Summing values in R based on column value with dplyr - r

Question

I have a data set that has the following information:
Subject Value1 Value2 Value3 UniqueNumber
001 1 0 1 3
002 0 1 1 2
003 1 1 1 1
If the value of UniqueNumber > 0, I would like to sum the values with dplyr for each subject from rows 1 through UniqueNumber and calculate the mean. So for Subject 001, sum = 2 and mean = .67.
total = 0;
average = 0;
for(i in 1:length(Data$Subject)){
for(j in 1:ncols(Data)){
if(Data$UniqueNumber[i] > 0){
total[i] = sum(Data[i,1:j])
average[i] = mean(Data[i,1:j])
}
}
Edit: I am only looking to sum through the number of columns listed in the 'UniqueNumber' column. So this is looping through every row and stopping at column listed in 'UniqueNumber'.
Example: Row 2 with Subject 002 should sum up the values in columns 'Value1' and 'Value2', while Row 3 with Subject 003 should only sum the value in column 'Value1'.

Answer 1

Not a tidyverse fan/expert, but I would try this using long format. Then, just filter by row index per group and then run any functions you want on a single column (much easier this way).
library(tidyr)
library(dplyr)
Data %>%
gather(variable, value, -Subject, -UniqueNumber) %>% # long format
group_by(Subject) %>% # group by Subject in order to get row counts
filter(row_number() <= UniqueNumber) %>% # filter by row index
summarise(Mean = mean(value), Total = sum(value)) %>% # do the calculations
ungroup()
## A tibble: 3 x 3
# Subject Mean Total
# <int> <dbl> <int>
# 1 1 0.667 2
# 2 2 0.5 1
# 3 3 1 1
A very similar way to achieve this could be filtering by the integers in the column names. The filter step comes before the group_by so it could potentially increase performance (or not?) but it is less robust as I'm assuming that the cols of interest are called "Value#"
Data %>%
gather(variable, value, -Subject, -UniqueNumber) %>% #long format
filter(as.numeric(gsub("Value", "", variable, fixed = TRUE)) <= UniqueNumber) %>% #filter
group_by(Subject) %>% # group by Subject
summarise(Mean = mean(value), Total = sum(value)) %>% # do the calculations
ungroup()
## A tibble: 3 x 3
# Subject Mean Total
# <int> <dbl> <int>
# 1 1 0.667 2
# 2 2 0.5 1
# 3 3 1 1
Just for fun, adding a data.table solution
library(data.table)
data.table(Data) %>%
melt(id = c("Subject", "UniqueNumber")) %>%
.[as.numeric(gsub("Value", "", variable, fixed = TRUE)) <= UniqueNumber,
.(Mean = round(mean(value), 3), Total = sum(value)),
by = Subject]
# Subject Mean Total
# 1: 1 0.667 2
# 2: 2 0.500 1
# 3: 3 1.000 1

Answer 2

Here is another method that uses tidyr::nest to collect the Values columns into a list so that we can iterate through the table with map2. In each row, we select the correct values from the Values list-col and take the sum or mean respectively.
library(tidyverse)
tbl <- read_table2(
"Subject Value1 Value2 Value3 UniqueNumber
001 1 0 1 3
002 0 1 1 2
003 1 1 1 1"
)
tbl %>%
filter(UniqueNumber > 0) %>%
nest(starts_with("Value"), .key = "Values") %>%
mutate(
sum = map2_dbl(UniqueNumber, Values, ~ sum(.y[1:.x], na.rm = TRUE)),
mean = map2_dbl(UniqueNumber, Values, ~ mean(as.numeric(.y[1:.x], na.rm = TRUE))),
)
#> # A tibble: 3 x 5
#> Subject UniqueNumber Values sum mean
#> <chr> <dbl> <list> <dbl> <dbl>
#> 1 001 3 <tibble [1 × 3]> 2 0.667
#> 2 002 2 <tibble [1 × 3]> 1 0.5
#> 3 003 1 <tibble [1 × 3]> 1 1
Created on 2019-02-14 by the reprex package (v0.2.1)

