## Hello there,
## I have an issue where I need to use the value of column names to
multiply with the individual values in a column and I have many
columns to do this over. I have data like this where the column names
are numbers:
mydf <- data.frame(`2.72`=runif(20, 0, 125),
`3.2`=runif(20, 50, 75),
`3.78`=runif(20, 0, 100),
yy= head(letters,2), check.names=FALSE)
## I had been doing something like this but this seems rather tedious
and clunky. These append the correct values to my dataframe but is
there any way that I can do this generally over each column, also
using each column name as the multiplier for that column?
mydf$vd2.72 <- mydf$'2.72'*2.72
mydf$vd3.2 <- mydf$'3.2'*3.2
mydf$vd3.78 <- mydf$'3.78'*3.78
## So can I get to this point with a more generalized solution? For
now, I would also prefer to keep this in wide format and I am aware
(thanks to the list!) that I could use melt() to get the values I
want.
mydf
## Thanks so much in advance!
Sam
David Winsemius
2011-Aug-16 20:13 UTC
[R] Utilizing column names to multiply over all columns
On Aug 16, 2011, at 3:37 PM, Sam Albers wrote:> ## Hello there, > ## I have an issue where I need to use the value of column names to > multiply with the individual values in a column and I have many > columns to do this over. I have data like this where the column names > are numbers: > > mydf <- data.frame(`2.72`=runif(20, 0, 125), > `3.2`=runif(20, 50, 75), > `3.78`=runif(20, 0, 100), > yy= head(letters,2), check.names=FALSE)> mydf 2.72 3.2 3.78 yy 1 31.07874 74.48555 89.306591 a 2 123.68290 74.30030 11.943576 b 3 89.64024 68.26378 97.627211 a 4 81.46604 59.79607 91.005217 b> > ## I had been doing something like this but this seems rather tedious > and clunky. These append the correct values to my dataframe but is > there any way that I can do this generally over each column, also > using each column name as the multiplier for that column? > > mydf$vd2.72 <- mydf$'2.72'*2.72 > mydf$vd3.2 <- mydf$'3.2'*3.2 > mydf$vd3.78 <- mydf$'3.78'*3.78 > > ## So can I get to this point with a more generalized solution? For > now, I would also prefer to keep this in wide format and I am aware > (thanks to the list!) that I could use melt() to get the values I > want.You will get the warning that last last column is not "going right" but otherwise this returns what you asked for: sapply(1:length(mydf), function(i) mydf[[i]]* as.numeric(names(mydf) [i]) ) [,1] [,2] [,3] [,4] [1,] 84.53416 238.3538 337.57891 NA [2,] 336.41748 237.7610 45.14672 NA [3,] 243.82145 218.4441 369.03086 NA [4,] 221.58762 191.3474 343.99972 NA [5,] 81.78911 213.0770 97.90072 NA snipped remainder -- David Winsemius, MD West Hartford, CT