Count occurences in table and show in rmarkdown, scalability

Question

Let's take the following dataframe as an example:

df <- data.frame(state = c('up', 'up', 'up', 'down', 'down', 'up', 'down', 'down'), type = c('A', 'A', 'A', 'A', 'B', 'B', 'B', 'B'))

To count how many times each of the possible combinations of A/B and up/down occur I would do as follows:

a1 <- nrow(subset(subset(df, state == 'up'), type == 'A'))
a2 <- nrow(subset(subset(df, state == 'down'), type == 'A'))
a3 <- nrow(subset(subset(df, state == 'up'), type == 'B'))
a4 <- nrow(subset(subset(df, state == 'down'), type == 'B'))

And to show that in a Rmarkdown table I would do as follows:

| State/type     |   A    |    B   | 
|:--------------:|:------:|:------:|
| Up             | `r a1` | `r a3` |
| Down           | `r a2` | `r a4` |

My question is: is there a better way to do this? This approach becomes very tedious when you have many factor combinations and scales poorly.

To do what? Count occurrences or show them in table? Those are two very different questions. — pogibas, May 18 '18 at 08:07
To count occurrences, the `table` solution posted below seems to be the correct approach. Sorry for the ambiguity. — Marc P, May 18 '18 at 08:15

akrun · Accepted Answer · 2018-05-18T08:44:24.940

We can just use table

tbl <- table(df)

and convert it to a data.frame

as.data.frame.matrix(tbl)

Or if we need the 'state' column as well (instead of row names in the above)

library(tidyverse)
count(df, state, type) %>% 
              spread(type, n)

In the rmarkdown, the above can be used as

```{r echo = FALSE, include = FALSE, cache=FALSE}
df <- data.frame(state = c('up', 'up', 'up', 'down', 'down', 'up', 
   'down', 'down'), type = c('A', 'A', 'A', 'A', 'B', 'B', 'B', 'B'))
library(dplyr)
library(tidyr)
library(kableExtra)
out <- count(df, state, type) %>% 
              spread(type, n)

```



```{r code1, results = 'asis', echo = FALSE}
kable(out, "html") %>%
  kable_styling()
```

Count occurences in table and show in rmarkdown, scalability

1 Answers1