Home > Blockchain >  Count the number of non-zero elements of each column- Piping Friendly
Count the number of non-zero elements of each column- Piping Friendly

Time:03-18

I am a behavioral ecologist using R. I am trying to count the non-zero elements across several columns. Normally when I do this I have successfully employed colSums with the !=0 operator, as has been suggested in many posts here and elsewhere i.e.(Count the number of non-zero elements of each column).

However in this particular case I would really prefer to using piping - as this is simply one step of building out a much larger data frame- and I cannot seem to get colSums with the !=0 to play nicely with the piping. Is there anyway to get this to work or is there a more elegant alternative to counting non-zero values across columns living in the tidy-verse somewhere?

I have put some example code below to demonstrate. Thanks very much!

'''

#make example dataframe
example<- as.data.frame(matrix(sample(c(0,0,0,100),size=70,replace=T),ncol=7))
example<-cbind(Indentifier1="character a",Indentifier2="character b", example)
example


#works great but isn't piping friendly:
colSums(example[-c(1:2)] !=0)  

#Runs but because it lacks the !=0 operator it gives sums not counts of non-zero elements, can't figure out how to employ !=0:
example %>% select(-c(1:2)) %>% colSums() 


#gives me counts of all values, not sure how to call just non-zero values:
example %>% summarise(across(where(is.numeric), length))
 

'''

CodePudding user response:

example %>% summarise(across(where(is.numeric), ~sum(. != 0)))

CodePudding user response:

If you want to go the full magrittr way, read about ?alias.

library(magrittr)
example %>% 
  extract(-c(1, 2)) %>%
  equals(0) %>%
  not() %>%
  colSums()
# V1 V2 V3 V4 V5 V6 V7 
# 2  3  1  4  3  4  1 

data

set.seed(42)
example<- as.data.frame(matrix(sample(c(0,0,0,100),size=70,replace=T),ncol=7))
example<-cbind(Indentifier1="character a",Indentifier2="character b", example)
example


#works great but isn't piping friendly:
colSums(example[-c(1:2)] !=0) 
# V1 V2 V3 V4 V5 V6 V7 
# 2  3  1  4  3  4  1 

CodePudding user response:

Get the logical matrix you'd like to compute on, and pipe it to colSums()

(example |> select(-(1:2)) != 0) |> 
    colSums()

Of course this is tricky, because it relies on the implicit order of operations rule that forces example |> select(-(1:2)) to be evaluated before comparison to 0, and the explicit grouping with () required to evaluate the inequality before piping to colSums().

CodePudding user response:

You can still use piping with base R code. For example, you could do something like this:

example %>%
  .[-c(1:2)] %>%
  replace(., . != 0, 1) %>%
  colSums()

#V1 V2 V3 V4 V5 V6 V7 
# 2  3  1  2  4  1  1 

CodePudding user response:

Just using base R (>= 4.1) you may do

example |>
  subset(select=-(1:2)) |>
  {\(.) colSums(. != 0)}()
# V1 V2 V3 V4 V5 V6 V7 
#  4  4  3  3  3  2  2 
  • Related