我有下表排序的第一，第二和名称组。
myData <- structure(list(first = c(120L, 120L, 126L, 126L, 126L, 132L, 132L), second = c(1.33, 1.33, 0.36, 0.37, 0.34, 0.46, 0.53),
Name = structure(c(5L, 5L, 3L, 3L, 4L, 1L, 2L), .Label = c("Benzene",
"Ethene._trichloro-", "Heptene", "Methylamine", "Pentanone"
), class = "factor"), Area = c(699468L, 153744L, 32913L,
4948619L, 83528L, 536339L, 105598L), Sample = structure(c(3L,
2L, 3L, 3L, 3L, 1L, 1L), .Label = c("PO1:1", "PO2:1", "PO4:1"
), class = "factor")), .Names = c("first", "second", "Name",
"Area", "Sample"), class = "data.frame", row.names = c(NA, -7L))
在每组中，我想提取对应于特定样本的区域。有几个组没有来自样本的区域，因此如果未检测到样本，则应返回“NA”。理想情况下，最终输出应为每个样本的列。
我已尝试使用ifelse函数为每个样本创建一列：
PO1<-ifelse(myData$Sample=="PO1:1",myData$Area, "NA")
然而，这并未考虑群组分布。我想这样做，但在小组内。如果sample = PO1：1，则为每个组（一组为first，second和Name列的值相等），否则为NA。
对于第一组：
structure(list(first = c(120L, 120L), second = c(1.33, 1.33),
Name = structure(c(1L, 1L), .Label = "Pentanone", class = "factor"),
Area = c(699468L, 153744L), Sample = structure(c(2L, 1L), .Label = c("PO2:1",
"PO4:1"), class = "factor")), .Names = c("first", "second", "Name",
"Area", "Sample"), class = "data.frame", row.names = c(NA, -2L))
输出应为：
structure(list(PO1.1 = NA, PO2.1 = 153744L, PO3.1 = NA, PO4.1 = 699468L), .Names =c("PO1.1", "PO2.1", "PO3.1", "PO4.1"), class = "data.frame", row.names = c(NA, -1L))
有什么建议吗？
答案 0 :(得分：1)
正如问题中的示例所示，我假设Sample是一个因素。如果不是这种情况，请考虑这样做。
Sample以使其成为合法名称，否则可能会导致错误levels(myData$Sample) <- make.names(levels(myData$Sample))
## DEFINE THE CUTS##
# Adjust these as necessary
#--------------------------
max.second <- 3 # max & nin range of myData$second
min.second <- 0 #
sprd <- 0.15 # with spread for each group
#--------------------------
# we will cut the myData$second according to intervals, cut(myData$second, intervals)
intervals <- seq(min.second, max.second, sprd*2)
# Next, lets create a group column to split our data frame by
myData$group <- paste(myData$first, cut(myData$second, intervals), myData$Name, sep='-')
groups <- split(myData, myData$group)
samples <- levels(myData$Sample) ## I'm assuming not all samples are present in the example. Manually adjusting with: samples <- sort(c(samples, "PO3.1"))
# Apply over each group, then apply over each sample
myOutput <-
t(sapply(groups, function(g) {
#-------------------------------
# NOTE: If it's possible that within a group there is more than one Area per Sample, then we have to somehow allow for thi. Hence the "paste(...)"
res <- sapply(samples, function(s) paste0(g$Area[g$Sample==s], collapse=" - ")) # allowing for multiple values
unlist(ifelse(res=="", NA, res))
## If there is (or should be) only one Area per Sample, then remove the two lines aboce and uncomment the two below:
# res <- sapply(samples, function(s) g$Area[g$Sample==s]) # <~~ This line will work when only one value per sample
# unlist(ifelse(res==0, NA, res))
#-------------------------------
}))
# Cleanup names
rownames(myOutput) <- paste("Group", 1:nrow(myOutput), sep="-") ## or whichever proper group name
# remove dummy column
myData$group <- NULL
myOutput
PO1.1 PO2.1 PO3.1 PO4.1
Group-1 NA "153744" NA "699468"
Group-2 NA NA NA "32913 - 4948619"
Group-3 NA NA NA "83528"
Group-4 "536339" NA NA NA
Group-5 "105598" NA NA NA
答案 1 :(得分：1)
你真的不能指望R直觉PO2和PO4之间有第四个因子水平，现在可以。
> reshape(inp, direction="wide", idvar=c('first','second','Name'), timevar="Sample")
first second Name Area.PO4:1 Area.PO2:1 Area.PO1:1
1 120 1.3 Pentanone 699468 153744 NA
3 126 0.4 Heptene 32913 NA NA
4 126 0.4 Heptene 4948619 NA NA
5 126 0.3 Methylamine 83528 NA NA
6 132 0.5 Benzene NA NA 536339
7 132 0.5 Ethene._trichloro- NA NA 105598