Counting not NA's for values of some column for each value of another row [duplicate]

Question

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• dplyr count non-NA value in group by [duplicate] 3 answers

In R language - I have lets say I have a DF with two columns Fam and Prop both categorical, now Fam has repeated names like Algea, Fungi, etc and column Prop has categorical numbers and NA's. How can I get a table/output that for each value of A it tells me how many values are not. NA example:

    Fam     Prop
    -------------
    Algea   one
    Fungi   two
    Algea   NA
    Algea   three
    Fungi   one
    Fungi   NA

Output:

Algea 2
Fungi 2

I know using the count function should be a direction for the solution but can't seem to solve it, because the Fam column has repeating values.

Answer 1

Maybe something like this.

library(dplyr)
data %>% group_by(Fam) %>% summarise(sum(!is.na(Prop)))

Answer 2

Four solutions:

Base R frame:

aggregate(DF$Prop, by=list(Fam=DF$Fam), FUN=function(a) sum(!is.na(a)))
#   Fam x
# 1   A 5
# 2   B 6
# 3   C 4

Base R, "table" (which is not a frame, see as.data.frame(xtabs(...)) to see the frame variant ... a little different):

xtabs(~ Fam + is.na(Prop), data=DF)
#    is.na(Prop)
# Fam FALSE TRUE
#   A     5    1
#   B     6    1
#   C     4    3

dplyr:

library(dplyr)
DF %>%
  group_by(Fam) %>%
  summarize(n = sum(!is.na(Prop)))
# # A tibble: 3 x 2
#   Fam       n
#   <fct> <int>
# 1 A         5
# 2 B         6
# 3 C         4

data.table

library(data.table)
# data.table 1.11.4  Latest news: http://r-datatable.com
# Attaching package: 'data.table'
# The following objects are masked from 'package:dplyr':
#     between, first, last
DT <- as.data.table(DF)
DT[,sum(!is.na(Prop)),keyby=.(Fam)]
#    Fam V1
# 1:   A  5
# 2:   B  6
# 3:   C  4

---

Data:

DF <- data.frame(Fam=sample(c('A','B','C'), size=20, replace=TRUE), Prop=sample(c('one','two','three'), size=20, replace=TRUE))
DF$Prop[sample(20,size=5)] <- NA
DF
#    Fam  Prop
# 1    B   one
# 2    B three
# 3    C  <NA>
# 4    A  <NA>
# 5    C   one
# 6    A   two
# 7    B   one
# 8    A three
# 9    B   two
# 10   C   one
# 11   C   two
# 12   B three
# 13   C  <NA>
# 14   C  <NA>
# 15   A   one
# 16   A   one
# 17   B three
# 18   A   two
# 19   C   two
# 20   B  <NA>

Answer 3

Some dplyr possibilities:

df %>%
  add_count(Fam, miss = !is.na(Prop)) %>%
  group_by(Fam) %>%
  summarise(Non_miss = first(n[miss = TRUE]))

df %>%
  filter(!is.na(Prop)) %>%
  group_by(Fam) %>%
  tally()

df %>%
  filter(!is.na(Prop)) %>%
  group_by(Fam) %>%
  summarise(Non_miss = n())

Answer 4

Base R shortest (and fastest?) solution

number.of.not.NAs <- table(df$Fam[!is.na(df$Prop)])

It takes df$Fam but chooses only elements which have not NA in the df$Prop vector positions. And then using the table function which you mentioned.

Base R usual solution

Alternatively, you can split the data frame into a list of data frame by df$Fam, and then count for each data frame, how many non-NA elements are in the second column - the usual split-apply-combine way. (But I guess, the table method above is faster).

dfsList <- split(df, df$Fam)
number.of.not.NAs <- sapply(dfsList, function(df) sum(!is.na(df$Prop)))