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I have a dataframe with some NA values:
dfa <- data.frame(a=c(1,NA,3,4,5,NA),b=c(1,5,NA,NA,8,9),c=c(7,NA,NA,NA,2,NA)) dfa I would like to replace the NAs with values in the same position in another dataframe:
dfrepair <- data.frame(a=c(2:7),b=c(6:1),c=c(8:3)) dfrepair I tried:
dfa1 <- dfa dfa1 <- ifelse(dfa == NA, dfrepair, dfa) dfa1 but this did not work.
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To replace NA with 0 in an R data frame, use is.na() function and then select all those values with NA and assign them to 0.
How to replace NA (missing values) with blank space or an empty string in an R dataframe? You can replace NA values with blank space on columns of R dataframe (data. frame) by using is.na() , replace() methods.
In R (or R Studio), NA stands for Not Available. Each cell of your data that displays NA is a missing value. Not available values are sometimes enclosed by < and >, i.e. <NA>. That happens when the vector or column that contains the NA is a factor.
How to replace NA values in columns of an R data frame form the mean of that column? In the whole world, the first step people teach to impute missing values is replacing them with the relevant mean. That means if we have a column which has some missing values then replace it with the mean of the remaining values.
You can use the following syntax to replace one of several values in a data frame with a new value: df [df == 'Old Value 1' | df == 'Old Value 2'] <- 'New value' And you can use the following syntax to replace a particular value in a specific column of a data frame with a new value: df ['column1'] [df ['column1'] == 'Old Value'] <- 'New value'
How to Replace NAs with Strings in R (With Examples) You can use the replace_na () function from the tidyr package to replace NAs with specific strings in a column of a data frame in R: #replace NA values in column x with "missing" df$x %>% replace_na('none')
replace_na (data, replace, ...) data: It is a data frame or Vector. replace: If the data is a Vector, the replace takes a single value. If the data is a data frame, the replace takes a list of values, with one value for each column that has NA values to be replaced. If the input data is a data frame, the replace_na () method returns a data frame.
You can do:
dfa <- data.frame(a=c(1,NA,3,4,5,NA),b=c(1,5,NA,NA,8,9),c=c(7,NA,NA,NA,2,NA)) dfrepair <- data.frame(a=c(2:7),b=c(6:1),c=c(8:3)) dfa[is.na(dfa)] <- dfrepair[is.na(dfa)] dfa a b c 1 1 1 7 2 3 5 7 3 3 4 6 4 4 3 5 5 5 8 2 6 7 9 3 In the tidyverse, you can use purrr::map2_df, which is a strictly bivariate version of mapply that simplifies to a data.frame, and dplyr::coalesce, which replaces NA values in its first argument with the corresponding ones in the second.
library(tidyverse) dfrepair %>% mutate_all(as.numeric) %>% # coalesce is strict about types map2_df(dfa, ., coalesce) ## # A tibble: 6 × 3 ## a b c ## <dbl> <dbl> <dbl> ## 1 1 1 7 ## 2 3 5 7 ## 3 3 4 6 ## 4 4 3 5 ## 5 5 8 2 ## 6 7 9 3 If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
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