[英]calculate median from data.table columns in R
I am trying to calculate a median value across a number of columns, however my data is a bit funky. 我试图计算多个列的中值,但我的数据有点时髦。 It looks like the following example. 它看起来像下面的例子。
library(data.table)
dt <- data.table("ID" = c(1,2,3,4),"none" = c(0,5,5,3),
"ten" = c(3,2,5,4),"twenty" = c(0,2,3,1))
ID none ten twenty
1: 1 0 3 0
2: 2 5 2 2
3: 3 5 5 3
4: 4 3 4 1
In the table to column represents the number of occurrences of that value. 在表中,列表示该值的出现次数。 I am wanting to calculate the median occurrence. 我想计算中位数。
For example for ID = 1 例如,对于ID = 1
median(c(10, 10, 10))
is the calculation I am wanting to create. 是我想要创建的计算。
for ID = 2 对于ID = 2
median(c(0, 0, 0, 0, 0, 10, 10, 20, 20))
I have tried using rep()
and lapply()
with very limited success and am after some clear guidance on how this might be achieved. 我尝试过使用rep()
和lapply()
取得了非常有限的成功,并且已经明确指出了如何实现这一目标。 I understand for the likes of rep()
I would be having to hard code my value to be repeated (eg rep(0,2)
or rep(10,2)
) and this is what I expect. 我理解为rep()
我将不得不硬编码我的值重复(例如rep(0,2)
或rep(10,2)
),这就是我所期望的。 I am just struggling to create a list or vector with the repetitions from each column. 我正在努力创建一个包含每列重复的列表或向量。
Here's another data.table
way (assuming unique ID
): 这是另一种data.table
方式(假设唯一ID
):
dt[, median(rep(c(0, 10, 20), c(none, ten, twenty))), by=ID]
# ID V1
# 1: 1 10
# 2: 2 0
# 3: 3 10
# 4: 4 10
This is just an attempt to get @eddi's answer without reshaping (which I tend to use as a last resort). 这只是试图获得@ eddi的答案而不重塑(我倾向于使用它作为最后的手段)。
You need a dictionary to translate column names to corresponding numbers, and then it's fairly straightforward: 您需要一个字典来将列名转换为相应的数字,然后它非常简单:
dict = data.table(name = c('none', 'ten', 'twenty'), number = c(0, 10, 20))
melt(dt, id.var = 'ID')[
dict, on = c(variable = 'name')][, median(rep(number, value)), by = ID]
# ID V1
#1: 1 10
#2: 2 0
#3: 3 10
#4: 4 10
Here's a way that avoids by-row operations and reshaping: 这是一种避免行间操作和重新整形的方法:
dt[, m := {
cSD = Reduce(`+`, .SD, accumulate=TRUE)
k = floor(cSD[[length(.SD)]]/2)
m = integer(.N)
for(i in seq_along(cSD)) {
left = m == 0L
if(!any(left)) break
m[left] = i * (cSD[[i]][left] >= k[left])
}
names(.SD)[m]
}, .SDcols=none:twenty]
which gives 这使
ID none ten twenty m
1: 1 0 3 0 ten
2: 2 5 2 2 none
3: 3 5 5 3 ten
4: 4 3 4 1 ten
For the loop, I'm borrowing @alexis_laz' style, eg https://stackoverflow.com/a/30513197/ 对于循环,我借用@alexis_laz'样式,例如https://stackoverflow.com/a/30513197/
I've skipped translation of the column names, but that's pretty straightforward. 我已经跳过了列名的翻译,但这非常简单。 You could use c(0,10,20)
instead of names(.SD)
at the end. 您最后可以使用c(0,10,20)
而不是names(.SD)
。
Here is a rowwise
dplyr
way: 这里是一个rowwise
dplyr
方式:
dt %>% rowwise %>%
do(med = median(c(rep(0, .$none), rep(10, .$ten), rep(20, .$twenty)))) %>%
as.data.frame
med
1 10
2 0
3 10
4 10
Inspired by @Arun's answer, this is also working: 受@ Arun的回答启发,这也有效:
dt %>% group_by(ID) %>%
summarise(med = median(rep(c(0, 10, 20), c(none, ten, twenty))))
Source: local data table [4 x 2]
ID med
(dbl) (dbl)
1 1 10
2 2 0
3 3 10
4 4 10
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