search for: 900001

...I guess these days it is standard to use DF with millions of entries so > I was wondering how people are doing that in the faster way. I did this to generate and search 40 million unique strings > grams <- as.character(1:4e7) ## a long time passes... > system.time(grep("^900001", grams)) ## similar times to grepl user system elapsed 10.384 0.168 10.543 Is that the basic task you're trying to accomplish? grep(l) goes quickly to C, so I don't think data.table or other will be markedly faster if you're looking for an arbitrary regular expression...

Hi Duncan, > Didn't you post the same question yesterday? Perhaps nobody answered > because your question is unanswerable. sorry, I got a email that my message was waiting for approval and when I look at the forum I didn't see my message and this is why I sent it again and this time I did check that the format of my message was text only. Sorry for the noise. > You need to

...?????Feature GroupA_3 1: ? ? ? ????????chr1 521369 ?750000 ????chr1-0001 ? ?????0.170 2: ? ? ? ????????chr1 750001 ?800000 ????chr1-0002 ? ????-0.086 3: ? ? ? ????????chr1 800001 ?850000 ????chr1-0003 ? ?????0.006 4: ? ? ? ????????chr1 850001 ?900000 ????chr1-0004 ? ?????0.050 5: ? ? ? ????????chr1 900001 ?950000 ????chr1-0005 ? ?????0.062 6: ? ? ? ????????chr1 950001 1000000 ? ?chr1-0006 ? ????-0.016 data_2: > dim(data_2) [1] 470870 5 > head(data_2) ? ?Chromosome ????Start ? End????????????Feature ????GroupA_3 1: ? ? ? ????????chr1 15864 15865 ????cg13869341 ? ?????????0.207 2: ? ? ? ??????...

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