替换NA，否则 <NA> 在数据框的列中有其他东西

Question

I have read what seem to be related posts, but am evidently much too much a Noob to understand or make anything work...

> df
  ID Area Address
1 NA    1    lane
2 11   NA    road
3 12    2    blvd
4 13    5    <NA>

> str(df)
'data.frame':   4 obs. of  3 variables:
 $ ID     : int  NA 11 12 13
 $ Area   : int  1 NA 2 5
 $ Address: Factor w/ 3 levels "blvd","lane",..: 2 3 1 NA

I want to be able -- not just for the data frame above, but for larger data frames having many more rows and many more columns -- to replace in whatever columns I choose (which I reference by column names) all occurences of

<NA>

with an element of my choosing from

<NA> , NA, "foo", "", 0

and whatever performs the replacement does not break or give error(s) when there is no

<NA>

to replace. Likewise, I want to perform an analogous replacement for

NA

in whatever columns I choose without breakage or errors.

If there are technical reasons as to why I cannot do what I propose, then what can I do to come as close as possible to the above (while sticking to data frames -- converting to and fro with something else is ok if the answer is very explicit as to how exactly to manage the conversions -- and preserving factors in the sense that, for example, the Address column is a factor so after the replacement it should still be a factor).

I expect there are technical reasons as to why I cannot do what I propose (I am confused to the point of asking the impossible), so I am hoping to come as close as reality permits, and that some kind soul will explain the extent to which I can come close to the above as well as how exactly to get however close is possible.

Please help (do not assume I can possibly understand without a detailed explicit answer).

Thanks

Answer 1

A character string cannot be inserted into a numeric or integer vector without making the entire vector character but we can insert a zero in place of NA and we do that below. Also we insert fill having default "foo" as a new level in place of NA for factors of the sort shown in the question.

1) Looking at df.orig shown reproducibly at the end it has integer and factor columns and the following works for those as well as numeric columns which are double. For numeric (double and integer) we assign 0L so that integer columns are not changed to double. The 0L will automatically be coerced to double for double columns. For factors having NA values, we add the NA as the last level and then change its label to fill . We also check if there are any NA levels and, if so, replace them with fill . One would not ordinarily find both situations. You will need to extend the code below if it is necessary to convert other classes not shown in the question.

df <- df.orig

# numeric (integer and double)
isNum <- sapply(df, is.numeric)
na2zero <- function(v, ...) replace(v, is.na(v), 0L)
df[isNum] <- lapply(df[isNum], na2zero)

# factor
isFactor <- sapply(df, is.factor)
na2fill <- function(v, fill = "foo", ...) { 
      # handle NA values
      if (any(is.na(v))) {
         v <- addNA(v)
         levels(v)[nlevels(v)] <- fill
      }
      # handle NA levels
      if (any(is.na(levels(v)))) levels[is.na(levels(v))] <- fill
      v 
}
df[isFactor] <- lapply(df[isFactor], na2fill)

giving:

> df
  ID Area Address
1  0    1    lane
2 11    0    road
3 12    2    blvd
4 13    5     foo

2) Alternately, we could use S3 to do it more compactly where na2zero and na2fill are from (1).

rmNA <- function(v, ...) UseMethod("rmNA")
rmNA.numeric <- na2zero
rmNA.factor <- na2fill
rmNA.default <- function(v, ...) v # do not process other classes

df <- df.orig
df[] <- lapply(df, rmNA)

Note: df in reproducible form is:

df.orig <- 
structure(list(ID = c(NA, 11L, 12L, 13L), Area = c(1L, NA, 2L, 
5L), Address = structure(c(2L, 3L, 1L, NA), .Label = c("blvd", 
"lane", "road"), class = "factor")), .Names = c("ID", "Area", 
"Address"), class = "data.frame", row.names = c("1", "2", "3", 
"4"))

Answer 2

If your data frame is named df like you show in your question just type:

df[is.na(df)] <- 0

Just be sur of the name of your data frame and if it's not df just replace df with the name how you asign to your data frame.