如何计算 R 中的多个组的条件熵 ..我在哪里 go 错了

Question

I've read many questions related to mine, however I can't figure out what's wrong with my code

The package I use is "dplyr" & "infotheo"

Usage of infotheo here is condentropy(time2, time1)

my data is like

id <- c("1", "1", "1", "1", "2", "2", "2", "2", "3", "3", "3", "3")
cond <- c("1", "2", "1", "2", "1", "2", "1", "2", "1", "2", "1", "2")
time1 <- c("1", "3", "3", "2", "3", "3", "1", "1", "1", "2", "2", "1")
time2 <- c("3", "3", "2", "3", "3", "1", "1", "1", "2", "2", "1" ,"1")
df <- data.frame(id, cond, time1, time2)

I want to calculate it by id & condition, which means I'll get 6 entropy values from 3 person with two conditions. Here is my code

df %>%
group_by(df$id, df$cond) %>%
summarize(condentropy(df$time2, df$time1))

why I only got one value for all the groups?

Thanks for the help in advance!

Answer 1

Something like this

First, convert you data to numics

df <- df %>% type_convert()

-- Column specification ------------------------------------------
cols(
  id = col_double(),
  cond = col_double(),
  time1 = col_double(),
  time2 = col_double()
)

Second, get at finding relevant means,

df  %>%
    group_by(id, cond) %>%
    summarise(mean = mean(id))
`summarise()` has grouped output by 'id'. You can override using the `.groups` argument.
# A tibble: 6 x 3
# Groups:   id [3]
     id  cond  mean
  <dbl> <dbl> <dbl>
1     1     1     1
2     1     2     1
3     2     1     2
4     2     2     2
5     3     1     3
6     3     2     3

Third, study this page for addition examples.

Answer 2

Convert the time columns to numeric, perform the grouping and summarize. Do not use df$ with dplyr verbs and be sure to assign the value of condentropy(...) to a column name. The subject of the question refers to mean but the code suggests you want to calculate the conditional entropy so we provide both.

library(dplyr)
library(infotheo)

df %>%
  mutate(time1 = as.numeric(time1), time2 = as.numeric(time2)) %>%
  group_by(id, cond) %>%
  summarize(cond_ent = condentropy(time2, time1), 
            mean1 = mean(time1), mean2 = mean(time2), .groups = "drop")

Answer 3

Use type.convert(as.is = TRUE) to get numeric variables and then summarise with across : You don't have to use $

This one:

library(dplyr)
library(infotheo)
df %>% 
  as_tibble() %>% 
  type.convert(as.is=TRUE) %>% 
  group_by(id, cond) %>% 
  summarise(mean = mean(c(time1, time2)))

Output:

     id  cond  mean
  <int> <int> <dbl>
1     1     1  2.25
2     1     2  2.75
3     2     1  2   
4     2     2  1.5 
5     3     1  1.5 
6     3     2  1.5 

OR

library(dplyr)
df %>% 
  as_tibble() %>% 
  type.convert(as.is=TRUE) %>% 
  group_by(id, cond) %>% 
  summarise(across(starts_with("time"), mean))

Output:

     id  cond time1 time2
  <int> <int> <dbl> <dbl>
1     1     1   2     2.5
2     1     2   2.5   3  
3     2     1   2     2  
4     2     2   2     1  
5     3     1   1.5   1.5
6     3     2   1.5   1.5