Answer 3

Check this solution:
df %>%
gather(key, val, Value1:Value3) %>%
group_by(Subject) %>%
mutate(
Sum = sum(val[c(1:(UniqueNumber[1]))]),
Mean = mean(val[c(1:(UniqueNumber[1]))]),
) %>%
spread(key, val)
Output:
Subject UniqueNumber Sum Mean Value1 Value2 Value3
<chr> <int> <dbl> <dbl> <dbl> <dbl> <dbl>
1 001 3 2 0.667 1 0 1
2 002 2 1 0.5 0 1 1
3 003 1 1 1 1 1 1

Answer 4

OP might be interested only for dplyr solution but for comparison purposes and for future readers a base R option using mapply
cols <- grep("^Value", names(df))
cbind(df, t(mapply(function(x, y) {
if (y > 0) {
vals = as.numeric(df[x, cols[1:y]])
c(Sum = sum(vals, na.rm = TRUE), Mean = mean(vals, na.rm = TRUE))
}
else
c(0, 0)
},1:nrow(df), df$UniqueNumber)))
# Subject Value1 Value2 Value3 UniqueNumber Sum Mean
#1 1 1 0 1 3 2 0.667
#2 2 0 1 1 2 1 0.500
#3 3 1 1 1 1 1 1.000
Here we subset each row based on its respective UniqueNumber and then calculate it's sum and mean if the UniqueNumber value is greater than 0 or else return only 0.

Answer 5

A solution that uses purrr::map_df(which is from the same author as dplyr).
library(dplyr)
library(purrr)
l_dat <- split(dat, dat$Subject) # first we need to split in a list
map_df(l_dat, function(x) {
n_cols <- x$UniqueNumber # finds the number of columns
x <- as.numeric(x[2:(n_cols+1)]) # subsets x and converts to numeric
mean(x, na.rm=T) # mean to be returned
})
# output:
# # A tibble: 1 x 3
# `1` `2` `3`
# <dbl> <dbl> <dbl>
# 1 0.667 0.5 1
Another option (output format closer to a dplyr solution):
map_df(l_dat, function(x) {
n_cols <- x$UniqueNumber
id <- x$Subject
x <- as.numeric(x[2:(n_cols+1)])
tibble(id=id, mean_values=mean(x, na.rm=T))
})
# # A tibble: 3 x 2
# id mean_values
# <int> <dbl>
# 1 1 0.667
# 2 2 0.5
# 3 3 1
Just as an example I added a sum() then divided by length(x)-1:
map_df(l_dat, function(x) {
n_cols <- x$UniqueNumber
id <- x$Subject
x <- as.numeric(x[2:(n_cols+1)])
tibble(id=id,
mean_values=sum(x, na.rm=T)/(length(x)-1)) # change here
})
# # A tibble: 3 x 2
# id mean_values
# <int> <dbl>
# 1 1 1.
# 2 2 1.
# 3 3 Inf #beware of this case where you end up dividing by 0
Data:
tt <- "Subject Value1 Value2 Value3 UniqueNumber
001 1 0 1 3
002 0 1 1 2
003 1 1 1 1"
dat <- read.table(text=tt, header=T)

Answer 6

I think the easiest way is to set to NA the zeros that really should be NA, then use rowSums and rowMeans on the appropriate subset of columns.
Data[2:4][(col(dat[2:4])>dat[[5]])] <- NA
Data
# Subject Value1 Value2 Value3 UniqueNumber
# 1 1 1 0 1 3
# 2 2 0 1 NA 2
# 3 3 1 NA NA 1
library(dplyr)
Data%>%
mutate(sum = rowSums(.[2:4], na.rm = TRUE),
mean = rowMeans(.[2:4], na.rm = TRUE))
# Subject Value1 Value2 Value3 UniqueNumber sum mean
# 1 1 1 0 1 3 2 0.6666667
# 2 2 0 1 NA 2 1 0.5000000
# 3 3 1 NA NA 1 1 1.0000000
or transform(Data, sum = rowSums(Data[2:4],na.rm = TRUE), mean = rowMeans(Data[2:4],na.rm = TRUE)) to stay in base R.
data
Data <- structure(
list(Subject = 1:3,
Value1 = c(1L, 0L, 1L),
Value2 = c(0L, 1L, NA),
Value3 = c(1L, NA, NA),
UniqueNumber = c(3L, 2L, 1L)),
.Names = c("Subject","Value1", "Value2", "Value3", "UniqueNumber"),
row.names = c(NA, 3L), class = "data.frame")