问题描述
我似乎在使用正确的语法时遇到了问题，该语法使我可以将现有列复制到data.table原始数据集中具有新名称的新列中。
请参考以下示例：
R data.table - new column with ':=' and keep existing column 这些示例如下：
如果您想使用：=进行替换，则可以 做
BertTokenizer.save_pretrained("OUTPUT_DIR")如果我们只需要一个步骤，则无需执行：=来修改 原始数据集，请使用=内部列表或。（
dt[,x:=mpg][,setdiff(colnames(dt),c('x','mpg')) := NULL]或者如果有必要在原始数据集中创建列，则可以 管道
dt[,.(x = mpg,mpg)]
将data.table与以下示例数据一起使用：
dt[,x := mpg][,.(x,mpg)]
> head(sample_dt)
date_of_birth date_employed Total AVC numtrans firstfundingdate minAccountingdte
1 10/01/1988 16/08/2013 490909.6 0 61 25/11/2014 31/10/2014
2 26/12/1971 08/01/2001 4400292.1 0 175 19/08/2006 28/02/2006
3 15/10/1979 14/01/2005 92240.0 0 44 25/10/2006 31/01/2005
4 04/04/1973 30/04/2002 1594627.9 0 158 18/09/2012 30/04/2007
5 22/02/1972 22/02/1996 627662.7 0 126 27/02/2007 31/10/2006
6 07/06/1976 01/03/2010 3735319.2 0 129 13/05/2010 31/03/2010
gender client_status Balance
1 F C 626567.9
2 M C 9955518.3
3 F C 385284.5
4 M C 3097565.4
5 M C 1815569.6
6 M C 7132986.0
其他数据结构信息：
> head(sample_dt)
date_of_birth date_employed Total AVC numtrans firstfundingdate minAccountingdte
1 10/01/1988 16/08/2013 490909.6 0 61 25/11/2014 31/10/2014
2 26/12/1971 08/01/2001 4400292.1 0 175 19/08/2006 28/02/2006
3 15/10/1979 14/01/2005 92240.0 0 44 25/10/2006 31/01/2005
4 04/04/1973 30/04/2002 1594627.9 0 158 18/09/2012 30/04/2007
5 22/02/1972 22/02/1996 627662.7 0 126 27/02/2007 31/10/2006
6 07/06/1976 01/03/2010 3735319.2 0 129 13/05/2010 31/03/2010
gender client_status Balance dob employment_dte firstfunding_dte minAccounting_dte
1 F C 626567.9 1988-01-11 2013-08-16 2014-11-25 2014-10-31
2 M C 9955518.3 1971-12-26 2001-01-08 2006-08-19 2006-02-28
3 F C 385284.5 1979-10-15 2005-01-14 2006-10-25 2005-01-31
4 M C 3097565.4 1973-04-04 2002-04-30 2012-09-18 2007-04-30
5 M C 1815569.6 1972-02-22 1996-02-22 2007-02-27 2006-10-31
6 M C 7132986.0 1976-06-07 2010-03-01 2010-05-13 2010-03-31
但是，当我运行以下命令时，出现错误：
> dput(head(sample_dt))
structure(list(date_of_birth = c("10/01/1988","26/12/1971","15/10/1979","04/04/1973","22/02/1972","07/06/1976"),date_employed = c("16/08/2013","08/01/2001","14/01/2005","30/04/2002","22/02/1996","01/03/2010"
),Total = c(490909.59,4400292.09,92240,1594627.95,627662.74,3735319.25),AVC = c(0,0),numtrans = c(61L,175L,44L,158L,126L,129L),firstfundingdate = c("25/11/2014","19/08/2006","25/10/2006","18/09/2012","27/02/2007","13/05/2010"),minAccountingdte = c("31/10/2014","28/02/2006","31/01/2005","30/04/2007","31/10/2006","31/03/2010"
),gender = c("F","M","F","M"),client_status = c("C","C","C"),Balance = c(626567.94,9955518.35,385284.46,3097565.35,1815569.61,7132985.99)),row.names = c(NA,6L),class = c("data.table","data.frame"))
和
> sample_dt[,dob:=date_of_birth][,setdiff(colnames(sample_dt),c('dob','date_of_birth')) := NULL]
Error in `:=`(dob,date_of_birth) : Could not find function ":="
> sample_dt[,c("dob","date_of_birth")) := NULL]
Error in `:=`(dob,date_of_birth) : Could not find function ":="
也
> sample_dt[,.(dob = date_of_birth,date_of_birth)]
Error in .(dob = date_of_birth,date_of_birth) :
Could not find function "."
如果可能的话...我希望能够同时运行所有这些列副本：
> sample_dt[,dob := date_of_birth][,.(dob,date_of_birth)]
Error in `:=`(dob,date_of_birth) : Could not find function ":="
...在继续进行操作之前，将它们转换为带有以下内容的日期列：
sample_dt[,employment_dte := date_employed][,.(employment_dte,date_employed)]
sample_dt[,firstfunding_dte := firstfundingdate][,.(firstfunding_dte,firstfundingdate)]
sample_dt[,minAccounting_dte := minAccountingdte][,.(minAccounting_dte,minAccountingdte)]
我在做什么错了？
任何帮助将不胜感激！
解决方法
我无法重现您的示例中的错误：
library("data.table")
sample_dt <-
data.table(
date_of_birth = c("10/01/1988","26/12/1971","15/10/1979","04/04/1973","22/02/1972","07/06/1976"),date_employed = c("16/08/2013","08/01/2001","14/01/2005","30/04/2002","22/02/1996","01/03/2010"),Total = c(490909.59,4400292.09,92240,1594627.95,627662.74,3735319.25),AVC = c(0,0),numtrans = c(61L,175L,44L,158L,126L,129L),firstfundingdate = c("25/11/2014","19/08/2006","25/10/2006","18/09/2012","27/02/2007","13/05/2010"),minAccountingdte = c("31/10/2014","28/02/2006","31/01/2005","30/04/2007","31/10/2006","31/03/2010"),gender = c("F","M","F","M"),client_status = c("C","C","C"),Balance = c(626567.94,9955518.35,385284.46,3097565.35,1815569.61,7132985.99))
###
sample_dt[,dob:=date_of_birth][,setdiff(colnames(sample_dt),c('dob','date_of_birth')) := NULL]
sample_dt[,.(dob = date_of_birth,date_of_birth)]
sample_dt[,dob := date_of_birth][,.(dob,date_of_birth)] ## warning
只有最后一行会发出警告。
最终您正在寻找这样的东西：
library("data.table")
Iris <- iris; setDT(Iris); head(Iris)
Iris[,':='(S=Species,SL=Sepal.Length)][]