tidyr replace_na() in R: Replace NA With a Value

The replace_na() function in tidyr replaces NA values in a vector or data frame with a specified value. For a vector, pass one replacement; for a data frame, pass a named list with column-specific replacements.

By Selva Prabhakaran · Published May 11, 2026 · Last updated May 11, 2026

⚡ Quick Answer

replace_na(x, 0)                                  # vector: NA -> 0
replace_na(df, list(x = 0, y = "missing"))        # data frame: per-col
mutate(df, x = replace_na(x, 0))                  # inside mutate
mutate(df, across(where(is.numeric), ~ replace_na(., 0)))  # all numerics
coalesce(x, 0)                                    # alternative: first non-NA
ifelse(is.na(x), 0, x)                            # base R alternative
df |> mutate(x = if_else(is.na(x), 0, x))         # dplyr if_else

Need explanation? Read on for examples and pitfalls.

📊 Is replace_na() the right tool?

What replace_na() does in one sentence

replace_na() swaps NA values in a vector or data frame for a value you provide. It is type-strict (replacement must match the column type) and pipeline-friendly.

For more flexible NA handling (e.g., conditional replacement, multi-column fallbacks, mean imputation), use coalesce(), case_when(), or mutate() with is.na() instead.

Syntax

For vectors: replace_na(vec, value). For data frames: replace_na(df, list(col1 = val1, col2 = val2)).

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RReplace NA in a vector

library(tidyr) library(dplyr) library(tibble) x <- c(1, NA, 3, NA, 5) replace_na(x, 0) #> [1] 1 0 3 0 5

For a data frame, you pass a named list specifying per-column replacements:

RReplace NAs differently per column

df <- tibble::tribble( ~name, ~age, ~city, "Alice", 30, "NYC", "Bob", NA, NA, "Carol", 25, "SF" ) df |> replace_na(list(age = 0, city = "unknown")) #> # A tibble: 3 x 3 #> name age city #> <chr> <dbl> <chr> #> 1 Alice 30 NYC #> 2 Bob 0 unknown #> 3 Carol 25 SF

Tip

Replacement values must match the column's TYPE. replace_na(df, list(x = "zero")) errors if x is numeric. R is strict here; the replacement type must be compatible with the column type.

Five common patterns

1. Replace NA in a vector

RVector: NA -> 0

replace_na(c(1, NA, 3, NA, 5), 0) #> [1] 1 0 3 0 5

The simplest case. Single replacement value for a single vector.

2. Replace NA per-column in a data frame

RDifferent value per column

df |> replace_na(list(age = 0, city = "unknown"))

Pass a named list. Columns NOT in the list keep their NAs.

3. Replace NA across many columns at once

RAll numeric columns -> 0

df_num <- tibble::tibble( id = 1:3, a = c(1, NA, 3), b = c(NA, 2, 4), name = c("x", NA, "z") ) df_num |> dplyr::mutate(dplyr::across(where(is.numeric), ~ replace_na(., 0))) #> # A tibble: 3 x 4 #> id a b name #> <int> <dbl> <dbl> <chr> #> 1 1 1 0 x #> 2 2 0 2 NA #> 3 3 3 4 z

across(where(is.numeric), ~ replace_na(., 0)) applies the replacement to every numeric column. Non-numeric columns (name) are untouched.

4. coalesce: first non-NA across columns

RPick first available value

df_co <- tibble::tibble( primary = c(1, NA, 3), backup = c(NA, 2, NA), default = c(99, 99, 99) ) df_co |> dplyr::mutate(value = dplyr::coalesce(primary, backup, default)) #> # A tibble: 3 x 4 #> primary backup default value #> <dbl> <dbl> <dbl> <dbl> #> 1 1 NA 99 1 #> 2 NA 2 99 2 #> 3 3 NA 99 3

coalesce() returns the FIRST non-NA value across the listed vectors. Useful for fallback chains: try column A, then B, then a default.

5. Conditional replacement with case_when

RMore than just constant replacement

df |> dplyr::mutate(age_cleaned = dplyr::case_when( is.na(age) & city == "NYC" ~ 35, # NYC NAs default to 35 is.na(age) ~ 30, # other NAs default to 30 TRUE ~ age ))

case_when() lets you replace NAs based on OTHER columns or conditions. More flexible than replace_na().

