How do I replace NA values with zeros in an R dataframe?

Question

I have a data frame and some columns have NA values.

How do I replace these NA values with zeroes?

Answer 1

The cleaner package has an na_replace() generic, that at default replaces numeric values with zeroes, logicals with FALSE, dates with today, etc.:

starwars %>% na_replace()
na_replace(starwars)

It even supports vectorised replacements:

mtcars[1:6, c("mpg", "hp")] <- NA
na_replace(mtcars, mpg, hp, replacement = c(999, 123))

Documentation: https://msberends.github.io/cleaner/reference/na_replace.html

Answer 2

To replace all NAs in a dataframe you can use:

df %>% replace(is.na(.), 0)

Answer 3

The dplyr hybridized options are now around 30% faster than the Base R subset reassigns. On a 100M datapoint dataframe mutate_all(~replace(., is.na(.), 0)) runs a half a second faster than the base R d[is.na(d)] <- 0 option. What one wants to avoid specifically is using an ifelse() or an if_else(). (The complete 600 trial analysis ran to over 4.5 hours mostly due to including these approaches.) Please see benchmark analyses below for the complete results.

If you are struggling with massive dataframes, data.table is the fastest option of all: 40% faster than the standard Base R approach. It also modifies the data in place, effectively allowing you to work with nearly twice as much of the data at once.

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A clustering of other helpful tidyverse replacement approaches

Locationally:

• index mutate_at(c(5:10), ~replace(., is.na(.), 0))
• direct reference mutate_at(vars(var5:var10), ~replace(., is.na(.), 0))
• fixed match mutate_at(vars(contains("1")), ~replace(., is.na(.), 0))
  • or in place of contains(), try ends_with(),starts_with()
• pattern match mutate_at(vars(matches("\\d{2}")), ~replace(., is.na(.), 0))

Conditionally:
(change just single type and leave other types alone.)

• integers mutate_if(is.integer, ~replace(., is.na(.), 0))
• numbers mutate_if(is.numeric, ~replace(., is.na(.), 0))
• strings mutate_if(is.character, ~replace(., is.na(.), 0))

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The Complete Analysis -

Updated for dplyr 0.8.0: functions use purrr format ~ symbols: replacing deprecated funs() arguments.

Approaches tested:

# Base R: 
baseR.sbst.rssgn   <- function(x) { x[is.na(x)] <- 0; x }
baseR.replace      <- function(x) { replace(x, is.na(x), 0) }
baseR.for          <- function(x) { for(j in 1:ncol(x))
    x[[j]][is.na(x[[j]])] = 0 }

# tidyverse
## dplyr
dplyr_if_else      <- function(x) { mutate_all(x, ~if_else(is.na(.), 0, .)) }
dplyr_coalesce     <- function(x) { mutate_all(x, ~coalesce(., 0)) }

## tidyr
tidyr_replace_na   <- function(x) { replace_na(x, as.list(setNames(rep(0, 10), as.list(c(paste0("var", 1:10)))))) }

## hybrid 
hybrd.ifelse     <- function(x) { mutate_all(x, ~ifelse(is.na(.), 0, .)) }
hybrd.replace_na <- function(x) { mutate_all(x, ~replace_na(., 0)) }
hybrd.replace    <- function(x) { mutate_all(x, ~replace(., is.na(.), 0)) }
hybrd.rplc_at.idx<- function(x) { mutate_at(x, c(1:10), ~replace(., is.na(.), 0)) }
hybrd.rplc_at.nse<- function(x) { mutate_at(x, vars(var1:var10), ~replace(., is.na(.), 0)) }
hybrd.rplc_at.stw<- function(x) { mutate_at(x, vars(starts_with("var")), ~replace(., is.na(.), 0)) }
hybrd.rplc_at.ctn<- function(x) { mutate_at(x, vars(contains("var")), ~replace(., is.na(.), 0)) }
hybrd.rplc_at.mtc<- function(x) { mutate_at(x, vars(matches("\\d+")), ~replace(., is.na(.), 0)) }
hybrd.rplc_if    <- function(x) { mutate_if(x, is.numeric, ~replace(., is.na(.), 0)) }

# data.table   
library(data.table)
DT.for.set.nms   <- function(x) { for (j in names(x))
    set(x,which(is.na(x[[j]])),j,0) }
DT.for.set.sqln  <- function(x) { for (j in seq_len(ncol(x)))
    set(x,which(is.na(x[[j]])),j,0) }
DT.nafill        <- function(x) { nafill(df, fill=0)}
DT.setnafill     <- function(x) { setnafill(df, fill=0)}

The code for this analysis:

library(microbenchmark)
# 20% NA filled dataframe of 10 Million rows and 10 columns
set.seed(42) # to recreate the exact dataframe
dfN <- as.data.frame(matrix(sample(c(NA, as.numeric(1:4)), 1e7*10, replace = TRUE),
                            dimnames = list(NULL, paste0("var", 1:10)), 
                            ncol = 10))
# Running 600 trials with each replacement method 
# (the functions are excecuted locally - so that the original dataframe remains unmodified in all cases)
perf_results <- microbenchmark(
    hybrid.ifelse    = hybrid.ifelse(copy(dfN)),
    dplyr_if_else    = dplyr_if_else(copy(dfN)),
    hybrd.replace_na = hybrd.replace_na(copy(dfN)),
    baseR.sbst.rssgn = baseR.sbst.rssgn(copy(dfN)),
    baseR.replace    = baseR.replace(copy(dfN)),
    dplyr_coalesce   = dplyr_coalesce(copy(dfN)),
    tidyr_replace_na = tidyr_replace_na(copy(dfN)),
    hybrd.replace    = hybrd.replace(copy(dfN)),
    hybrd.rplc_at.ctn= hybrd.rplc_at.ctn(copy(dfN)),
    hybrd.rplc_at.nse= hybrd.rplc_at.nse(copy(dfN)),
    baseR.for        = baseR.for(copy(dfN)),
    hybrd.rplc_at.idx= hybrd.rplc_at.idx(copy(dfN)),
    DT.for.set.nms   = DT.for.set.nms(copy(dfN)),
    DT.for.set.sqln  = DT.for.set.sqln(copy(dfN)),
    times = 600L
)

