I 开发者_JAVA百科have a data frame containing (in random places) a character value (say "foo"
) that I want to replace with a NA
.
What's the best way to do so across the whole data frame?
This:
df[df == "foo"] <- NA
One way to nip this in the bud is to convert that character to NA when you read the data in in the first place.
df <- read.csv("file.csv", na.strings = c("foo", "bar"))
Using dplyr::na_if
, you can replace specific values with NA
. In this case, that would be "foo"
.
library(dplyr)
set.seed(1234)
df <- data.frame(
id = 1:6,
x = sample(c("a", "b", "foo"), 6, replace = T),
y = sample(c("c", "d", "foo"), 6, replace = T),
z = sample(c("e", "f", "foo"), 6, replace = T),
stringsAsFactors = F
)
df
#> id x y z
#> 1 1 a c e
#> 2 2 b c foo
#> 3 3 b d e
#> 4 4 b d foo
#> 5 5 foo foo e
#> 6 6 b d e
na_if(df$x, "foo")
#> [1] "a" "b" "b" "b" NA "b"
If you need to do this for multiple columns, you can pass "foo"
through from mutate
with across
(updated for dplyr
v1.0.0+).
df %>%
mutate(across(c(x, y, z), na_if, "foo"))
#> id x y z
#> 1 1 a c e
#> 2 2 b c <NA>
#> 3 3 b d e
#> 4 4 b d <NA>
#> 5 5 <NA> <NA> e
#> 6 6 b d e
Another option is is.na<-
:
is.na(df) <- df == "foo"
Note that its use may seem a bit counter-intuitive, but it actually assigns NA
values to df
at the index on the right hand side.
This could be done with dplyr::mutate_all()
and replace
:
library(dplyr)
df <- data_frame(a = c('foo', 2, 3), b = c(1, 'foo', 3), c = c(1,2,'foobar'), d = c(1, 2, 3))
> df
# A tibble: 3 x 4
a b c d
<chr> <chr> <chr> <dbl>
1 foo 1 1 1
2 2 foo 2 2
3 3 3 foobar 3
df <- mutate_all(df, funs(replace(., .=='foo', NA)))
> df
# A tibble: 3 x 4
a b c d
<chr> <chr> <chr> <dbl>
1 <NA> 1 1 1
2 2 <NA> 2 2
3 3 3 foobar 3
Another dplyr
option is:
df <- na_if(df, 'foo')
Assuming you do not know the column names or have large number of columns to select, is.character()
might be of use.
df <- data.frame(
id = 1:6,
x = sample(c("a", "b", "foo"), 6, replace = T),
y = sample(c("c", "d", "foo"), 6, replace = T),
z = sample(c("e", "f", "foo"), 6, replace = T),
stringsAsFactors = F
)
df
# id x y z
# 1 1 b d e
# 2 2 a foo foo
# 3 3 a d foo
# 4 4 b foo foo
# 5 5 foo foo e
# 6 6 foo foo f
df %>%
mutate_if(is.character, list(~na_if(., "foo")))
# id x y z
# 1 1 b d e
# 2 2 a <NA> <NA>
# 3 3 a d <NA>
# 4 4 b <NA> <NA>
# 5 5 <NA> <NA> e
# 6 6 <NA> <NA> f
One alternate way to solve is below:
for (i in 1:ncol(DF)){
DF[which(DF[,i]==""),columnIndex]<-"ALL"
FinalData[which(is.na(FinalData[,columnIndex])),columnIndex]<-"ALL"
}
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