Key Insight

replace_na() is for CONSTANT replacement. For dynamic or conditional replacement, use case_when() or mutate(if_else(is.na(x), ...)) instead. Replace_na fits a narrow but very common case: "fill NA with a specific value per column".

replace_na() vs coalesce() vs ifelse(is.na(), ...)

Approach	Best for	Notes
`replace_na(x, val)`	Constant value	Simple, type-strict
`coalesce(x, y, z)`	Fallback chain across cols	First non-NA wins
`ifelse(is.na(x), val, x)`	One-off in base R	Works without packages
`dplyr::if_else(is.na(x), val, x)`	One-off in dplyr	Type-strict like replace_na
`case_when()`	Conditional replacement	Flexible, multi-condition

When to use which:

Use replace_na() for clean per-column constants.
Use coalesce() when the replacement comes from another column.
Use case_when() when replacement depends on conditions.

Common pitfalls

Pitfall 1: type mismatch errors. replace_na(df, list(x = 0)) errors if x is character. Cast first: mutate(x = as.numeric(x)) or use a string replacement: list(x = "0").

Pitfall 2: replacing with mean / median naively. replace_na(df, list(x = mean(df$x, na.rm = TRUE))) works but is brittle. For mean imputation: mutate(x = ifelse(is.na(x), mean(x, na.rm = TRUE), x)). Better: use the mice package for principled imputation.

Warning

Replacing NA with 0 changes the meaning of "missing" to "zero". This is appropriate for COUNTS where missing usually means "didn't happen" but WRONG for measurements where missing means "we did not record it". Pick the replacement value that matches your domain semantics.

Try it yourself

Try it: Replace NAs in airquality$Ozone with the mean Ozone value. Save the modified vector to ex_filled.

RYour turn: mean impute Ozone

mean_ozone <- mean(airquality$Ozone, na.rm = TRUE) ex_filled <- # your code here sum(is.na(ex_filled)) #> Expected: 0 (no more NAs)

Click to reveal solution

RSolution

mean_ozone <- mean(airquality$Ozone, na.rm = TRUE) ex_filled <- replace_na(airquality$Ozone, mean_ozone) sum(is.na(ex_filled)) #> [1] 0

Explanation: Compute the mean (excluding NAs) first, then replace_na(Ozone, mean_ozone) fills NA with that mean. The result has zero NAs. For real analysis, mean imputation is a starting point; consider mice::mice() for proper missing-data treatment.

After mastering replace_na, look at:

coalesce(): pick first non-NA across multiple columns
fill(): forward/backward-fill NAs
complete(): fill in missing combinations
case_when(), if_else(): conditional replacement
drop_na(): remove rows with NA (alternative to replacement)
mice::mice(): principled multiple imputation

For time series, zoo::na.locf() (last observation carried forward) and imputeTS package handle NAs more sophisticatedly.

FAQ

How do I replace NA with 0 in R?

For a vector: replace_na(x, 0). For a data frame column: mutate(x = replace_na(x, 0)). For all numeric columns: mutate(across(where(is.numeric), ~ replace_na(., 0))).

What is the difference between replace_na and coalesce in R?

replace_na(x, value) replaces NA with a CONSTANT. coalesce(x, y, z) takes the FIRST non-NA across multiple inputs. Use replace_na for fixed defaults; coalesce for fallback chains.

How do I replace NA values per column in a data frame?

Pass a named list to replace_na: replace_na(df, list(x = 0, name = "unknown", date = as.Date("1970-01-01"))). Each column gets its own replacement value.

Can I impute the mean for NA values with replace_na?

Yes: mean_x <- mean(df$x, na.rm = TRUE); replace_na(df$x, mean_x). But this is naive. For real analysis with missing data, use mice for multiple imputation; mean replacement understates variability.

What replaces dplyr's deprecated replace_na?

replace_na() is from tidyr (not deprecated). dplyr re-exports it for convenience. Both tidyr::replace_na and dplyr::replace_na refer to the same function.

tidyr replace_na() in R: Replace NA With a Value

What replace_na() does in one sentence

Syntax

Five common patterns

1. Replace NA in a vector

2. Replace NA per-column in a data frame

3. Replace NA across many columns at once

4. coalesce: first non-NA across columns

5. Conditional replacement with case_when

replace_na() vs coalesce() vs ifelse(is.na(), ...)

Common pitfalls

Try it yourself

Related tidyr / dplyr functions

FAQ