在列表列中设置操作 [英] set operation within a list column

问题描述

我想在设置操作 -with-a-list-column> this 。

I am trying to do set operations between the vectors stored in a list column like this.

DT  <- data.table(exp = c("exp1", "exp2", "exp2"), 
                  sample = c(1L, 1L, 2L), 
                  listdata = list(c(2L,5L), c(2L,3L,5L,7L), c(1L,2L,6L)))

> DT
    exp sample listdata
1: exp1      1      2,5
2: exp2      1  2,3,5,7
3: exp2      2    1,2,6

虽然很繁琐，但我可以做

while very cumbersome, I can do

DT$inc = list(setdiff(unlist(DT$listdata[2]), unlist(DT$listdata[1])))

并获取值为 c（3,7）的新列表列。但如果我尝试使用

and obtain a new list column with the value c(3,7). But if I try to calculate the difference between the current row and the first row using

DT$inc = list(list(setdiff(unlist(DT$listdata, recursive = FALSE), unlist(DT$listdata[1]))))

预期新列inc

0
c(3,7)
c(1,6)

我得到 c（3,7,1,6） 。显然， unlist 将整个列表列放在一起。任何想法发生了什么？

I get c(3,7,1,6). Apparently unlist flattened the whole list column together. Any idea what's going on?

我也学习了dplyr和data.table。所以，如果你能提供解决方案使用其中之一，真的会有帮助。

I am also learning dplyr and data.table. So it would really help if you can provide solutions using one of them.

推荐答案

[...]我尝试计算当前行和第一个

[...] I try to calculate the difference between the current row and the first row

好吧，你可以...

DT[, inc := .(Map(setdiff, listdata, listdata[1L]))]

#     exp sample listdata inc
# 1: exp1      1      2,5    
# 2: exp2      1  2,3,5,7 3,7
# 3: exp2      2    1,2,6 1,6

但我认为只是不能使用列表列表更好。

But I think it's far better to just not work with list columns.

不能使用列表列可能看起来像...

Not working with list columns might look like...

DT[, r := .I]
DT2 = DT[,c(.SD[rep(.I, lengths(listdata))], .(v = unlist(listdata))), .SDcols=!"listdata"]

#     exp sample r v
# 1: exp1      1 1 2
# 2: exp1      1 1 5
# 3: exp2      1 2 2
# 4: exp2      1 2 3
# 5: exp2      1 2 5
# 6: exp2      1 2 7
# 7: exp2      2 3 1
# 8: exp2      2 3 2
# 9: exp2      2 3 6

然后我们只使用这个数据集， p>

Then we just work with this data set, and can do

DT2[!DT2[r==1L], on="v"]

#     exp sample r v
# 1: exp2      1 2 3
# 2: exp2      1 2 7
# 3: exp2      2 3 1
# 4: exp2      2 3 6

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