Got it! Thank you Rui!
cp
On Tue, Jul 3, 2012 at 10:14 AM, Rui Barradas <[email protected]> wrote:
> Hello,
>
> I'm glad it helped. See answer inline.
>
> Em 03-07-2012 17:09, Claudia Penaloza escreveu:
>
> Thank you Rui and Jim, both 'i1' and 'i1new' worked perfectly
>> because there are no instances of 'Dd' or 'dD' in the data set (that I
>> would/not want to include/exclude)... but I understand that 'i1new'
>> targets precisely what I want.
>> Why isn't a leader of zero's required for either 'i1' or 'i1new', as so?
>> i1newer <- grepl("^0{0,}[D]*$|^0{0,}[d]*$**", dd$ch)
>>
>>
> Because both 'i1' and 'i1new' test from beginning to end of string,
> allowing only '0' and either 'd' or 'D', but not both (i1new).
>
> So, there's no need to explicitly test for a string that begins with '0'.
>
> Rui Barradas
>
> Thank you again,
>> Claudia
>> On Tue, Jul 3, 2012 at 2:06 AM, Rui Barradas <[email protected]
>> <mailto:[email protected]>> wrote:
>>
>> Hello,
>>
>> Inline.
>>
>> Em 03-07-2012 01:15, jim holtman escreveu:
>>
>> You will have to change the 'i1' expression as follows:
>>
>> i1 <- grepl("^([0D]|[0d])*$", 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
>>
>>
>>
>> Right, apparently, I forgot that grep is greedy, and the test cases
>> were not complete.
>>
>>
>>
>> I put a 'd' and 'D' in the second string and the original regular
>> expression is equivalent to
>>
>> grepl("^[0dD]*$", dd$ch)
>>
>>
>> This is only for the first request, and does not solve cases where
>> there are characters other than '0', 'd' or 'D', but 'd' or 'D' are
>> the first non-zero. This is the case of my 4th row, changed from the
>> OP's data example.
>>
>> My regexpr for 'i2' is equivalent to this one, that I believe is
>> more readable:
>>
>>
>> i2b <- grepl("^0{0,}[Dd]", dd$ch)
>>
>>
>> First a zero, that might occur zero or more times, then a 'd' or
>> 'D', then and til the end, irrelevant.
>>
>>
>> which will match strings containing d, D and 0. If you only
>> want 'd'
>> or 'D' (and not both), then you will have to use the one in
>> 'i1new'.
>>
>>
>> To the OP: bottom line, use Jim's 'i1new' and my 'i2' or 'i2b'.
>>
>> Rui Barradas
>>
>>
>> On Mon, Jul 2, 2012 at 7:24 PM, Rui Barradas
>> <[email protected] <mailto:[email protected]>> wrote:
>>
>> Hello,
>>
>> Try regular expressions instead.
>> In this data.frame, I've changed row nr.4 to have a row with
>> 'D' as first
>> non-zero character.
>>
>> dd <- read.table(text="
>>
>> ch count
>> 1 0000000000D0000000000000000000**__000000000000000000
>> 0.007368
>> 2 0000000000d0000000000000000000**__000000000000000000
>> 0.002456
>> 3 000000000T00000000000000000000**__000000000000000000
>> 0.007368
>> 4 000000000DT0000000000000000000**__000000000000000000
>> 0.007368
>>
>> 5 000000000T00000000000000000000**__000000000000000000
>> 0.002456
>> 6 000000000Td0000000000000000000**__000000000000000000
>> 0.002456
>> 7 00000000T000000000000000000000**__000000000000000000
>> 0.007368
>> 8 00000000T0D0000000000000000000**__000000000000000000
>> 0.007368
>> 9 00000000T000000000000000000000**__000000000000000000
>> 0.002456
>> 10 00000000T0d0000000000000000000**__000000000000000000
>> 0.002456
>>
>> ", header=TRUE)
>> dd
>>
>> i1 <- grepl("^([0D]|[0d])*$", 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$ch character
>> string (I want to
>> remove rows with only "0" & "D" or "0" & "d").
>> Alternatively, I would like
>> to remove rows if the first non-zero element is "D" or
>> "d".
>>
>>
>> ch
>> count
>> 1 0000000000D0000000000000000000**__000000000000000000
>> 0.007368;
>> 2 0000000000d0000000000000000000**__000000000000000000
>> 0.002456;
>> 3 000000000T00000000000000000000**__000000000000000000
>> 0.007368;
>> 4 000000000TD0000000000000000000**__000000000000000000
>> 0.007368;
>> 5 000000000T00000000000000000000**__000000000000000000
>> 0.002456;
>> 6 000000000Td0000000000000000000**__000000000000000000
>> 0.002456;
>> 7 00000000T000000000000000000000**__000000000000000000
>> 0.007368;
>> 8 00000000T0D0000000000000000000**__000000000000000000
>> 0.007368;
>> 9 00000000T000000000000000000000**__000000000000000000
>> 0.002456;
>> 10 00000000T0d0000000000000000000**__000000000000000000
>>
>> 0.002456;
>>
>>
>> I tried the following but it doesn't work if there is
>> more than one
>> character per string:
>>
>> df <- df[!df$ch %in% c("0","D"),]
>> df <- df[!df$ch %in% c("0","d"),]
>>
>>
>>
>> Any help greatly appreciated,
>> Claudia
>>
>> [[alternative HTML version deleted]]
>>
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