Using loop in functions

Question

I have a data set where I wanted to compute some parameters repeatedly depending on a range of variable (time). So my data set looks like this:

structure(list(A = c(25L, 25L, 25L, 25L, 25L, 25L), T = 56:61, 
    X = c(481.07, 487.04, 490.03, 499, 504.97, 507.96), Y = c(256.97, 
    256.97, 256.97, 256.97, 256.97, 256.97), V = c(4.482, 5.976, 
    7.47, 4.482, 5.976, 7.47), thetarad = c(0.164031585831919, 
    0.169139558949956, 0.171661200692621, 0.179083242584008, 
    0.183907246800473, 0.186289411097781), thetadeg = c(9.39831757286096, 
    9.69098287432395, 9.83546230358968, 10.2607139792383, 10.537109061132, 
    10.6735970214433), bin = structure(c(1L, 1L, 1L, 1L, 1L, 
    1L), .Label = c("binA", "binB", "binC", "outbin"), class = "factor")), .Names = c("A", 
"T", "X", "Y", "V", "thetarad", "thetadeg", "bin"), row.names = c(NA, 
6L), class = "data.frame")

And this code works perfectly for computing my parameters:

NT <- data.table(st1binned [st1binned$T<31, ], key="bin")
alox1=NT[, list(ang=length(unique(thetadeg)), len=length(T), Vm=mean(V), T=c("30s")), 
         by=c("A", "bin")]

I can repeatedly use this code if I want to and just changed the subset on my data to st1binned and bind it to my data set:

NT <- data.table(st1binned [st1binned$T>30 & T<61,  ], key="bin")
alox1=rbind(alox1, NT[, list(ang=length(unique(thetadeg)), len=length(T), Vm=mean(V), T=c("60s")), 
         by=c("A", "bin")])

But is there a way to loop these functions where I can say that it should consider every 30sec of data and to change also the T variable?

edited:

Resulting data set should look like this:

structure(list(A = c(38L, 45L, 115L, 118L, 121L, 692L), bin = structure(c(1L, 
1L, 1L, 1L, 1L, 1L), .Label = c("binA", "binB", "binC", "outbin"
), class = "factor"), ang = c(2L, 7L, 4L, 4L, 11L, 1L), len = c(30L, 
30L, 24L, 23L, 11L, 30L), Vm = c(0.3984, 1.07771666666667, 0.465545833333333, 
0.760526086956522, 4.27607272727273, 0), T = c("30s", "30s", 
"30s", "30s", "30s", "30s")), .Names = c("A", "bin", "ang", "len", 
"Vm", "T"), class = c("data.table", "data.frame"), row.names = c(NA, 
-6L), .internal.selfref = <pointer: 0x0000000000170788>)

Take a look on cut()s help. I never used data.table, so I don't know if it'll help. — Rcoster
– Rcoster, Commented Feb 3, 2014 at 16:22

BrodieG · Accepted Answer · 2014-02-03 17:13:58Z

2

I think you want your data to be binned in 30 second increments. If that's what you want you can do it with cut. In order to demonstrate, I had to create some data that had more than one 30 second interval. I just copied your data frame three times and added 30 second increments to each copy. This is the result:

st1binned2$bin <- cut(st1binned2$T, rep(1:6 * 30), include.lowest=T)
NT <- data.table(st1binned2, key="bin")
NT[, list(ang=length(unique(thetadeg)), len=length(T), Vm=mean(V), T=paste(range(T), collapse="-")), 
         by=c("A", "bin")]
#     A       bin ang len     Vm       T
# 1: 25   [30,60]   5   5 5.6772   56-60
# 2: 25   (60,90]   6   6 5.9760   61-90
# 3: 25  (90,120]   6   6 5.9760  91-120
# 4: 25 (120,150]   6   6 5.9760 121-150
# 5: 25 (150,180]   1   1 7.4700 151-151

And this is how I created the data:

st1binned2 <- do.call(rbind, replicate(4, st1binned, simplify=F))
st1binned2$T <- st1binned2$T + rep(0:3 * 30, each=nrow(df))

EDIT, to modify labels, an example:

bins <- rep(1:6 * 30)
st1binned2$bin <- cut(
  st1binned2$T, bins, include.lowest=T,
  labels=paste(head(bins, -1L), tail(bins, -1L), sep="-"), 
)

edited Feb 3, 2014 at 17:13

answered Feb 3, 2014 at 16:23

BrodieG

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3 Comments

Kaye11 Over a year ago

Yes you were right that I want to bin my data into 30sec increments. However, I have also a variable named bin so I changed the st1binned2$bin to st1binned2$time. Since I also want the time to be a variable I edited your script to NT[, list(ang=length(unique(thetadeg)), len=length(T), Vm=mean(V), time=time), by=c("A", "bin")]. However, there is this brackets and parenthesis on time (which could also be seen on your answer). Any idea how to remove this? Thanks @BrodieG

BrodieG Over a year ago

@Kaye11, You can use the labels argument to cut to make your own labels, see update above.

Kaye11 Over a year ago

I just observed one problem, the resulting data set should just have one unique A (as I have used some functions to compute some parameters but now its showing all the A, i.e. for your answer above, it was showing 5 25's). Sorry for not pointing this out earlier!

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