"For" loop and "if" question

Wed, Dec 19, 2012 3:16 PM

Hello,

Thanks for the data example.
Try the following.


fun <- function(DF){
     f <- function(x, y){
         if(ncol(y) == 3) rbind(x, y) else x
     }

     p <- grep("^p[[:digit:]]+$", names(DF))
     lat <- grep("^lat[[:digit:]]+$", names(DF))
     long <- grep("^long[[:digit:]]+$", names(DF))
     p <- sapply(DF[p], as.logical)
     lat <- sapply(DF[lat], as.numeric)
     long <- sapply(DF[long], as.numeric)

     tmp <- sapply(seq_len(ncol(p)), function(j)
         cbind(p = j, lat = lat[p[, j], j], long = long[p[, j], j]))
     tmp <- na.omit(Reduce(f, tmp))
     data.frame(tmp, row.names = seq_len(nrow(tmp)))
}

fun(dat)  # 'dat' is your data example, dat <- structure(...)


Hope this helps,

Rui Barradas
Em 19-12-2012 20:56, Steven Ranney escreveu:

A friendly r-helper asked me to use dput() to output some of my data.

The results are below:

structure(list(ID = 2004:2008, p1 = c(1L, 1L, 1L, 1L, 0L), p2 = c(0L,
0L, 0L, 0L, 1L), p3 = c(0L, 0L, 0L, 0L, 0L), p4 = c(0L, 0L, 1L,
1L, 0L), p5 = c(0L, 0L, 0L, 0L, 0L), p6 = c(1L, 1L, 0L, 1L, 1L
), p7 = c(0L, 0L, 1L, 1L, 0L), p8 = c(1L, 0L, 0L, 0L, 0L), p9 = c(0L,
0L, 1L, 1L, 0L), p10 = c(1L, 0L, 1L, 0L, 0L), p11 = c(0L, 0L,
0L, 1L, 0L), p12 = c(1L, 0L, 0L, 1L, 0L), p13 = c(0L, 0L, 1L,
0L, 0L), p14 = c(0L, 0L, 0L, 0L, 1L), p15 = c(0L, 0L, 1L, 1L,
0L), p16 = c(0L, 0L, 0L, 0L, 0L), p17 = c(0L, 1L, 0L, 0L, 0L),
     p18 = c(0L, 0L, 0L, 0L, 0L), p19 = c(0L, 0L, 0L, 0L, 1L),
     p29 = c(0L, 0L, 0L, 1L, 0L), p21 = c(0L, 1L, 0L, 1L, 0L),
     lat1 = c(29.266, 29.27057, 29.24012, 29.22798, NA), lat2 = c(NA,
     NA, NA, NA, 29.27332), lat3 = c(NA, NA, NA, NA, NA), lat4 = c(NA,
     NA, 29.23984, 29.26174, NA), lat5 = c(NA, NA, NA, NA, NA),
     lat6 = c(29.21016, 29.27799, NA, 29.24824, 29.27873), lat7 = c(NA,
     NA, 29.24511, 29.26614, NA), lat8 = c(29.27555, NA, NA, NA,
     NA), lat9 = c(NA, NA, 29.24437, 29.24437, NA), lat10 = c(29.26266,
     NA, 29.2633, NA, NA), lat11 = c(NA, NA, NA, 29.26547, NA),
     lat12 = c(29.26146, NA, NA, 29.2581, NA), lat13 = c(NA, NA,
     29.24212, NA, NA), lat14 = c(NA, NA, NA, NA, 29.27507), lat15 = c(NA,
     NA, 29.27403, 29.27403, NA), lat16 = c(NA, NA, NA, NA, NA
     ), lat17 = c(NA, 29.26448, NA, NA, NA), lat18 = c(NA, NA,
     NA, NA, NA), lat19 = c(NA, NA, NA, NA, 29.27167), lat20 = c(NA,
     NA, NA, 29.24208, NA), lat21 = c(NA, 29.27493, NA, 29.26212,
     NA), long1 = c(-89.96268, -89.95791, -90.00651, -90.02192,
     NA), long2 = c(NA, NA, NA, NA, -89.9543), long3 = c(NA, NA,
     NA, NA, NA), long4 = c(NA, NA, -90.00668, -89.98277, NA),
     long5 = c(NA, NA, NA, NA, NA), long6 = c(-90.0387, -89.95317,
     NA, -89.98473, -89.94817), long7 = c(NA, NA, -90.00278, -89.95956,
     NA), long8 = c(-89.9545, NA, NA, NA, NA), long9 = c(NA, NA,
     -90.02349, -90.02349, NA), long10 = c(-89.96309, NA, -89.96299,
     NA, NA), long11 = c(NA, NA, NA, -89.97777, NA), long12 = c(-89.97981,
     NA, NA, -89.99077, NA), long13 = c(NA, NA, -90.00163, NA,
     NA), long14 = c(NA, NA, NA, NA, -89.95477), long15 = c(NA,
     NA, -89.95444, -89.95444, NA), long16 = c(NA, NA, NA, NA,
     NA), long17 = c(NA, -89.96214, NA, NA, NA), long18 = c(NA,
     NA, NA, NA, NA), long19 = c(NA, NA, NA, NA, -89.95428), long20 = c(NA,
     NA, NA, -90.06522, NA), long21 = c(NA, -89.95477, NA, -89.98407,
     NA)), .Names = c("ID", "p1", "p2", "p3", "p4", "p5", "p6",
"p7", "p8", "p9", "p10", "p11", "p12", "p13", "p14", "p15", "p16",
"p17", "p18", "p19", "p29", "p21", "lat1", "lat2", "lat3", "lat4",
"lat5", "lat6", "lat7", "lat8", "lat9", "lat10", "lat11", "lat12",
"lat13", "lat14", "lat15", "lat16", "lat17", "lat18", "lat19",
"lat20", "lat21", "long1", "long2", "long3", "long4", "long5",
"long6", "long7", "long8", "long9", "long10", "long11", "long12",
"long13", "long14", "long15", "long16", "long17", "long18", "long19",
"long20", "long21"), row.names = c(NA, 5L), class = "data.frame")



