Question

我正在尝试向data.frame添加新列（插槽），具体取决于id的行值。简短的代表性数据如下。

        id                Cap         Rs       R_inv
  257   464A485SSEE3    1.41E-10    736665.125  1.36E-06
  258   464A485SSEE3    1.30E-10    364822.6875 2.74E-06
  289   464A485TSEB2    1.44E-10    111996.1016 8.93E-06
  290   464A485TSEB2    1.33E-10    108541      9.21E-06

我认为我们可以编写一个循环函数来添加新列。但我希望学习简单的功能，如cbind或raw $ slot＆lt; - ???

Answer 1

尝试

 raw$slot <- with(raw, paste0("slot",as.numeric(factor(id))))
 raw
 #              id      Cap       Rs    R_inv  slot
 #257 464A485SSEE3 1.41e-10 736665.1 1.36e-06 slot1
 #258 464A485SSEE3 1.30e-10 364822.7 2.74e-06 slot1
 #289 464A485TSEB2 1.44e-10 111996.1 8.93e-06 slot2
 #290 464A485TSEB2 1.33e-10 108541.0 9.21e-06 slot2

或者如果数据集按id排序，您也可以

 raw$slot <- paste0("slot",cumsum(c(TRUE,raw$id[-1]!=raw$id[-nrow(raw)])))

更新

如果您需要一些自定义标签，可以将id转换为factor（如果不是）并指定您想要的labels

raw$slot <- with(raw,  as.character(factor(id, labels=c('split6', 'split9'))) )
raw$slot
#[1] "split6" "split6" "split9" "split9"

或者只需将numeric转换为factor即可使用numeric索引，并在该索引上使用names的向量。在这里，您需要在执行此操作之前了解级别的顺序。

 with(raw, c('split6', 'split9')[as.numeric(factor(id))])
 #[1] "split6" "split6" "split9" "split9"

数据

 raw <- structure(list(id = c("464A485SSEE3", "464A485SSEE3", "464A485TSEB2", 
 "464A485TSEB2"), Cap = c(1.41e-10, 1.3e-10, 1.44e-10, 1.33e-10
 ), Rs = c(736665.125, 364822.6875, 111996.1016, 108541), R_inv = c(1.36e-06, 
 2.74e-06, 8.93e-06, 9.21e-06)), .Names = c("id", "Cap", "Rs", 
 "R_inv"), class = "data.frame", row.names = c("257", "258", "289", "290"))

Answer 2

如果你有一个长桌，可能会考虑使用“dplyr”包的“join”功能。

# First, here is your data:
df <- structure(list(id = structure(c(1L, 1L, 2L, 2L), .Label = c("464A485SSEE3", 
"464A485TSEB2"), class = "factor"), Cap = c(1.41e-10, 1.3e-10, 
1.44e-10, 1.33e-10), Rs = c(736665.125, 364822.6875, 111996.1016, 
108541), R_inv = c(1.36e-06, 2.74e-06, 8.93e-06, 9.21e-06)), .Names = c("id", 
"Cap", "Rs", "R_inv"), class = "data.frame", row.names = c("257", 
"258", "289", "290"))

#               id      Cap       Rs    R_inv
# 257 464A485SSEE3 1.41e-10 736665.1 1.36e-06
# 258 464A485SSEE3 1.30e-10 364822.7 2.74e-06
# 289 464A485TSEB2 1.44e-10 111996.1 8.93e-06
# 290 464A485TSEB2 1.33e-10 108541.0 9.21e-06


# Then create a matching table similar to below:
match_table <- structure(list(id = structure(1:2, .Label = c("464A485SSEE3", 
"464A485TSEB2"), class = "factor"), slot_no = structure(1:2, .Label = c("slot_1", 
"slot_2"), class = "factor")), .Names = c("id", "slot_no"), class = "data.frame", row.names = c(NA, 
-2L))

#             id slot_no
# 1 464A485SSEE3  slot_1
# 2 464A485TSEB2  slot_2

# Do joining
library(dplyr)
left_join(df, match_table)
#             id      Cap       Rs    R_inv slot_no
# 1 464A485SSEE3 1.41e-10 736665.1 1.36e-06  slot_1
# 2 464A485SSEE3 1.30e-10 364822.7 2.74e-06  slot_1
# 3 464A485TSEB2 1.44e-10 111996.1 8.93e-06  slot_2
# 4 464A485TSEB2 1.33e-10 108541.0 9.21e-06  slot_2

如何添加新列取决于R中的行值？

2 个答案:

更新

数据