考取多個可變參數的函數與申請家庭

Question

我有以下的矢量和變量

x<-c(1,2,3,4,5,7,6,7,8,9,8,7,6,4,5,3,2,2,3,4,5,6,67,7,7) 

wavelet<-c("d2","s2") 
n.level<-c(1,2) 
schirnkfun <- c("soft", "hard", "mid")  
threshfun<-c("universal", "adaptive", "minimax") 
threshscale<-c(0.25,1.25) 

#I select one variable from above and use lapply which yields results per variable. 

library(wmtsa) 
ws <- lapply(threshfun, function(k,x) 
wavShrink(x,wavelet= "s2",n.level=2,shrink.fun="hard", thresh.fun= k,threshold=NULL,thresh.scale=1, xform="modwt", noise.variance=-1, reflect=TRUE),x=x) 

我的問題是，如果它有可能傳遞的變量的所有更多鈔票組合與x向量該使用應用系列的最後一個功能？例如

a）爲不同變量的所有可能組合創建一個數據框，並將該數據框傳遞給函數，以便以類似的方式將該函數作用於數據框的行，該方法如何將應用應用於原始數據在下面的簡單表達式中，確保函數還必須考慮向量x。

A<-expand.grid(c(1,2), c(3,6), c(4,2), c(2,5,3), c(0.25,1.25)) 
apply(A[,c('Var1' , 'Var5' , 'Var2' , 'Var3' , 'Var4')], 1, function (x) sum(x)) 

b）通過利用mapply在矢量（不知道的輸出將是每個變量更多鈔票組合和所有組合將被傳遞給函數）

# so here is my attempt to use apply to the combination of variables. First I create a data frame of variables. 
A <- expand.grid(c("d2","s2"), c(1,2),c("soft", "hard"),c("universal", "adaptive", "minimax"),c(0.25,1.25)) 

# and write the function 
function(k,l,m,n,e,x) (wavShrink(x,wavelet= k,n.level= l,shrink.fun= m, thresh.fun= n,threshold=NULL,thresh.scale= e, xform="modwt", noise.variance=-1, reflect=TRUE) 

#Attempt to pass the data frame for lapply without success possibly because lapply is for vectors or lists 

A<-expand.grid(c("d2","s2"), c(1,2),c("soft", "hard"),c("universal", "adaptive", "minimax"),c(0.25,1.25)) 

ws<-lapply(A[,c('Var1' , 'Var2' , 'Var3' , 'Var4' , 'Var5')],1, function(k,l,m,n,e,x) 
wavShrink(x,wavelet= k,n.level= l,shrink.fun= m, thresh.fun= n, threshold=NULL,thresh.scale= e, xform="modwt", noise.variance=-1, reflect=TRUE),x=x) 
#Error en match.fun(FUN) : '1' is not a function, character or symbol 

#and for mapply which take multiple lists or vectors I pass the variables as vectors, however it fails as it seems I can not pass the x vector to the function 

FUN<-function(x,k,l,m,n,e) (wavShrink(x,wavelet= k,n.level= l,shrink.fun= m, thresh.fun= n, threshold=NULL,thresh.scale= e, xform="modwt", noise.variance=-1, reflect=TRUE)) 
mapply(FUN, c("d2","s2"), c(1,2),c("soft", "hard"),c("universal", "adaptive", "minimax"),c(0.25,1.25),x=x) 
#Error en wavShrink(x, wavelet = k, n.level = l, shrink.fun = m, thresh.fun = n, : 
#Time series must contain more than one point 

我的結果對R進行全新排序並對應用函數新增我可能在嘗試使用應用程序或函數時犯了基本錯誤，但在嘗試完成兩天後，我決定編寫一個可能很簡單的問題，並且歡迎任何幫助。

Answer 1

我做出undertanding進步一點點mapply所以我相信我能回答我的問題

wavelet<-c("d2","s2") 
n.level<-c(1,2) 
schirnkfun <- c("soft", "hard", "mid")  
threshfun<-c("universal", "adaptive", "minimax") 
threshscale<-c(0.25,1.25) 
x<-c(1,2,3,4,5,7,6,7,8,9,8,7,6,4,5,3,2,2,3,4,5,6,6,7,7,7,5,6,7,7,8,8,9,0,9,0,8,7,5,4,3,3,4,4,4,4,3,2,2,1,2,3,4,5,6,5,7,8,8,9,9,0,2,3,4,2,3,5,5,2,4,6,7) 

w1<- expand.grid(wavelet=wavelet,n.level=n.level,schirnkfun= schirnkfun,threshfun= threshfun,threshscale= threshscale, stringsAsFactors=FALSE) 

result<-mapply(function(m,k,p,u,l,x) (wavShrink(x, wavelet= m,n.level=k,shrink.fun= p,thresh.fun=u, threshold=NULL,thresh.scale= l, xform="modwt", noise.variance=-1, reflect=TRUE)), w1$wavelet, w1$n.level , w1$schirnkfun, w1$ threshfun, w1$threshscale ,MoreArgs=list(x=x)) 

colnames(result)=c(rownames(w1)) 

柱的側向承載力的數量等於從擴大每一行W1的行數，網格評估由功能。