Bug about a wrong row (or column) was selected in data.frame with character index (rowname or colnam-CodePudding

when I want to select a row with row names as 'CYP3A4', a different row with similar name was selected. Is it a bug for R or a default function? How can I avoid this?

CodePudding user response：

The single quotation mark does not belong to the name, it is just to indicate that you have an object of class character representing text instead of a symbol named CPY3A4.

x <- data.frame(var = seq(4))
rownames(x) <- c("foo", "'foo'", "`foo`", "\"foo\"")
x
#>       var
#> foo     1
#> 'foo'   2
#> `foo`   3
#> "foo"   4
x["foo",]
#> [1] 1
x["'foo'",]
#> [1] 2
x["`foo",]
#> [1] 3
x["\"foo\"",]
#> [1] 4

^{Created on 2022-04-08 by the reprex package (v2.0.0)}

CodePudding user response：

What you're seeing is "partial matching" that occurs when your data is in a dataframe rather than a matrix. Compare for example the behaviour of a matrix and dataframe with:

matrix.a3 <- matrix(1:10, 2, 5, dimnames=list(c("CYP3A33", "CYP3A43"),
                                     paste0("P01",1:5)))

a3 <- data.frame(matrix.a3)
a3
#         P011 P012 P013 P014 P015
# CYP3A33    1    3    5    7    9
# CYP3A43    2    4    6    8   10

a3['CYP3A4',1:5]
#        P011 P012 P013 P014 P015
#CYP3A43    2    4    6    8   10

matrix.a3['CYP3A4',1:5]
#Error in matrix.a3["CYP3A4", 1:5] : subscript out of bounds

This is mentioned in the help file for [.data.frame saying:

Both [ and [[ extraction methods partially match row names. By default neither partially match column names, but [[ will if exact = FALSE (and with a warning if exact = NA). If you want to exact matching on row names use match, as in the examples.

The easiest solution would be to store your data in a matrix if it is all one type, but if that's not suitable you can work around by selecting with a logical index, rather than by string:

a3[rownames(a3) == "CYP3A4", 1:5]
#[1] P011 P012 P013 P014 P015
#<0 rows> (or 0-length row.names)

or by using match:

a3[match("CYP3A4", row.names(a3)), ] 
#   P011 P012 P013 P014 P015
#NA   NA   NA   NA   NA   NA

Note the different results that they give, depending on which is more useful for you