Summary of Results

> print(perf_results)
Unit: milliseconds
              expr       min        lq     mean   median       uq      max neval
      hybrd.ifelse 6171.0439 6339.7046 6425.221 6407.397 6496.992 7052.851   600
     dplyr_if_else 3737.4954 3877.0983 3953.857 3946.024 4023.301 4539.428   600
  hybrd.replace_na 1497.8653 1706.1119 1748.464 1745.282 1789.804 2127.166   600
  baseR.sbst.rssgn 1480.5098 1686.1581 1730.006 1728.477 1772.951 2010.215   600
     baseR.replace 1457.4016 1681.5583 1725.481 1722.069 1766.916 2089.627   600
    dplyr_coalesce 1227.6150 1483.3520 1524.245 1519.454 1561.488 1996.859   600
  tidyr_replace_na 1248.3292 1473.1707 1521.889 1520.108 1570.382 1995.768   600
     hybrd.replace  913.1865 1197.3133 1233.336 1238.747 1276.141 1438.646   600
 hybrd.rplc_at.ctn  916.9339 1192.9885 1224.733 1227.628 1268.644 1466.085   600
 hybrd.rplc_at.nse  919.0270 1191.0541 1228.749 1228.635 1275.103 2882.040   600
         baseR.for  869.3169 1180.8311 1216.958 1224.407 1264.737 1459.726   600
 hybrd.rplc_at.idx  839.8915 1189.7465 1223.326 1228.329 1266.375 1565.794   600
    DT.for.set.nms  761.6086  915.8166 1015.457 1001.772 1106.315 1363.044   600
   DT.for.set.sqln  787.3535  918.8733 1017.812 1002.042 1122.474 1321.860   600

Boxplot of Results

ggplot(perf_results, aes(x=expr, y=time/10^9)) +
    geom_boxplot() +
    xlab('Expression') +
    ylab('Elapsed Time (Seconds)') +
    scale_y_continuous(breaks = seq(0,7,1)) +
    coord_flip()

Color-coded Scatterplot of Trials (with y-axis on a log scale)

qplot(y=time/10^9, data=perf_results, colour=expr) + 
    labs(y = "log10 Scaled Elapsed Time per Trial (secs)", x = "Trial Number") +
    coord_cartesian(ylim = c(0.75, 7.5)) +
    scale_y_log10(breaks=c(0.75, 0.875, 1, 1.25, 1.5, 1.75, seq(2, 7.5)))

A note on the other high performers

When the datasets get larger, Tidyr''s replace_na had historically pulled out in front. With the current collection of 100M data points to run through, it performs almost exactly as well as a Base R For Loop. I am curious to see what happens for different sized dataframes.

Additional examples for the mutate and summarize _at and _all function variants can be found here: https://rdrr.io/cran/dplyr/man/summarise_all.html Additionally, I found helpful demonstrations and collections of examples here: https://blog.exploratory.io/dplyr-0-5-is-awesome-heres-why-be095fd4eb8a

Attributions and Appreciations

With special thanks to:

• Tyler Rinker and Akrun for demonstrating microbenchmark.
• alexis_laz for working on helping me understand the use of local(), and (with Frank's patient help, too) the role that silent coercion plays in speeding up many of these approaches.
• ArthurYip for the poke to add the newer coalesce() function in and update the analysis.
• Gregor for the nudge to figure out the data.table functions well enough to finally include them in the lineup.
• Base R For loop: alexis_laz
• data.table For Loops: Matt_Dowle
• Roman for explaining what is.numeric() really tests.

(Of course, please reach over and give them upvotes, too if you find those approaches useful.)

Note on my use of Numerics: If you do have a pure integer dataset, all of your functions will run faster. Please see alexiz_laz's work for more information. IRL, I can't recall encountering a data set containing more than 10-15% integers, so I am running these tests on fully numeric dataframes.

Hardware Used 3.9 GHz CPU with 24 GB RAM

Answer 4

Another example using imputeTS package:

library(imputeTS)
na.replace(yourDataframe, 0)

Answer 5

With dplyr 0.5.0, you can use coalesce function which can be easily integrated into %>% pipeline by doing coalesce(vec, 0). This replaces all NAs in vec with 0:

Say we have a data frame with NAs:

library(dplyr)
df <- data.frame(v = c(1, 2, 3, NA, 5, 6, 8))

df
#    v
# 1  1
# 2  2
# 3  3
# 4 NA
# 5  5
# 6  6
# 7  8

df %>% mutate(v = coalesce(v, 0))
#   v
# 1 1
# 2 2
# 3 3
# 4 0
# 5 5
# 6 6
# 7 8

Answer 6

This simple function extracted from Datacamp could help:

replace_missings <- function(x, replacement) {
  is_miss <- is.na(x)
  x[is_miss] <- replacement

  message(sum(is_miss), " missings replaced by the value ", replacement)
  x
}

Then

replace_missings(df, replacement = 0)