Dear R users,
The best in this new year 2011.
I am dealing with a character vector (xx) whose nchar are not the same.
Ex.
nchar(xx)
[1] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
[38] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 3 3 3 3 3 4 4 4 4 4 4 4 4
[75] 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 ....... 9
I need xx to be nchar = 9
My best guest was to paste 0's. Then I need substring (xx, 6, 9).
I came with:
xx[1:61]<-paste("00000000", xx[1:61], sep="")
xx[62:66]<-paste("000000", xx[62:66], sep="")
xx[67:100]<-paste("00000", xx[67:100], sep="")
......
> nchar(xx)
[1] 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9
[38] 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9
[75] 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 .... 9
xx<-substring(xx, 6, 9)
This is a solution for one data set would be sufficient but not if I will
continuously deal with this same issue.
Furthermore, I am trying to automate the process but I have not be able to came
with adequate solution.
I was thinking to create a character vector of 0's 9-nchar(xx).
Then paste it to xx.
9-nchar(xx)
[1] 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8
[38] 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 6 6 6 6 6 5 5 5 5 5 5 5 5
[75] 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 ......1
Nevertheless, I have not been able to create this vector nor I do not know if
this is the best option.
Another way I thought was to create an if statement, but this will be long and
not efficient (I think).
Any suggestion, will be appreciated.
Jose
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