使用data.table的X [Y]语法在R中交错连接/新分组/连接

Question

我有两个data.tables:samples, resources

resources与samplesvia primary和secondaryid 连接.我想首先通过主id将来自资源的信息与sample-table相结合,并且只有当它产生NA时,我才想要从同一个表(在一个data.table命令链中)中使用辅助资源.

# resources:
   primary secondary info
1:      17        42  "I"
2:      18        NA  "J"
3:      19        43  "K"

# samples:
   name primary secondary
1:  "a"      17        55
2:  "b"       0        42
3   "c"      18        42

期望的结果是:

# joined tables:
   name info  # primary secondary
1:  "a"  "I"
2:  "b"  "I"
3:  "c"  "J"

第一个连接通道primary很容易,它产生

# Update:
samples <- data.table(name = letters[1:3], 
                      primary = c(17, 0, 18), 
                      secondary = c(55, 42, 42))
resources <- data.table(primary = 17:19, 
                        secondary = c(42, NA, 43), 
                        info = LETTERS[9:11])
# first join:
setkey(samples, primary)
setkey(resources, primary)
samples[resources]

   name info  # primary secondary
1:  "a"  "I"
2:  "b"   NA
3:  "c"  "J"

但是之后？我需要重新关键样品setkey(samples, secondary),对吗？然后将子集仅限于那些产生NA的行.但是在一个命令链中所有这一切都不可能实现(并且假设有两个以上的标准......).我怎样才能更简洁地实现这一目标？

...使用data.tables的代码更新.

Answer 1

虽然你可以在一条线上完成它,但我认为这会掩盖你所做的事情的意义,让你难以理解的阅读/理解/调试/记住你在一个月内做了什么,这简直是一个坏主意.

更小,更容易消化的块是imo的方式:

setkey(samples, primary)
setkey(resources, primary)
samples[resources, info := i.info]

setkey(samples, secondary)
setkey(resources, secondary)
samples[resources, info := ifelse(is.na(info), i.info, info)]

samples
#   name primary secondary info
#1:    b       0        42    I
#2:    c      18        42    J
#3:    a      17        55    I

# keep going with tertiary and so on if you like

正如@nachti在评论中指出的那样,您可能需要allow.cartesian=TRUE在1.9.5之前添加版本,具体取决于您的数据.