), 
                 Four=c("AAA","","CCC"), 
                 stringsAsFactors = FALSE)

df
#   One Two Three Four
# 1   A   A        AAA
# 2       B    BA     
# 3   C        CC  CCC

Replace blank with NA in R

What type of variable are we talking about? Numeric? Character?
A better formulated question makes it easier to give a better answer.

This could help:

DT[DT == ""] <- NA

Do not try so hard. R should be fun!

How to replace blank strings with NA?

To show that the code works:

data <- data.frame( col1= c("", letters[1:4]), col2=c(letters[1:4], ""))
 is.na(data) <- data==''
 data
 #  col1 col2
 #1 <NA>    a
 #2   a    b
 #3    b    c
 #4    c    d
 #5    d <NA>

Suppose, if you have '' along with spaces ' ', this won't work

 data <- data.frame( col1= c("", letters[1:4]), col2=c(letters[1:4], " "))
 data1 <- data
 is.na(data) <- data==''
  data
  col1 col2
 #1 <NA>    a
 #2    a    b
 #3    b    c
 #4    c    d
 #5    d     

In such cases, you could use str_trim

  library(stringr)
  data1[] <- lapply(data1, str_trim)
  is.na(data1) <- data1==''
  data1
  #  col1 col2
  #1 <NA>    a
  #2    a    b
  #3    b    c
  #4    c    d
  #5    d <NA>

How to replace NA's with blank value?

If your column is of type double (numbers), you can't replace NAs (which is the R internal for missings) by a character string. And "" IS a character string even though you think it's empty, but it is not.

So you need to choose: converting you whole column to type character or leave the missings as NA.

EDIT:

• If you really want to covnert your numeric column to character, you can just use as.character(MYCOLUMN). But I think what you really want is:
• Telling your exporting function how to treat NA'S, which is easy, e.g. write.csv(df, na = ""). Also check the help function with ?write.csv.

Fast way to replace all blanks with NA in R data.table

Here's probably the generic data.table way of doing this. I'm also going to use your regex which handles several types of blanks (I havn't seen other answers doing this). You probably shouldn't run this over all your columns rather only over the factor or character ones, because other classes won't accept blank values.

For factors

indx <- which(sapply(data, is.factor))
for (j in indx) set(data, i = grep("^$|^ $", data[[j]]), j = j, value = NA_integer_) 

For characters

indx2 <- which(sapply(data, is.character)) 
for (j in indx2) set(data, i = grep("^$|^ $", data[[j]]), j = j, value = NA_character_)

How can I replace empty cells with NA in R?

The result you get from readHTMLTable is giving you a list of two tables, so you need to work on each list element, which can be done using lapply

table <- lapply(table, function(x){
     x[x == ""] <- NA
     return(x)
})

table$team_stats
Player  PF  Yds  Ply  Y/P TO FL 1stD  Cmp Att  Yds TD Int NY/A 1stD Att  Yds TD Y/A 1stD  Pen  Yds 1stPy
1      Team Stats 442 6268 1021  6.1 25 14  350  339 483 4302 35  11  8.1  209 493 1966 14 4.0  124  109  922    17
2      Opp. Stats 253 4618  979  4.7 37 16  283  316 564 3235 15  21  5.3  178 372 1383  9 3.7   76   75  581    29
3 Lg Rank Offense   1    1 <NA> <NA>  2 10    1 <NA>  20    2  1   1    1 <NA>  13   10 12  13 <NA> <NA> <NA>  <NA>
4 Lg Rank Defense   3    4 <NA> <NA> 11  9    9 <NA>  25   11  3   9    5 <NA>   1    3  3   8 <NA> <NA> <NA>  <NA>

How to replace empty strings in a dataframe with NA (missing value) not NA string

By specifying just NA, according to ?NA -"NA is a logical constant of length 1 which contains a missing value."

The class can be checked

class(NA)
#[1] "logical"
class(NA_character_) 
#[1] "character"

and both of them is identified by standard functions such as is.na

is.na(NA)
#[1] TRUE
is.na(NA_character_)
#[1] TRUE

The if_else is type sensitive, so instead of specifying as NA which returns a logical output, it can specified as either NA_real_, NA_integer_, NA_character_ depending on the type of the 'boat' column. Assuming that the 'boat' is character class, we may need NA_character_

titanic %>% 
       mutate(boat = if_else(boat=="", NA_character_ ,boat))

Replace strings containing only blanks with NA

One dplyr possibility could be:

df %>%
 mutate_all(~ ifelse(nchar(trimws(.)) == 0, NA_character_, .))

        Q1            Q2
1 Test test Sample sample
2      Test          <NA>
3      <NA>        Sample
4      <NA>        Sample

Or the same with base R:

df[] <- lapply(df, function(x) ifelse(nchar(trimws(x)) == 0, NA_character_, x))

Or:

df %>%
 mutate_all(~ trimws(.)) %>%
 na_if(., "")

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