在数据帧中用 NA 替换字符值

小开

最佳答案

This:

df[df == "foo"] <- NA

小开

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"))

小开

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"
}

小开

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.

小开

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')

小开

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

小开

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