CODEWITHSUNDEEP
r
Luca H.
Luca H.

Reputation: 13

R function that counts rows where conditions are met

I am trying to create a new column that counts each column where criteria are met. That is because I want to summarize the number of correct answers by each participant in my master thesis. I am really new to R and in desperate need for help, even on easy tasks.

For Example:

(Participant, Task1, Task2, Task3; COUNT)

1 4 8 1 ; 1|

2 3 8 7 ; 1|

3 1 3 4 ; 2|

4 5 6 4 ; 1|

5 1 8 4 ; 3

The column COUNT should count all correct answers of the rows Task1-Task3. If the correct answers are (1, 8, 4), the COUNT row should result in the numbers shown in the example above.

Can anybody tell me how to create such a variable?

Really appreciated, thanks Luca

Upvotes: 1

Views: 1351

Answers (2)

akrun
akrun

Reputation: 887971

We can use rowSums by making the vector c(1, 8, 4) length same as the 'Task' columns length and do a ==, and get the rowSums

i1 <- startsWith(names(df1), 'Task')
df1$COUNT <-   rowSums(df1[i1] ==  c(1, 8, 4)[col(df1[i1])])
df1$COUNT
#[1] 1 1 2 1 3

Or with sweep

rowSums(sweep(df1[i1], 2, c(1, 8, 4), `==`))

Or another option is apply

df1$COUNT <- apply(df1[i1], 1, function(x) sum(x == c(1, 8, 4)))

NOTE: None of the solutions require any external package

data

df1 <- data.frame(Participant = 1:5, Task1 = c(4, 3, 1, 5, 1),
   Task2 = c(8, 8, 3, 6, 8), Task3 = c(1, 7, 4, 4, 4))

Upvotes: 2

Ronak Shah
Ronak Shah

Reputation: 389325

We can use pmap_int from purrr to count number of correct answers. 

library(dplyr)

df %>% mutate(COUNT = purrr::pmap_int(select(., starts_with('Task')), 
                               ~sum(c(...) == c(1, 8, 4))))


#  Participant Task1 Task2 Task3 COUNT
#1           1     4     8     1     1
#2           2     3     8     7     1
#3           3     1     3     4     2
#4           4     5     6     4     1
#5           5     1     8     4     3

Another option is to get data in long format, calculate the number of correct answers for each Participant and join the data back. 

df1 %>%
  tidyr::pivot_longer(cols = starts_with('Task')) %>%
  group_by(Participant) %>%
  summarise(COUNT = sum(value == c(1, 8, 4))) %>%
  left_join(df1, by = 'Participant')

Upvotes: 0

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