On Aug 10, 2011, at 7:34 PM, Ward, Michael Patrick wrote:
I used this service several months ago and was very pleased with the
response.
I have a dataframe with several thousand lines and to each line I
need to apply a series of "if else" statements. For each row I need
either a value or a blank/NA. Below is the series of if else
statements I have been trying without success to integrate into a
function such as "apply".
Try this:
dfrm <-
structure(list(t1.secondstrongest = c(-10682L, -11154L, -10930L,
-10997L, -11244L, -12744L), Ant_test = c(60L, 240L, 300L, 240L,
180L, 60L), value.str1 = c(240L, 0L, 0L, 300L, 180L, 180L),
value.str1_adj = c(242L,
2L, 2L, 302L, 182L, 182L), value.str2 = c(180L, 240L, 300L, 60L,
0L, 240L), value.str2_adj = c(182L, 242L, 302L, 62L, 2L, 242L
), Noise = c(-12344L, -13444L, -14022L, -13456L, -14209L, -14134L
), bearingdiff = c(11.23, 27.23, 27.55, 14.23, 25.22, 8.13)), .Names =
c("t1.secondstrongest",
"Ant_test", "value.str1", "value.str1_adj", "value.str2",
"value.str2_adj",
"Noise", "bearingdiff"), row.names = c("1", "2", "3", "4", "5",
"6"), class = "data.frame")
ddfrm$newval <- NA
dfrm$newval <-apply(dfrm, 1, function(x) ifelse(
any( x["t1.secondstrongest"]< -13000,
x["Noise"] > -13000 ,
abs(x["value.str1"]-x["value.str2"])==120 ,
abs(x["value.str1"]-x["value.str2"])==180 ,
abs(x["value.str1"]-x["value.str2"])==240 ), NA,
ifelse (x["value.str1"] == 300 & x["value.str2"] ==0,
x["value.str1_adj"] + x["bearingdiff"],
ifelse (x["value.str1"] == 0 & x["value.str2"] == 360 ,
x["value.str1_adj"] + 360 - x["bearingdiff"] ,
ifelse (x["value.str2"] < x["value.str1"] , x["value.str1_adj"] -
x["bearingdiff"],
ifelse ( x["value.str2"] > x["value.str1"] , x["value.str1_adj"] +
x["bearingdiff"], NA ) ) ) ) ) )
Below is an example of the dataframe
t1.secondstrongest Ant_test
value.str1 value.str1_adj
value.str2 value.str2_adj
Noise bearingdiff
1 -10682
60 240
242
180
182 -12344 11.23
2 -11154
240 0
2
240
242 -13444 27.23
3 -10930
300 0
2
300
302 -14022 27.55
4 -10997
240 300
302
60
62 -13456 14.23
5 -11244
180 180
182
0
2 -14209 25.22
6 -12744
60 180
182
240
242 -14134 8.13
The answer to the examples should be...
1 NA
2 NA
3 334.45
4 NA
5 NA
6 190.13
I get
> as.matrix(dfrm$newval)
[,1]
[1,] NA
[2,] NA
[3,] 29.55
[4,] NA
[5,] NA
[6,] 190.13
Given the number of syntactic errors that needed to be corrected, I
trust R more than your "hand" calculations. Not jsut the erroneous use
of "&&" but in particular your unfortunate practice of not using
spaces created a bug where your wrote:
t1.secondstrongest<-13000
Notice the warning message:
Warning messages:
1: In any(x["t1.secondstrongest"] <- 13000, x["Noise"] > -13000,
abs(x["value.str1"] - :
coercing argument of type 'double' to logical
That attempted to assign a value rather than testing whether a number
is less than negative 13000.
MORAL: USE SPACES TO IMPROVE READABILITY AND AVOID SYNTACTIC ERRORS.
THANKS!
Mike
Department of Natural Resources and Environmental Sciences
University of Illinois
David Winsemius, MD
West Hartford, CT
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