Count NAs per row in dataframe [duplicate]
You can count the NAs in each row with this command: rowSums(is.na(dat)) where dat is the name of your data frame.
na
Omit rows containing specific column of NA
Use is.na DF <- data.frame(x = c(1, 2, 3), y = c(0, 10, NA), z=c(NA, 33, 22)) DF[!is.na(DF$y),]
How to omit NA values while pasting numerous column values together?
You could try na.omit() to omit the values, then paste. Also, you could use toString(), as it is the equivalent of paste(…, collapse = “, “). apply(dd2, 1, function(x) toString(na.omit(x))) # [1] “A, AK2, PPT” “B, HFM1, PPT” “C, TRR” # [4] “D, TRR, RTT, GGT” “E, RTT” If you have specific columns you are … Read more
Subsetting R data frame results in mysterious NA rows
Wrap the condition in which: df[which(df$number1 < df$number2), ] How it works: It returns the row numbers where the condition matches (where the condition is TRUE) and subsets the data frame on those rows accordingly. Say that: which(df$number1 < df$number2) returns row numbers 1, 2, 3, 4 and 5. As such, writing: df[which(df$number1 < df$number2), … Read more
Coalesce two string columns with alternating missing values to one
You may try pmax df$c <- pmax(df$a, df$b) df # a b c # 1 dog <NA> dog # 2 mouse <NA> mouse # 3 <NA> cat cat # 4 bird <NA> bird …or ifelse: df$c <- ifelse(is.na(df$a), df$b, df$a) For more general solutions in cases with more than two columns, you find several ways … Read more
How to delete rows from a dataframe that contain n*NA
Use rowSums. To remove rows from a data frame (df) that contain precisely n NA values: df <- df[rowSums(is.na(df)) != n, ] or to remove rows that contain n or more NA values: df <- df[rowSums(is.na(df)) < n, ] in both cases of course replacing n with the number that’s required
Replace NA in column with value in adjacent column
It didn’t work because status was a factor. When you mix factor with numeric then numeric is the least restrictive. By forcing status to be character you get the results you’re after and the column is now a character vector: TEST$UNIT[is.na(TEST$UNIT)] <- as.character(TEST$STATUS[is.na(TEST$UNIT)]) ## UNIT STATUS TERMINATED START STOP ## 1 ACTIVE ACTIVE 1999-07-06 2007-04-23 … Read more
How to replace NA values in a table for selected columns
You can do: x[, 1:2][is.na(x[, 1:2])] <- 0 or better (IMHO), use the variable names: x[c(“a”, “b”)][is.na(x[c(“a”, “b”)])] <- 0 In both cases, 1:2 or c(“a”, “b”) can be replaced by a pre-defined vector.
Collapsing rows where some are all NA, others are disjoint with some NAs
Here’s a data table approach that uses na.omit() across the columns, grouped by ID. library(data.table) setDT(df)[, lapply(.SD, na.omit), by = ID] # ID Col1 Col2 Col3 Col4 # 1: 1 5 10 15 20 # 2: 2 25 30 35 40
suppress NAs in paste()
For the purpose of a “true-NA”: Seems the most direct route is just to modify the value returned by paste2 to be NA when the value is “” paste3 <- function(…,sep=”, “) { L <- list(…) L <- lapply(L,function(x) {x[is.na(x)] <- “”; x}) ret <-gsub(paste0(“(^”,sep,”|”,sep,”$)”),””, gsub(paste0(sep,sep),sep, do.call(paste,c(L,list(sep=sep))))) is.na(ret) <- ret==”” ret } val<- paste3(c(“a”,”b”, “c”, … Read more