我有一个大数据帧（这里减少到前5行），包括来自多个天线的无线电遥测读数。通常每两周有10,000多行这样的数据。
structure(list(freq.id = c(13, 13, 13, 13, 13), DT = structure(c(1393835337,
1393921137, 1393879437, 1393881387, 1393920987), class = c("POSIXct",
"POSIXt"), tzone = "America/Bogota"), S1 = c(-13624L, -12866L,
-13291L, -13415L, -13002L), N1 = c(-13969L, -13824L, -13868L,
-13881L, -13911L), S2 = c(-14114L, -14026L, -13957L, -13969L,
-14052L), N2 = c(-14211L, -14238L, -14168L, -14148L, -14211L),
S3 = c(-13245L, -13113L, -12801L, -12860L, -13133L), N3 = c(-13816L,
-13832L, -13878L, -14001L, -13706L), S4 = c(-13479L, -12702L,
-12388L, -12501L, -12692L), N4 = c(-13872L, -13820L, -13992L,
-13905L, -13798L), S5 = c(-12516L, -11485L, -10871L, -10900L,
-11452L), N5 = c(-13884L, -13995L, -13804L, -13840L, -13929L
), S6 = c(-12661L, -12168L, -10982L, -11112L, -12164L), N6 = c(-13911L,
-13914L, -13078L, -13778L, -13911L), PW = c(20L, 20L, 20L,
20L, 21L), PI = c(1078L, 1078L, 1080L, 2156L, 1078L), aru.unk = c(2072L,
2058L, 2014L, 2052L, 2047L), msrfreq = c(164421600L, 164421700L,
164421400L, 164421300L, 164421800L), TOWERID = structure(c(1L,
1L, 1L, 1L, 1L), .Label = c("TOWER4", "TOWER5", "TOWER6",
"TOWER7"), class = "factor"), prog.freq = structure(c(9L,
9L, 9L, 9L, 9L), .Label = c("162.7920", "162.9774", "163.0780",
"163.6804", "163.8600", "164.0309", "164.2930", "164.3950",
"164.4220", "164.4350", "164.5040", "164.5430", "164.5620",
"164.7026", "164.7840", "164.8230", "164.8430", "164.9338",
"165.5000"), class = "factor")), .Names = c("freq.id", "DT",
"S1", "N1", "S2", "N2", "S3", "N3", "S4", "N4", "S5", "N5", "S6",
"N6", "PW", "PI", "aru.unk", "msrfreq", "TOWERID", "prog.freq"
), row.names = 40615:40619, class = "data.frame")
列S1，S2 ... S6是来自不同天线的信号值，N1，N2 ... N6是相应的噪声值
我试图提取每行的最大和第二大信号值及其相应的噪声值。我可以得到信号值，以及它＆＃34;索引＆＃34;只是信号列。
maxn <- function(n) function(x) order(x, decreasing = TRUE)[n]
mydata$strongest<-apply(mydata[,c(3,5,7,9,11,13)],1,function(x) x[maxn(1)(x)])
#columns 3,5,6,11,13 are the subset of columns containing signal values
mydata$secondstrongest<-apply(mydata[,c(3,5,7,9,11,13)],1,function(x) x[maxn(2)(x)])
mydata$strongestantenna<-apply(mydata[,c(3,5,7,9,11,13)],1,maxn(1))
# returns 5 because in the first 5 rows, the strongest signal is the 5th antenna (S5)
mydata$secondstrongestantenna<-apply(mydata[,c(3,5,7,9,11,13)],1,maxn(2))
#returns a 6
我试图创建2个新列来提取具有第1和第2个最强信号的天线的噪声值。我希望使用每个天线的位置索引（1-6）来提取这样的正确噪声值，但它不起作用。它会提取正确的值，但重复次数与mydata $ strongantenna的值相同
mydata$strongantennanoise<-mydata[c(4,6,8,10,12,14)][mydata$strongestantenna],
#Columns 4,6,8,10,and 12 are the noise values
最强和最强的天线在这里不会发生变化，但数据会随着被跟踪的动物移动而变化。
我觉得我忽略了一些简单的事情，但我无法弄明白。我感谢您提供的任何帮助。
答案 0 :(得分：1)
也许你也可以尝试：
dat1 <- dat[,grep("S", colnames(dat))]
Strongest <- do.call(`pmax`, dat1)
Strongest
#[1] -12516 -11485 -10871 -10900 -11452
indx1 <-which(dat1==Strongest,arr.ind=T)
indx11 <- unique(indx1[,2])
SecondStrongest <- do.call(`pmax`, dat1[,-indx])
SecondStrongest
#[1] -12661 -12168 -10982 -11112 -12164
indx2 <- which(SecondStrongest ==dat1,arr.ind=TRUE)
dat2 <- dat[,grep("N", colnames(dat))]
MatchingNoise <- dat2[indx1]
MatchingSecondNoise <- dat2[indx2]
答案 1 :(得分：1)
# Get names of the strongest and second strongest antennas by row:
strongest <- apply(mydata[,c(3,5,7,9,11,13)],1, function(x) names(x[maxn(1)(x)]))
secondstrongest <- apply(mydata[,c(3,5,7,9,11,13)],1, function(x) names(x[maxn(2)(x)]))
# Get column index for associated noise columns
biggest.noise.col <- sapply(seq_along(mydata[,1]),
function(x) which(colnames(mydata) == strongest[x]) +1)
second.biggest.noise.col <- sapply(seq_along(mydata[,1]),
function(x) which(colnames(mydata) == secondstrongest[x]) +1)
# Use the indices to extract relevant noise values:
mydata$strongestantennanoise <- sapply(seq_along(mydata[,1]),
function(x) mydata[x, biggest.noise.col[x]])
mydata$secondstrongestantennanoise <- sapply(seq_along(mydata[,1]),
function(x) mydata[x, second.biggest.noise.col[x]])