On Tue, May 10, 2011 at 3:09 PM, David Winsemius <dwinsem...@comcast.net>wrote:
> > On May 10, 2011, at 3:18 AM, noxyp...@gmail.com wrote: > > On Fri, May 6, 2011 at 7:41 PM, David Winsemius <dwinsem...@comcast.net> >> wrote: >> >>> >>> On May 6, 2011, at 11:35 AM, Pete Pete wrote: >>> >>> >>>> Gabor Grothendieck wrote: >>>> >>>>> >>>>> On Tue, Dec 7, 2010 at 11:30 AM, Pete Pete <noxyp...@gmail.com> >>>>> wrote: >>>>> >>>>>> >>>>>> Hi, >>>>>> consider the following two dataframes: >>>>>> x1=c("232","3454","3455","342","13") >>>>>> x2=c("1","1","1","0","0") >>>>>> data1=data.frame(x1,x2) >>>>>> >>>>>> y1=c("232","232","3454","3454","3455","342","13","13","13","13") >>>>>> y2=c("E1","F3","F5","E1","E2","H4","F8","G3","E1","H2") >>>>>> data2=data.frame(y1,y2) >>>>>> >>>>>> I need a new column in dataframe data1 (x3), which is either 0 or 1 >>>>>> depending if the value "E1" in y2 of data2 is true while x1=y1. The >>>>>> result >>>>>> of data1 should look like this: >>>>>> x1 x2 x3 >>>>>> 1 232 1 1 >>>>>> 2 3454 1 1 >>>>>> 3 3455 1 0 >>>>>> 4 342 0 0 >>>>>> 5 13 0 1 >>>>>> >>>>>> I think a SQL command could help me but I am too inexperienced with it >>>>>> to >>>>>> get there. >>>>>> >>>>>> >>>>> Try this: >>>>> >>>>> library(sqldf) >>>>>> sqldf("select x1, x2, max(y2 = 'E1') x3 from data1 d1 left join data2 >>>>>> d2 >>>>>> on (x1 = y1) group by x1, x2 order by d1.rowid") >>>>>> >>>>> >>>>> x1 x2 x3 >>>>> 1 232 1 1 >>>>> 2 3454 1 1 >>>>> 3 3455 1 0 >>>>> 4 342 0 0 >>>>> 5 13 0 1 >>>>> >>>>> >>>>> snipped Gabor's sig >>> >>>> >>>> That works pretty cool but I need to automate this a bit more. Consider >>>> the >>>> following example: >>>> >>>> list1=c("A01","B04","A64","G84","F19") >>>> >>>> x1=c("232","3454","3455","342","13") >>>> x2=c("1","1","1","0","0") >>>> data1=data.frame(x1,x2) >>>> >>>> y1=c("232","232","3454","3454","3455","342","13","13","13","13") >>>> y2=c("E13","B04","F19","A64","E22","H44","F68","G84","F19","A01") >>>> data2=data.frame(y1,y2) >>>> >>>> I want now to creat a loop, which creates for every value in list1 a new >>>> binary variable in data1. Result should look like: >>>> x1 x2 A01 B04 A64 G84 F19 >>>> 232 1 0 1 0 0 0 >>>> 3454 1 0 0 1 0 1 >>>> 3455 1 0 0 0 0 0 >>>> 342 0 0 0 0 0 0 >>>> 13 0 1 0 0 1 1 >>>> >>> >>> Loops!?! We don't nee no steenking loops! >>> >>> xtb <- with(data2, table(y1,y2)) >>>> cbind(data1, xtb[match(data1$x1, rownames(xtb)), ] ) >>>> >>> x1 x2 A01 A64 B04 E13 E22 F19 F68 G84 H44 >>> 232 232 1 0 0 1 1 0 0 0 0 0 >>> 3454 3454 1 0 1 0 0 0 1 0 0 0 >>> 3455 3455 1 0 0 0 0 1 0 0 0 0 >>> 342 342 0 0 0 0 0 0 0 0 0 1 >>> 13 13 0 1 0 0 0 0 1 1 1 0 >>> >>> I am guessing that you were to ... er, busy? ... to complete the table? >>> >>> -- >>> >>> David Winsemius, MD >>> West Hartford, CT >>> >>> >>> >> Thanks a lot! Pretty simple. I am so much used to SQLDF right now. >> >> So how would you handle more complicated strings like that: >> y1=c("232","232", "232", "3454","3454","3455","342","13","13","13","13") >> y2=c("E13","B04 A01 F19","B04","F19","A64 G84 A05","E22","H44 >> C35","F68","G84","F19","A01") >> data2=data.frame(y1,y2) >> >> Where you want to extract for instance all "A01" from the strings? >> > > I think you need either to explain what you want in more words of the > English language or to offer an example of the desired output. I suspect you > did not want something as simple as this: > > > A01.instances <- grep("A01" , data2$y2) > > A01.instances > [1] 2 11 > > data2[A01.instances, ] > y1 y2 > 2 232 B04 A01 F19 > 11 13 A01 > > Or maybe you did? > > -- > David Winsemius, MD > West Hartford, CT > > With sqldf I could do it manually: > > data1=sqldf("SELECT data1.*, max(data2.y2 LIKE '% A01%') OR max(data2.y2 > LIKE 'A01%') A01 FROM data1 left join data2 on (data1.x1 = data2.y1) group > by data1.x1, data2.y1") > > data1=sqldf("SELECT data1.*, max(data2.y2 LIKE '% B04%') OR max(data2.y2 > LIKE 'B04%') B04 FROM data1 left join data2 on (data1.x1 = data2.y1) group > by data1.x1, data2.y1") > > data1 > x1 x2 A01 B04 > 1 13 0 1 0 > 2 232 1 1 1 > 3 342 0 0 0 > 4 3454 1 0 0 > 5 3455 1 0 0 > > > But I need to automate this for some thousand "substrings". Any suggestion? [[alternative HTML version deleted]] ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.