[r] How to replace NA values in a table for selected columns

There are a lot of posts about replacing NA values. I am aware that one could replace NAs in the following table/frame with the following:

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

But, what if I want to restrict it to only certain columns? Let's me show you an example.

First, let's start with a dataset.

set.seed(1234)
x <- data.frame(a=sample(c(1,2,NA), 10, replace=T),
                b=sample(c(1,2,NA), 10, replace=T), 
                c=sample(c(1:5,NA), 10, replace=T))

Which gives:

    a  b  c
1   1 NA  2
2   2  2  2
3   2  1  1
4   2 NA  1
5  NA  1  2
6   2 NA  5
7   1  1  4
8   1  1 NA
9   2  1  5
10  2  1  1

Ok, so I only want to restrict the replacement to columns 'a' and 'b'. My attempt was:

x[is.na(x), 1:2]<-0

and:

x[is.na(x[1:2])]<-0

Which does not work.

My data.table attempt, where y<-data.table(x), was obviously never going to work:

y[is.na(y[,list(a,b)]), ]

I want to pass columns inside the is.na argument but that obviously wouldn't work.

I would like to do this in a data.frame and a data.table. My end goal is to recode the 1:2 to 0:1 in 'a' and 'b' while keeping 'c' the way it is, since it is not a logical variable. I have a bunch of columns so I don't want to do it one by one. And, I'd just like to know how to do this.

Do you have any suggestions?

this works fine for me

DataTable DT = new DataTable();

DT = DT.AsEnumerable().Select(R =>
{
      R["Campo1"] = valor;
      return (R);
}).ToArray().CopyToDataTable();

For a specific column, there is an alternative with sapply

DF <- data.frame(A = letters[1:5],
             B = letters[6:10],
             C = c(2, 5, NA, 8, NA))

DF_NEW <- sapply(seq(1, nrow(DF)),
                    function(i) ifelse(is.na(DF[i,3]) ==
                                       TRUE,
                                       0,
                                       DF[i,3]))

DF[,3] <- DF_NEW
DF

Starting from the data.table y, you can just write:
y[, (cols):=lapply(.SD, function(i){i[is.na(i)] <- 0; i}), .SDcols = cols]
Don't forget to library(data.table) before creating y and running this command.

We can solve it in data.table way with tidyr::repalce_na function and lapply

library(data.table)
library(tidyr)
setDT(df)
df[,c("a","b","c"):=lapply(.SD,function(x) replace_na(x,0)),.SDcols=c("a","b","c")]

In this way, we can also solve paste columns with NA string. First, we replace_na(x,""),then we can use stringr::str_c to combine columns!

This is now trivial in tidyr with replace_na(). The function appears to work for data.tables as well as data.frames:

tidyr::replace_na(x, list(a=0, b=0))

Building on @Robert McDonald's tidyr::replace_na() answer, here are some dplyr options for controlling which columns the NAs are replaced:

library(tidyverse)

# by column type:
x %>%
  mutate_if(is.numeric, ~replace_na(., 0))

# select columns defined in vars(col1, col2, ...):
x %>%
  mutate_at(vars(a, b, c), ~replace_na(., 0))

# all columns:
x %>%
  mutate_all(~replace_na(., 0))

Not sure if this is more concise, but this function will also find and allow replacement of NAs (or any value you like) in selected columns of a data.table:

update.mat <- function(dt, cols, criteria) {
  require(data.table)
  x <- as.data.frame(which(criteria==TRUE, arr.ind = TRUE))
  y <- as.matrix(subset(x, x$col %in% which((names(dt) %in% cols), arr.ind = TRUE)))
  y
}

To apply it:

y[update.mat(y, c("a", "b"), is.na(y))] <- 0

The function creates a matrix of the selected columns and rows (cell coordinates) that meet the input criteria (in this case is.na == TRUE).

Edit 2020-06-15

Since data.table 1.12.4 (Oct 2019), data.table gains two functions to facilitate this: nafill and setnafill.

nafill operates on columns:

cols = c('a', 'b')
y[ , (cols) := lapply(.SD, nafill, fill=0), .SDcols = cols]

setnafill operates on tables (the replacements happen by-reference/in-place)

setnafill(y, cols=cols, fill=0)
# print y to show the effect
y[]

This will also be more efficient than the other options; see ?nafill for more, the last-observation-carried-forward (LOCF) and next-observation-carried-backward (NOCB) versions of NA imputation for time series.

This will work for your data.table version:

for (col in c("a", "b")) y[is.na(get(col)), (col) := 0]

Alternatively, as David Arenburg points out below, you can use set (side benefit - you can use it either on data.frame or data.table):

for (col in 1:2) set(x, which(is.na(x[[col]])), col, 0)

it's quite handy with {data.table} and {stringr}

library(data.table)
library(stringr)

x[, lapply(.SD, function(xx) {str_replace_na(xx, 0)})]

FYI