df <- data.frame(A = 1:10, B = 11:20, c = 21:30) head(df) ## A B c ## 1 1 11 21 ## 2 2 12 22 ## 3 3 13 23 ## 4 4 14 24 ## 5 5 15 25 ## 6 6 16 26 as.data.frame(lapply(df, function(cc) cc[ sample(c(TRUE, NA), prob = c(0.85, 0.15), … Read more
Another alternative: df <- sapply(df, as.character) # since your values are `factor` df[is.na(df)] <- 0 If you want blanks instead of zeroes > df <- sapply(df, as.character) > df[is.na(df)] <- ” ” > df class Year1 Year2 Year3 Year4 Year5 [1,] “classA” “A” “A” “A” “A” “A” [2,] ” ” ” ” ” ” ” … Read more
One way of doing it: df[, colSums(is.na(df)) != nrow(df)] If the count of NAs in a column is equal to the number of rows, it must be entirely NA. Or similarly df[colSums(!is.na(df)) > 0]
If we decide to use subset function, then we need to watch out: For ordinary vectors, the result is simply ‘x[subset & !is.na(subset)]’. So only non-NA values will be retained. If you want to keep NA cases, use logical or condition to tell R not to drop NA cases: subset(df1, Height < 40 | is.na(Height)) … Read more
The data: Itun <- data.frame(v1 = c(1,1,2,1,2,1), v2 = c(NA, 1, 2, 1, 2, NA)) This will remove all columns containing at least one NA: Itun[ , colSums(is.na(Itun)) == 0] An alternative way is to use apply: Itun[ , apply(Itun, 2, function(x) !any(is.na(x)))]
A dplyr::coalesce based solution could be as: data %>% mutate(mycol = coalesce(x,y,z)) %>% select(a, mycol) # a mycol # 1 A 1 # 2 B 2 # 3 C 3 # 4 D 4 # 5 E 5 Data data <- data.frame(‘a’ = c(‘A’,’B’,’C’,’D’,’E’), ‘x’ = c(1,2,NA,NA,NA), ‘y’ = c(NA,NA,3,NA,NA), ‘z’ = c(NA,NA,NA,4,5))
Use suppressWarnings(): suppressWarnings(as.numeric(c(“1”, “2”, “X”))) [1] 1 2 NA This suppresses warnings.
Good question, but in my opinion, this shouldn’t have caused a major debugging headache because it is documented quite clearly in multiple places in the manual page for aggregate. First, in the usage section: ## S3 method for class ‘formula’ aggregate(formula, data, FUN, …, subset, na.action = na.omit) Later, in the description: na.action: a function … Read more
Never use ==’NA’ to test for missing values. Use is.na() instead. This should do it: new_DF <- DF[rowSums(is.na(DF)) > 0,] or in case you want to check a particular column, you can also use new_DF <- DF[is.na(DF$Var),] In case you have NA character values, first run Df[Df==’NA’] <- NA to replace them with missing values.
You can use this mydf %>% group_by(col_1) %>% summarise(non_na_count = sum(!is.na(col_2))) # A tibble: 2 x 2 col_1 non_na_count <fctr> <int> 1 A 1 2 B 2