R: Replace Na with Item from Vector

R: replace NA with item from vector

ifelse is your friend.

Using Dirk's dataset

df <- within(df, X <- ifelse(is.na(X), Y, X))

Replace NA's in a dataframe with a vector of numbers in r

We need to replace the specific column

mydata$Freq2[is.na(mydata$Freq2)] <- L8Replace
mydata
#            Freq2     myDate  numDays
#2015-04-16  0.42 2015-04-16   0 days
#2015-04-21  0.60 2015-04-21   5 days
#2015-04-26  0.43 2015-04-26  10 days
#7           0.56 2015-07-11  86 days
#8           0.57 2015-07-27 102 days
#9           0.87 2015-08-12 118 days
#10          0.81 2015-08-28 134 days
#11          0.00 2015-09-13 150 days

Also, the length of 'L8Replace' should match the number of NA in 'Freq2'

data

L8Replace <-  c(0.56, 0.57, 0.87,0.81, 0)

Replace NA using a vector of column names

columns value should be string, you can then use is.na as -

columns<-c("col1","col2")

data[columns][is.na(data[columns])] <- 0
data

#  col1 col2 col3 col4
#1    9    9   NA    9
#2    0    5    3    1
#3   25   25   25   NA
#4   26   26   26   26
#5    0    0   NA   NA
#6   51   51   51   51

Or using tidyverse -

library(dplyr)
library(tidyr)

data <- data %>% mutate(across(all_of(columns), replace_na, 0))

Replace NA's with a vector

From what I read, you want to replace the NAs in each column of a matrix with a corresponding value of a vector.

Lets say this is your data set and your replacement vector

set.seed(1)
(m <- matrix(sample(c(NA, 1), 25, replace = TRUE), ncol = 5))
#      [,1] [,2] [,3] [,4] [,5]
# [1,]   NA    1   NA   NA    1
# [2,]   NA    1   NA    1   NA
# [3,]    1    1    1    1    1
# [4,]    1    1   NA   NA   NA
# [5,]   NA   NA    1    1   NA
(vec <- sample(5))
## [1] 2 1 5 3 4

Here's a quite simple vectorized way to replace the NAs with corresponding values

indx <- is.na(m)
m[indx] <- vec[col(m)][indx]
#      [,1] [,2] [,3] [,4] [,5]
# [1,]    2    1    5    3    1
# [2,]    2    1    5    1    4
# [3,]    1    1    1    1    1
# [4,]    1    1    5    3    4
# [5,]    2    1    1    1    4

This is basically takes the advantage is.na returning a logical matrix that can be used as a subsetting index and the col function which converts vec to a matrix.

Replace NA values in a data frame column with values with elements from a vector

We can do an assignment by replicating the vector

i1 <- is.na(df)
df[i1] <- v1[col(df)][i1]

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Or we can use

v1[col(df)]*is.na(df) + replace(df, is.na(df), 0)

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Or we can use Map to replace the corresponding columns NAs with the vector elements

df[] <- Map(function(x, y) replace(x, is.na(x), y), df, v1)

data

set.seed(24)
df <- as.data.frame(matrix(sample(c(NA, 1:5), 100*20, replace = TRUE), ncol = 20))
set.seed(48)
v1 <- sample(1:10, 20, replace = TRUE)

Replace NA with preceding character string in vector

tidyr has a function fill that fills in NAs with the closest non-missing value above it.

If you're fine filling in values in X in place:

library(dplyr)
library(tidyr)

d %>%
  fill(X)
#>       X Y
#> 1   one 1
#> 2   one 2
#> 3   two 3
#> 4   two 4
#> 5 three 5
#> 6 three 6

Or if you need to keep the original X with its missing values, copy it over to another column, and fill that one in:

d %>%
  mutate(X2 = X) %>%
  fill(X2)
#>       X Y    X2
#> 1   one 1   one
#> 2  <NA> 2   one
#> 3   two 3   two
#> 4  <NA> 4   two
#> 5 three 5 three
#> 6  <NA> 6 three

Is there a way to replace a character in a vector with a NULL value in R?

It's worth remembering the difference between NULL and NA. NA values are a dodgy value, NULL is no value whatsoever. In order to get the second output to be the same as the first output, you would have something the same as the following

column <- c("None", "Some", "NULL", "Many", "All")
column <- column[column != "NULL"]

This creates a shorter vector, which is why str_replace doesn't like it.

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