Removing rows if certain elements are found in character string

You will have to change the 'i1' expression as follows:

i1 <- grepl("^([0D]|[0d])*$", dd$ch)
i1  # matches strings with d & D in them

  [1]  TRUE  TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE

# second string had 'd' & 'D' in it so it was TRUE above and FALSE below
i1new <- grepl("^([0D]*$|[0d]*$)", dd$ch)
i1new

  [1]  TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE

I put a 'd' and 'D' in the second string and the original regular
expression is equivalent to

grepl("^[0dD]*$", dd$ch)

which will match strings containing d, D and 0.  If you only want 'd'
or 'D' (and not both), then you will have to use the one in 'i1new'.

On Mon, Jul 2, 2012 at 7:24 PM, Rui Barradas <ruipbarradas at sapo.pt> wrote:

Hello,

Try regular expressions instead.
In this data.frame, I've changed row nr.4 to have a row with 'D' as first
non-zero character.

dd <- read.table(text="

ch     count
1  0000000000D0000000000000000000000000000000000000 0.007368
2  0000000000d0000000000000000000000000000000000000 0.002456
3  000000000T00000000000000000000000000000000000000 0.007368
4  000000000DT0000000000000000000000000000000000000 0.007368

5  000000000T00000000000000000000000000000000000000 0.002456
6  000000000Td0000000000000000000000000000000000000 0.002456
7  00000000T000000000000000000000000000000000000000 0.007368
8  00000000T0D0000000000000000000000000000000000000 0.007368
9  00000000T000000000000000000000000000000000000000 0.002456
10 00000000T0d0000000000000000000000000000000000000 0.002456
", header=TRUE)
dd

i1 <- grepl("^([0D]|[0d])*$", dd$ch)
i2 <- grepl("^0*[Dd]", dd$ch)

dd[!i1, ]
dd[!i2, ]
dd[!(i1 | i2), ]


Hope this helps,

Rui Barradas

Em 02-07-2012 23:48, Claudia Penaloza escreveu:

I would like to remove rows from the following data frame (df) if there
are
only two specific elements found in the df$ch character string (I want to
remove rows with only "0" & "D" or "0" & "d"). Alternatively, I would like
to remove rows if the first non-zero element is "D" or "d".


                                                   ch     count
1  0000000000D0000000000000000000000000000000000000 0.007368;
2  0000000000d0000000000000000000000000000000000000 0.002456;
3  000000000T00000000000000000000000000000000000000 0.007368;
4  000000000TD0000000000000000000000000000000000000 0.007368;
5  000000000T00000000000000000000000000000000000000 0.002456;
6  000000000Td0000000000000000000000000000000000000 0.002456;
7  00000000T000000000000000000000000000000000000000 0.007368;
8  00000000T0D0000000000000000000000000000000000000 0.007368;
9  00000000T000000000000000000000000000000000000000 0.002456;
10 00000000T0d0000000000000000000000000000000000000 0.002456;


I tried the following but it doesn't work if there is more than one
character per string:

df <- df[!df$ch %in% c("0","D"),]
df <- df[!df$ch %in% c("0","d"),]

Any help greatly appreciated,
Claudia

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