我有一个带有列标题的 data.frame。
如何从 data.frame 中获取特定行作为列表(列标题作为列表的键)?
具体来说,我的 data.frame 是
美国广播公司 1 5 4.25 4.5 2 3.5 4 2.5 3 3.25 4 4 4 4.25 4.5 2.25 5 1.5 4.5 3
我想得到一个相当于
> c(a=5, b=4.25, c=4.5)
a b c
5.0 4.25 4.5
x[r,]
其中 r 是您感兴趣的行。试试这个,例如:
#Add your data
x <- structure(list(A = c(5, 3.5, 3.25, 4.25, 1.5 ),
B = c(4.25, 4, 4, 4.5, 4.5 ),
C = c(4.5, 2.5, 4, 2.25, 3 )
),
.Names = c("A", "B", "C"),
class = "data.frame",
row.names = c(NA, -5L)
)
#The vector your result should match
y<-c(A=5, B=4.25, C=4.5)
#Test that the items in the row match the vector you wanted
x[1,]==y
逻辑索引非常R-ish。尝试:
x[ x$A ==5 & x$B==4.25 & x$C==4.5 , ]
或者:
subset( x, A ==5 & B==4.25 & C==4.5 )
尝试:
> d <- data.frame(a=1:3, b=4:6, c=7:9)
> d
a b c
1 1 4 7
2 2 5 8
3 3 6 9
> d[1, ]
a b c
1 1 4 7
> d[1, ]['a']
a
1 1
如果您不知道行号,但知道一些值,那么您可以使用子集
x <- structure(list(A = c(5, 3.5, 3.25, 4.25, 1.5 ),
B = c(4.25, 4, 4, 4.5, 4.5 ),
C = c(4.5, 2.5, 4, 2.25, 3 )
),
.Names = c("A", "B", "C"),
class = "data.frame",
row.names = c(NA, -5L)
)
subset(x, A ==5 & B==4.25 & C==4.5)
10 年后 ---> 使用 tidyverse 我们可以简单地实现这一点,并从Christopher Bottoms那里借用一片叶子。为了更好地掌握,请参阅slice()
。
library(tidyverse)
x <- structure(list(A = c(5, 3.5, 3.25, 4.25, 1.5 ),
B = c(4.25, 4, 4, 4.5, 4.5 ),
C = c(4.5, 2.5, 4, 2.25, 3 )
),
.Names = c("A", "B", "C"),
class = "data.frame",
row.names = c(NA, -5L)
)
x
#> A B C
#> 1 5.00 4.25 4.50
#> 2 3.50 4.00 2.50
#> 3 3.25 4.00 4.00
#> 4 4.25 4.50 2.25
#> 5 1.50 4.50 3.00
y<-c(A=5, B=4.25, C=4.5)
y
#> A B C
#> 5.00 4.25 4.50
#The slice() verb allows one to subset data row-wise.
x <- x %>% slice(1) #(n) for the nth row, or (i:n) for range i to n, (i:n()) for i to last row...
x
#> A B C
#> 1 5 4.25 4.5
#Test that the items in the row match the vector you wanted
x[1,]==y
#> A B C
#> 1 TRUE TRUE TRUE
由reprex 包(v0.3.0)于 2020-08-06 创建