How Many Non-Na Values in Each Row for a Matrix

How many non-NA values in each row for a matrix?

Most Simply:

rowSums(!is.na(x)) (thanks to @Khashaa for this code).

Note the use of ! which equates to "not". This means that !is.na(x) is evaluating the statement "values that are not equal to "NA".

Alternatively:

To return not NA you can change the code as follows:

sum(is.na(x)==FALSE)

You can modify the code using apply to apply the code over the matrix as follows:

apply(d,2,function(x) sum(is.na(x))==TRUE))

where d is a matrix such as:

d=matrix(c(1,NA,NA,NA),ncol=2,nrow=2)

Count number of non-NA values for every column in a dataframe

You can also call is.na on the entire data frame (implicitly coercing to a logical matrix) and call colSums on the inverted response:

# make sample data
set.seed(47)
df <- as.data.frame(matrix(sample(c(0:1, NA), 100*5, TRUE), 100))

str(df)
#> 'data.frame':    100 obs. of  5 variables:
#>  $ V1: int  NA 1 NA NA 1 NA 1 1 1 NA ...
#>  $ V2: int  NA NA NA 1 NA 1 0 1 0 NA ...
#>  $ V3: int  1 1 0 1 1 NA NA 1 NA NA ...
#>  $ V4: int  NA 0 NA 0 0 NA 1 1 NA NA ...
#>  $ V5: int  NA NA NA 0 0 0 0 0 NA NA ...

colSums(!is.na(df))
#> V1 V2 V3 V4 V5 
#> 69 55 62 60 70

Select set of columns so that each row has at least one non-NA entry

Using a while loop, this should work to get the minimum set of variables with at least one non-NA per row.

best <- function(df){
  best <- which.max(colSums(sapply(df, complete.cases)))
  while(any(rowSums(sapply(df[best], complete.cases)) == 0)){
    best <- c(best, which.max(sapply(df[is.na(df[best]), ], \(x) sum(complete.cases(x)))))
  }
  best
}

testing

best(df)
#d c 
#4 3

df[best(df)]
#   d  c
#1  1  1
#2  1 NA
#3  1 NA
#4  1 NA
#5 NA  1

First, select the column with the least NAs (stored in best). Then, update the vector with the column that has the highest number of non-NA rows on the remaining rows (where best has still NAs), until you get every rows with a complete case.

How to identify which columns are not “NA” per row in a dataframe?

Loop through rows, get column names with for non-na columns, then paste:

d$myCol <- apply(d, 1, function(i) paste(colnames(d)[ !is.na(i) ], collapse = ","))

Sum the last n non NA values in each column of a matrix in R

You can use apply with tail to sum up the last non NA like:

apply(x, 2, function(x) sum(tail(x[!is.na(x)], 3)))
#x1 x2 x3 x4 x5 
#15 11  9  6  3 

R: function or similar to sum up number of non-NA values for columns that contain specific characters in large data set

Use rowSums :

library(dplyr)
p %>% mutate(n_fu =  rowSums(!is.na(select(., contains('fu_location')))))

Or in base :

p$n_fu <- rowSums(!is.na(p[grep('fu_location', names(p))]))

From an R dataframe: count non-NA values by column, grouped by one of the columns

We can use summarise_all

library(dplyr)
litmus %>% 
   group_by(grouping) %>% 
   summarise_all(funs(sum(!is.na(.))))

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