[英]How to calculate conditional entropy by multiple groups in R ..where did I go wrong
我已经阅读了许多与我相关的问题,但是我无法弄清楚我的代码有什么问题
我使用的 package 是“dplyr”和“infotheo”
这里使用 infotheo 是condentropy(time2, time1)
我的数据就像
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)
我想通过 id 和条件来计算它,这意味着我将从 3 个人的两个条件下获得 6 个熵值。 这是我的代码
df %>%
group_by(df$id, df$cond) %>%
summarize(condentropy(df$time2, df$time1))
为什么我对所有组只有一个值?
我在这里先向您的帮助表示感谢!
像这样的东西
首先,将您的数据转换为数字
df <- df %>% type_convert()
-- Column specification ------------------------------------------
cols(
id = col_double(),
cond = col_double(),
time1 = col_double(),
time2 = col_double()
)
其次,要设法寻找相关手段,
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
第三,研究此页面以获取其他示例。
将时间列转换为数字,执行分组和汇总。 不要将 df$ 与 dplyr 动词一起使用,并确保将 condentropy(...) 的值分配给列名。 问题的主题是指均值,但代码建议您要计算条件熵,因此我们提供两者。
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")
使用type.convert(as.is = TRUE)
来获取数值变量,然后用across
summarise
:你不必使用$
这个:
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
或者
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
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