Hi, Sarah,
There is a mistake in the column name and the desired output is pretty much
the same as the input data frames except for the replacement of
'some_string' by the word 'alcohol' in the column I specified above.
Yet, your code is what I am looking for and helped me to get started and
figure out how to move forward.
Thank you very much!
Andre
On Tue, Sep 29, 2020 at 5:58 PM Sarah Goslee <sarah.gos...@gmail.com> wrote:
> Your desired output seems to be the same as your desired input in your
> example, and your data frames have different column names.
>
> Nonetheless, this bit of code will find rows with "alcohol" in column
> 3, and for those rows replace the contents of column 4 with column 3.
> That may not be exactly what you're after, but should get you started.
>
> lapply(mylist, function(x){
> x[grepl("alcohol", x[, 3]), 4] <- x[grepl("alcohol", x[, 3]), 3]
> x
> })
>
> Sarah
>
> On Tue, Sep 29, 2020 at 4:24 PM André Luis Neves <andrl...@ualberta.ca>
> wrote:
> >
> > Dear all,
> >
> > I have had difficulties copying the word "alcohol" in the "vehicle"
> column
> > to replace the string in the column "accident_type". It is a huge list
> but
> > I have prepared a workable and simple example below and the desired
> > output. I am sure you guys can give me some advice on how to deal with
> > this.
> >
> > sapply( all_files, function(x) dim(x))
> > [,1] [,2] [,3] [,4]
> > [1,] 89563 69295 67446 39709
> > [2,] 33 33 33 33
> >
> >
> >
> > ###Simple List
> >
> > A= data.frame(c("1", "2", "3", "4"),c(4:7),c("car", "byke", "alcohol",
> > "motocycle"), c("alcohol", "cell_phone", "some_string", "fog"))
> > colnames(A) <- c("id", "km","vehicle","accident_cause")
> > A
> >
> >
> > B= data.frame(c("1", "2", "3", "4"),c(4:7),c("car", "alcohol", "car",
> > "motocycle"), c("alcohol", "some_string", "rain", "fog"))
> > colnames(B) <- c("id", "km", "vehicle", "accident_type")
> > B
> >
> >
> > C= data.frame(c("1", "2", "3", "4"),c(4:7),c("car", "alcohol", "car",
> > "alcohol"), c("alcohol", "some_string", "rain", "some_string"))
> > colnames(C) <- c("id", "km", "vehicle", "accident_type")
> > C
> >
> > mylist <- list(A=A,B=B,C=C)
> > mylist
> >
> >
> > ###Desired output
> >
> > A= data.frame(c("1", "2", "3", "4"),c(4:7),c("car", "byke", "alcohol",
> > "motocycle"), c("alcohol", "cell_phone", "alcohol", "fog"))
> > colnames(A) <- c("id", "km","vehicle","accident_cause")
> > A
> >
> >
> > B= data.frame(c("1", "2", "3", "4"),c(4:7),c("car", "alcohol", "car",
> > "motocycle"), c("alcohol", "alcohol", "rain", "fog"))
> > colnames(B) <- c("id", "km", "vehicle", "accident_type")
> > B
> >
> >
> > C= data.frame(c("1", "2", "3", "4"),c(4:7),c("car", "alcohol", "car",
> > "alcohol"), c("alcohol", "alcohol", "rain", "alcohol"))
> > colnames(C) <- c("id", "km", "vehicle", "accident_type")
> > C
> >
> > mylist <- list(A=A,B=B,C=C)
> > mylist
> >
> >
> > ##Thank you very much,
> >
> > --
> > Andre
> >
>
>
> --
> Sarah Goslee (she/her)
> http://www.numberwright.com
>
--
Andre
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