我可以将列表中的数据添加到数据框的列吗

Question

I've written code to extract data from a dataframe in a for loop and added it to a list. Like this:

  list_vraag1 <- list()        
  for (i in seq(1,nrow(questionanswer),16)){        
    if (questionanswer$answer[i] == 1){        
     antwoord1 <- "Ja"
      } else if (questionanswer$answer[i+1] == 1){        
    antwoord1 <- "Nee"
      } else{        
       antwoord1 <- NA
      }          
      df <- append(list_vraag1, antwoord1)
    }

It gives a correct list (I created 6 of those lists for the six different questions), now I've tried to add them to a new dataframe where I need to have 6 different columns and under each column there has to be one whole list. Is this possible? When I tried this, the whole list came on the column name. Does someone has a solution for this?

Answer 1

It's not entirely clear what you are after here. Taking every 16th row is a bit puzzling to me but I am sure it has utility in your application. I am assuming that questionanswer is either a vector or a tibble (data frame) containing a 1 or a 0 (or anything not equal to 1). Under those assumptions:

library(dplyr)
library(tibble)

questionanswer <- tibble(
  answer = sample(c(0, 1), replace = TRUE, size = 100)
)

out <- questionanswer %>%
  mutate(antwoord1 = case_when(answer == 1 ~ "Ja",
                               lead(answer, 1) == 1 ~ "Nee",
                               TRUE ~ "NA")) %>%
  slice(which(row_number() %% 16 == 1)) # slices every 16th row

Ok, out is not a list but a tibble. But we could easily coerce this to a list. lapply would probably not be useful here as you are using a lead function ( (questionanswer$answer[i+1] looks at the next element).

With a for loop this could look like this:

list_vraag1 <- list()        
index <- seq(1,nrow(questionanswer), by = 16)
for (i in seq_along(index)){   
  if (questionanswer$answer[index[i]] == 1){        
    antwoord1 <- "Ja"
  } else if (questionanswer$answer[index[i] + 1] == 1){        
    antwoord1 <- "Nee"
  } else{        
    antwoord1 <- NA
  }          
  list_vraag1[[i]] <- antwoord1
}

Which will return a list. Is this helpful?