Thanks again -

SR
Steven H. Ranney


On Wed, Dec 19, 2012 at 1:42 PM, Steven Ranney <steven.ranney at gmail.com> wrote:

All -

I have a large data frame that looks like

ID     p1     p2     p3...p20     Lat1     Lat2     Lat3...Lat20
Long1     Long2     Long3...Long20
1       0      0      1      0        NA       NA       29.xx NA
  NA          NA         -89.xx   NA
2       1      0      0      1        27.xx   NA       NA     29.00
  -88.00     NA         NA        -89.xx
3       0      0      0      0        NA      NA       NA      NA
   NA         NA          NA       NA
...
[truncated]

where length(ID) = 1300 and the zeroes and ones in the p1-p20 column
correspond to values in the Lat1-Lat20 and Long1-Long20 columns; a 0
indicates no corresponding value and 1 indicates there is a
corresponding value.

I'd like to create a dataframe that is

ID          p     Lat     Long
1           3     29.xx  -89.xx
2           1     27.xx  -88.xx
2           20   29.xx  -89.xx
...

and so on, such that for every ID I have, there is a corresponding row
with each Latn and Longn that may exist.

My problem is that I don't know how to do that in R.  I can set up
simple "for" loops, but I don't know how to set up something that has
a "for" and an "if/else" statement.  And frankly, with the number of
pX, LatX, and LongX, I think that my if/else statement would be huge
and unwieldly.

Could anyone offer any assistance?

Thanks for your help -

SR
Steven H. Ranney

______________________________________________
R-help at 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.

"For" loop and "if" question

Thread (5 messages)