我有一个data.table
对象,其中包含多个指定唯一案例的列。在下面的小示例中,变量" name
"," job
"和" sex
"指定唯一ID。我想添加缺失的行,以便每个案例对另一个变量的每个可能实例都有一行," from
" (类似于expand.grid
)。
library(data.table)
set.seed(1)
mydata <- data.table(name = c("john","john","john","john","mary","chris","chris","chris"),
job = c("teacher","teacher","teacher","teacher","police","lawyer","lawyer","doctor"),
sex = c("male","male","male","male","female","female","male","male"),
from = c("NYT","USAT","BG","TIME","USAT","BG","NYT","NYT"),
score = rnorm(8))
setkeyv(mydata, cols=c("name","job","sex"))
mydata[CJ(unique(name, job, sex), unique(from))]
这是当前的data.table对象:
> mydata
name job sex from score
1: john teacher male NYT -0.6264538
2: john teacher male USAT 0.1836433
3: john teacher male BG -0.8356286
4: john teacher male TIME 1.5952808
5: mary police female USAT 0.3295078
6: chris lawyer female BG -0.8204684
7: chris lawyer male NYT 0.4874291
8: chris doctor male NYT 0.7383247
以下是我想要的结果:
> mydata
name job sex from score
1: john teacher male NYT -0.6264538
2: john teacher male USAT 0.1836433
3: john teacher male BG -0.8356286
4: john teacher male TIME 1.5952808
5: mary police female NYT NA
6: mary police female USAT 0.3295078
7: mary police female BG NA
8: mary police female TIME NA
9: chris lawyer female NYT -NA
10: chris lawyer female USAT -NA
11: chris lawyer female BG -0.8204684
12: chris lawyer female TIME -NA
13: chris lawyer male NYT 0.4874291
14: chris lawyer male USAT NA
15: chris lawyer male BG NA
16: chris lawyer male TIME NA
17: chris doctor male NYT 0.7383247
18: chris doctor male USAT NA
19: chris doctor male BG NA
20: chris doctor male TIME NA
以下是我尝试的内容:
setkeyv(mydata, cols=c("name","job","sex"))
mydata[CJ(unique(name, job, sex), unique(from))]
但是我收到以下错误并添加fromLast = TRUE(或FALSE)并没有给我正确的解决方案:
Error in unique.default(name, job, sex) :
'fromLast' must be TRUE or FALSE
以下是我遇到的相关答案(但似乎没有一个处理多个键控列): add missing rows to a data table
答案 0 :(得分:4)
这里有几种可能性 - https://github.com/Rdatatable/data.table/pull/814
CJ.dt = function(...) {
rows = do.call(CJ, lapply(list(...), function(x) if(is.data.frame(x)) seq_len(nrow(x)) else seq_along(x)));
do.call(data.table, Map(function(x, y) x[y], list(...), rows))
}
setkey(mydata, name, job, sex, from)
mydata[CJ.dt(unique(data.table(name, job, sex)), unique(from))]
# name job sex from score
# 1: chris doctor male NYT 0.7383247
# 2: chris doctor male BG NA
# 3: chris doctor male TIME NA
# 4: chris doctor male USAT NA
# 5: chris lawyer female NYT NA
# 6: chris lawyer female BG -0.8204684
# 7: chris lawyer female TIME NA
# 8: chris lawyer female USAT NA
# 9: chris lawyer male NYT 0.4874291
#10: chris lawyer male BG NA
#11: chris lawyer male TIME NA
#12: chris lawyer male USAT NA
#13: john teacher male NYT -0.6264538
#14: john teacher male BG -0.8356286
#15: john teacher male TIME 1.5952808
#16: john teacher male USAT 0.1836433
#17: mary police female NYT NA
#18: mary police female BG NA
#19: mary police female TIME NA
#20: mary police female USAT 0.3295078
答案 1 :(得分:4)
tidyr的dev版本现在有一种优雅的方式来执行此操作,因为expand()
函数现在支持嵌套和交叉:
library(dplyr)
mydata <- data_frame(
name = c("john","john","john","john","mary","chris","chris","chris"),
job = c("teacher","teacher","teacher","teacher","police","lawyer","lawyer","doctor"),
sex = c("male","male","male","male","female","female","male","male"),
from = c("NYT","USAT","BG","TIME","USAT","BG","NYT","NYT"),
score = rnorm(8)
)
mydata %>%
expand(c(name, job, sex), from) %>%
left_join(mydata)
#> Joining by: c("name", "job", "sex", "from")
#> Source: local data frame [20 x 5]
#>
#> name job sex from score
#> 1 chris doctor male BG NA
#> 2 chris doctor male NYT 0.5448206
#> 3 chris doctor male TIME NA
#> 4 chris doctor male USAT NA
#> 5 chris lawyer female BG 1.2015173
#> 6 chris lawyer female NYT NA
#> 7 chris lawyer female TIME NA
#> 8 chris lawyer female USAT NA
#> 9 chris lawyer male BG NA
#> 10 chris lawyer male NYT -1.0930237
#> 11 chris lawyer male TIME NA
#> 12 chris lawyer male USAT NA
#> 13 john teacher male BG 1.1345461
#> 14 john teacher male NYT 1.3032946
#> 15 john teacher male TIME 2.4901830
#> 16 john teacher male USAT -1.6449096
#> 17 mary police female BG NA
#> 18 mary police female NYT NA
#> 19 mary police female TIME NA
#> 20 mary police female USAT -0.2443080
答案 2 :(得分:0)
一种可能性是paste
列name
,job
和sex
在一起,得到unique
值,然后{{1} CJ
值为unique
的{{1}}。之后,使用from
中的cSplit
将library(splitstackshape)
列拆分回三列,将这些列重命名为pasted
,将setnames
重命名为join
设置mydata
后。
key