R中的IDW參數

Question

我想使用包中的idw命令使用R執行IDW插值。我有這樣的數據：

#settings 
library(gstat) 
library(dplyr) 
library(sp) 
library(tidyr) 

id_rep <- rep(c(1,2), 20) 
f <- rep(c(930,930.2), each=20) 
perc <- rep(c(90, 80), each=10) 
x <- sample(1:50, 40) 
y <- sample(50:100, 40) 
E <- runif(40) 
df <- data.frame(id_rep, perc, x,y, f, E) 
df_split <- split(df, list(df$id_rep, df$perc, df$f), drop = TRUE, sep="_") 

#grid 
x.range <- range(df$x) 
y.range <- range(df$y) 

grid <- expand.grid(x = seq(x.range[1], x.range[2], by=1), 
         y = seq(y.range[1], y.range[2], by=1)) 
coordinates(grid) <- ~x + y 

#interpolation 
lst_interp_idw <- lapply(df_split, function(X) { 

    coordinates(X) <- ~x + y 
    E_idw <- idw(E~ 1, X, grid, idp=1, nmax=3) %>% as.data.frame() 

    df_interp <- select(E_idw, x,y,E_pred=var1.pred) 
    df_interp 
}) 

    df_interp_idw <- bind_rows(lst_interp_idw, .id = "interact") %>% 
    separate(interact, c("id_rep", "perc", "f"), sep = "\\_") 

現在我想要（0.5 IDP從1至3，並且由1 n最大3至6），以執行與特定值內的不同idp和nmax參數每次運行，並獲得了數據幀包含idp和nmax值的每個組合的列。我嘗試了兩個for循環，但它不起作用。

編輯 不工作的代碼是：

idp = seq(from = 1, to = 3, by = 0.5) 
nmax = seq(from = 3, to = 6, by = 1) 

... 
for(i in idp) { 
    for(j in nmax) 
{ E_idw= idw(E ~ 1, X, grid, nmax = i, idp = j) 
    } 
} 
... 

Answer 1

這是一種如何存儲列表中的每一次迭代的結果。

#settings 
#install.packages("gstat") 
library(gstat) 
library(dplyr) 
library(sp) 
library(tidyr) 

id_rep <- rep(c(1,2), 20) 
f <- rep(c(930,930.2), each=20) 
perc <- rep(c(90, 80), each=10) 
x <- sample(1:50, 40) 
y <- sample(50:100, 40) 
E <- runif(40) 
df <- data.frame(id_rep, perc, x,y, f, E) 
df_split <- split(df, list(df$id_rep, df$perc, df$f), drop = TRUE, sep="_") 

#grid 
x.range <- range(df$x) 
y.range <- range(df$y) 

grid <- expand.grid(x = seq(x.range[1], x.range[2], by=1), 
        y = seq(y.range[1], y.range[2], by=1)) 
coordinates(grid) <- ~x + y 

# ============================================== 
# NEW function 
# ============================================== 

idp = seq(from = 1, to = 3, by = 0.5) 
nmax = seq(from = 3, to = 6, by = 1) 

#interpolation 
lst_interp_idw <- lapply(df_split, function(X) { 

    coordinates(X) <- ~x + y 

    df_interp <- vector(length(idp)*length(nmax), mode = "list") 

    k <- 0 

    for(i in idp) { 

    for(j in nmax) { 

     print(paste(i, j)) 

     # Iterator 
     k <- k + 1 

     E_idw= idw(E ~ 1, X, grid, nmax = i, idp = j) %>% as.data.frame() 

     df_interp[[k]] <- select(E_idw, x,y,E_pred=var1.pred) 

    } 
    } 

    return(df_interp) 
}) 

# ============================================== 

一些合理性檢查（lapply應用於8個列表元素和20個變化計算）：

length(lst_interp_idw) # 8 
length(lst_interp_idw[[1]]) #20 
length(lst_interp_idw[[1]]) #20 

它應該很容易爲你去適應你的代碼的最後一行

df_interp_idw <- bind_rows(lst_interp_idw, .id = "interact") %>% 
    separate(interact, c("id_rep", "perc", "f"), sep = "\\_") 

以所需格式格式化輸出。這很大程度上取決於您想如何呈現不同的插值替代方法。