CODEWITHSUNDEEP
rdataframe
chimpsarehungry
chimpsarehungry

Reputation: 1821

Subsetting based on categorical row values in R

If I have a data frame with categorical rows describing the columns, how can I subset it based on those values? Here is my data:

     name1 name2 name3
cond   trt  ctrl  ctrl
hour     0     3     0
A        1     1     1
B        2     2     2
C        3     3     3
D        4     4     4
E        5     5     5
F        6     1     6
G        7     2     7
H        8     3     8
I        9     4     9
J       10     5    10

Recreate it with this line:

A <- structure(list(name1 = c("trt", "0", "1", "2", "3", "4", "5", 
"6", "7", "8", "9", "10"), name2 = c("ctrl", "3", "1", "2", "3", 
"4", "5", "1", "2", "3", "4", "5"), name3 = c("ctrl", "0", "1", 
"2", "3", "4", "5", "6", "7", "8", "9", "10")), .Names = c("name1", 
"name2", "name3"), row.names = c("cond", "hour", "A", "B", "C", 
"D", "E", "F", "G", "H", "I", "J"), class = "data.frame")

Here is what I've attempted:

subset(A, ( A[1,] == 'trt' & A[2,] == 0 ))

My unexplainable result. Why this?:

     name1 name2 name3
cond   trt  ctrl  ctrl
B        2     2     2
E        5     5     5
H        8     3     8

What I want:

     name1 
cond   trt  
hour     0  
A        1  
B        2  
C        3 
D        4  
E        5  
F        6  
G        7  
H        8  
I        9 
J       10

Thank you. This might be easier if the categories of 'cond' and 'hour' were factors. Or if the data was 'melted'. I'm open to any suggestions.

Upvotes: 1

Views: 1321

Answers (2)

CHP
CHP

Reputation: 17189

Alongwith direct subsetting with [ ], you can also use subset's select arguement to select columns

> subset(A, TRUE, select=( A[1,] == 'ctrl' & A[2,] == 3 ))
     name2
cond  ctrl
hour     3
A        1
B        2
C        3
D        4
E        5
F        1
G        2
H        3
I        4
J        5

Upvotes: 1

Simon O&#39;Hanlon
Simon O&#39;Hanlon

Reputation: 59970

Don't need to use subset. Just do:

A <- A[,1,drop=FALSE]

Or if you don't want to hard-code the column number:

A[,A[1,] == 'trt' & A[2,] == 0,drop = FALSE]

Upvotes: 3

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