我有數值向量的列表NList
像迭代數值向量列表中的
[[1]]
[1] 1959 9 4 62
[[2]]
[1] 2280 2 13
[[3]]
[1] 15 4 13
[[4]]
[1] 2902 178 13
,結構等
list(c(1959, 13), c(2280, 178, 13), c(2612, 178, 13), c(2902,
178, 13), c(2389, 178, 13), c(216, 736, 13), c(2337, 178, 13),
c(2639, 2126, 13), c(2924, 676, 178, 13), c(2416, 674, 178,
13), c(2223, 13), c(842, 178, 13), c(2618, 1570, 178, 13),
c(854, 178, 13), c(1847, 178, 13), c(2529, 178, 13), c(511,
178, 13), c(2221, 736, 13), c(415, 674, 178, 13), c(2438,
178, 13), c(2127, 178, 13), c(1910, 2126, 13), c(1904, 674,
178, 13), c(2310, 674, 178, 13), c(1732, 178, 13), c(1843,
178, 13), c(2539, 178, 13), c(1572, 676, 178, 13), c(1616,
876, 13).....)
欲迭代數值向量在此列表中,我想做點什麼:
sum<- 0
index<-1
list1 <- apply(NList,1,function (i){
#I want to get each of the numeric vector here
row <- NList[i]
#then I want to iterate the numeric vector for some calculation.
#I am expecting, for [[1]], I get f(1959,9)+f(9,4)+f(4,62), in which f is my customized function, below I use a simple multiple as example
for (j in (1:(length(row)-1)))
{
origin <- row[j]
dest <- row[j+1]
#a simple calculation example...I am expecting an array of sum which is the calculation result
sum[index] <- sum[index] + origin*dest
}
index <- index+1
})
但它不起作用並返回:
dim(X) must have a positive length
的lapply是不是爲我工作,並返回總和爲0 ...
listR1 <- lapply(NList,function (i){
row <- i
for (j in 1:length(row))
{origin <- row[j]
dest <- row[j+1]
sum[index] <- sum[index] + origin*dest
}
})
我錯過了什麼?我怎樣才能做到這一點?
謝謝!
讓一個名爲'list'的對象變得非常混亂e list()是一個函數。 – Seth 2012-07-30 20:31:42
要'申請'在你想要的'?lapply'列表上。除此之外,我無法效仿你的榜樣... – Justin 2012-07-30 20:31:42
你能告訴我們更多關於你喜歡的結果,而不是如何去做。你的方法看起來非常不像R。 – 2012-07-30 20:31:59