按组移动data.table中的列的列 [英] Shift a column of lists in data.table by group

问题描述

我在R中有一列列表：

DT <- data.table(foo = c(list(c("a","b","c")), list(c("b","c")), list(c("a","b")), list(c("a"))), id = c(1,1,2,2))
DT
     foo id
1: a,b,c  1
2:   b,c  1
3:   a,b  2
4:     a  2

想要做的是复制典型的移动行为来获得：

What I would like to do is replicate typical shift behavior to get:

     foo id
1:   b,c  1
2:    NA  1
3:     a  2
4:    NA  2

对于正常列，我将使用shift，但这将列表分成列并移动那些（并且标记一个警告）：

For a normal column I would use shift, but this splits the lists into columns and shifts those (and flags a warning):

DT[ , shift(foo,1,type = "lead"), by = id]
   id V1 V2
1:  1  b  c
2:  1  c NA
3:  1 NA  c
4:  2  b NA
5:  2 NA NA

如果我将shift调用包装到一个列表中，返回是一个列表，但只有向量元素已经移动：

If I wrap the shift call into a list, the return is a list but only the vector elements have been shifted:

DT[ , list(shift(foo,1,type = "lead")), by = id]
   id     V1
1:  1 b,c,NA
2:  1   c,NA
3:  2   b,NA
4:  2     NA

推荐答案

这不止一次。所以我已经开始添加这个功能。您现在必须使用开发版本，v1.9.7 ..请参阅安装说明 here 。

DT[, foo2 := shift(.(foo), type = "lead"), 
       by = id]
#      foo id foo2
# 1: a,b,c  1  b,c
# 2:   b,c  1   NA
# 3:   a,b  2    a
# 4:     a  2   NA

c $ c> foo 为列表中的每个组。注意，它返回一个列表，它可以很好地与：= 如上所示..如果你不添加/ data.table（这没有什么意义），那么你必须提取列表元素。

Just wrap foo for each group in a list. Note that it returns a list-of-list which works well with := as shown above.. If you're not adding/updating your data.table (which doesn't make much sense), then you'll have to extract the list element.

DT[, .(foo2 = shift(.(foo), type="lead")[[1L]]), 
        by = id]
#    id foo2
# 1:  1  b,c
# 2:  1   NA
# 3:  2    a
# 4:  2   NA

shift（）旨在与data.table的：= 语法它始终返回相同的行数。

shift() is designed to play nicely with data.table's := syntax, since it returns the same number of rows all the time.

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