我正在尝试在DF2$time_expected的输出中显示另一列DF$time/DF2$time_expected和DF[DF$Experiment=="A", ]。
显示某些命令输出的代码
library(data.table)
library('dplyr')
ow <- options("warn")
DF <- read.csv(text=
"Field,time,T,Experiment
Acute,0,0,A
An,9,120,A
En,15.6,2,A
Fo,9.2,2,A
Acute,8.3,1,B
An,7.7,26,B
En,12.9,1,B
Fo,0,0,B
Acute,7.5,1,C
An,7.9,43,C
En,0,0,C
Fo,5.4,1,C
Acute,8.6,2,D
An,7.8,77,D
En,0,0,D
Fo,0,0,D
Acute,0,0,E
An,7.9,60,E
En,14.3,1,E
Fo,0,0,E
Acute,8.3,4,F
An,8.2,326,F
En,14.6,4,F
Fo,7.9,3,F",
header=TRUE, sep=",")
# http://stackoverflow.com/a/43695774/54964
DF[DF$Experiment=="A", ]
# Field time T Experiment
#1 Acute 0.0 0 A
#2 An 9.0 120 A
#3 En 15.6 2 A
#4 Fo 9.2 2 A
# TODO integrate here relative values of DF$time/DF2$time_expected as another column
DF2 <- read.csv(text=
"Field,time_expected
Acute,6
An,6
En,6
Fo,5",
header=TRUE, sep=",")
#DF2[DF2$Field=="An", ]
## Now compare DF.time to DF2.time_expected by DF.time/DF2.time_expected
DF$time/DF2$time_expected
# [1] 0.000000 1.500000 2.600000 1.840000 1.383333 1.283333 2.150000 0.000000
# [9] 1.250000 1.316667 0.000000 1.080000 1.433333 1.300000 0.000000 0.000000
#[17] 0.000000 1.316667 2.383333 0.000000 1.383333 1.366667 2.433333 1.580000
两个新列（time_expected和time/time_expected）
Field time T Experiment time_expected time/time_expected
1 Acute 0.0 0 A 6 0.0
2 An 9.0 120 A 6 1.5
3 En 15.6 2 A 6 2.6
4 Fo 9.2 2 A 5 1.84
答案 0 :(得分：2)
将data.table和dplyr加在一起通常不是一个好主意。
使用dplyr，我们可以：
library(dplyr) # No need to quote
final_df <- DF %>%
filter(time < 8) %>%
full_join(DF2) %>%
mutate(`time/time_expected` = time / time_expected) %>%
select(Field, `time/time_expected`)
final_df
#> Joining, by = "Field"
#> Field time T Experiment time_expected time/time_expected
#> 1 Acute 0.0 0 A 6 0.000000
#> 2 An 7.7 26 B 6 1.283333
#> 3 Fo 0.0 0 B 5 0.000000
#> 4 Acute 7.5 1 C 6 1.250000
#> 5 An 7.9 43 C 6 1.316667
#> 6 En 0.0 0 C 6 0.000000
#> 7 Fo 5.4 1 C 5 1.080000
#> 8 An 7.8 77 D 6 1.300000
#> 9 En 0.0 0 D 6 0.000000
#> 10 Fo 0.0 0 D 5 0.000000
#> 11 Acute 0.0 0 E 6 0.000000
#> 12 An 7.9 60 E 6 1.316667
#> 13 Fo 0.0 0 E 5 0.000000
#> 14 Fo 7.9 3 F 5 1.580000
答案 1 :(得分：1)
我们可以与data.table进行联接并通过将:=）的输出分配给time/time_expected）来创建新列，以及时间N＆＃39;
setDT(DF)[DF2, time_expected := time_expected, on = .(Field)]
DF[, timeN := time/time_expected]
head(DF, 4)
# Field time T Experiment time_expected timeN
#1: Acute 0.0 0 A 6 0.00
#2: An 9.0 120 A 6 1.50
#3: En 15.6 2 A 6 2.60
#4: Fo 9.2 2 A 5 1.84
或者可以在on
setDT(DF)[DF2, c('time_expected', 'timeN') := .(time_expected,time/time_expected), on = .(Field)]
DF
# Field time T Experiment time_expected timeN
# 1: Acute 0.0 0 A 6 0.000000
# 2: An 9.0 120 A 6 1.500000
# 3: En 15.6 2 A 6 2.600000
# 4: Fo 9.2 2 A 5 1.840000
# 5: Acute 8.3 1 B 6 1.383333
# 6: An 7.7 26 B 6 1.283333
# 7: En 12.9 1 B 6 2.150000
# 8: Fo 0.0 0 B 5 0.000000
# 9: Acute 7.5 1 C 6 1.250000
#10: An 7.9 43 C 6 1.316667
#11: En 0.0 0 C 6 0.000000
#12: Fo 5.4 1 C 5 1.080000
#13: Acute 8.6 2 D 6 1.433333
#14: An 7.8 77 D 6 1.300000
#15: En 0.0 0 D 6 0.000000
#16: Fo 0.0 0 D 5 0.000000
#17: Acute 0.0 0 E 6 0.000000
#18: An 7.9 60 E 6 1.316667
#19: En 14.3 1 E 6 2.383333
#20: Fo 0.0 0 E 5 0.000000
#21: Acute 8.3 4 F 6 1.383333
#22: An 8.2 326 F 6 1.366667
#23: En 14.6 4 F 6 2.433333
#24: Fo 7.9 3 F 5 1.580000
对行进行子集化，其中＆＃39; time＆＃39;小于8
DF[time < 8]
# Field time T Experiment time_expected timeN
# 1: Acute 0.0 0 A 6 0.000000
# 2: An 7.7 26 B 6 1.283333
# 3: Fo 0.0 0 B 5 0.000000
# 4: Acute 7.5 1 C 6 1.250000
# 5: An 7.9 43 C 6 1.316667
# 6: En 0.0 0 C 6 0.000000
# 7: Fo 5.4 1 C 5 1.080000
# 8: An 7.8 77 D 6 1.300000
# 9: En 0.0 0 D 6 0.000000
#10: Fo 0.0 0 D 5 0.000000
#11: Acute 0.0 0 E 6 0.000000
#12: An 7.9 60 E 6 1.316667
#13: Fo 0.0 0 E 5 0.000000
#14: Fo 7.9 3 F 5 1.580000