Removing an element based on its content?

Question

df.cleaned <- df[-which(str_detect(df, "Not found")),]

"df" refers to a data frame, that consists of multiple columns and rows. A lot of the elements in this data frame have certain character words in them.

What I'm looking to do, is to remove all those values that contain the words "Not found" either as the whole element value, or part of it.

So far, the above command is what I've come up with, with the stringr package. However, this command seems to remove entire rows. I don't want to remove the entire row, I simply want to remove that specific element that contains "Not found".

Answer 1

How to get the behavior:

toy[toy == "Not found"] <- ""
toy
#    x y z  n
# 1  m   f  6
# 2  z t a  3
# 3    m    4
# 4    j    9
# 5  e      5
# 6  f n k  2
# 7  q f p  1
# 8      n  8
# 9  n k h  7
# 10 d u l 10

For matching vs. equality, you could try this. I'm not sure if it offers performance improvements over the @r2evans approach. EDIT: apparently, as @r2evans explains in the comments, the same conversion is done behind the scenes. In which case, it doesn't look as clean as the equality solution, but shouldn't drop in performance due to the conversion:

toy[matrix(grepl("Not found", as.matrix(toy)), nrow(toy))] <- ""
toy
#    x y z  n
# 1  m   f  6
# 2  z t a  3
# 3    m    4
# 4    j    9
# 5  e      5
# 6  f n k  2
# 7  q f p  1
# 8      n  8
# 9  n k h  7
# 10 d u l 10

Create the data:

toy <- data.frame(x = sample(letters, 10), y = sample(letters, 10), z = sample(letters, 10), stringsAsFactors = FALSE)
for (col in seq_along(toy)) toy[[col]][sample(10, 3)] <- "Not found"  
toy$n <- sample(10)
toy
#            x         y         z  n
# 1          m Not found         f  6
# 2          z         t         a  3
# 3  Not found         m Not found  4
# 4  Not found         j Not found  9
# 5          e Not found Not found  5
# 6          f         n         k  2
# 7          q         f         p  1
# 8  Not found Not found         n  8
# 9          n         k         h  7
# 10         d         u         l 10

Answer 2

It's often advantageous to write a simple function up front that does what you want, and then know how to apply that function to all of your columns.

For instance:

replace_notfound <- function(s, newstr="") s[grepl("Not found", s)] <- newstr

Now, let's apply that function to each column of your data:

# I'm assuming you want stringsAsFactors=FALSE
df.cleaned <- as.data.frame(lapply(df, replace_notfound), stringsAsFactors=FALSE)

It's not always the case that all columns of a frame are character, so you might want to conditionally do this:

ischr <- sapply(df, is.character)
df.cleaned <- df # just a copy
df.cleaned[ischr] <- lapply(df.cleaned[ischr], replace_notfound)

Answer 3

Your thought was in right direction. You need to try to apply it for each item. One option could be to use sapply. Check every item with str_detect and replace with "" or NA otherwise just return value of item.

library(stringr)
df.clean <- as.dataframe(sapply(df, 
                   function(x)ifelse(str_detect(x, "Not found"), "",x)))
df.clean
# 
# A    B
# 1  A Good
# 2  B     
# 3  C Good
# 4  D     
# 5  E Good
# 6  A     
# 7  B Good
# 8  C     
# 9  D Good
# 10 E   

Data

    df <- data.frame(A = rep(c("A", "B", "C", "D", "E"), 2), 
                     B = rep(c("Good","Bad with Not found"),5),
                      stringsAsFactors = FALSE)
df
# A                  B
# 1  A               Good
# 2  B Bad with Not found
# 3  C               Good
# 4  D Bad with Not found
# 5  E               Good
# 6  A Bad with Not found
# 7  B               Good
# 8  C Bad with Not found
# 9  D               Good
# 10 E Bad with Not found