使用 data.frame/list 作为函数的参数进行映射

Question

简而言之，我有一个更大的函数，它创建的 data.frames 是一个更大的 data.frame 的子集，并以函数的参数命名。它正在为原始数据以及 Holt-Winters 的输出和预测输出构建 data.frames……这意味着它正在创建多个 data.frames。一个小例子如下（虽然这里没有足够的时间间隔来实际生成一个 ts 类 data.frame）：

Group <- c("Primary_Group","Primary_Group","Primary_Group","Primary_Group","Primary_Group","Primary_Group","Secondary_Group","Secondary_Group","Secondary_Group","Secondary_Group","Secondary_Group","Secondary_Group","Tertiary_Group","Tertiary_Group","Tertiary_Group","Tertiary_Group","Tertiary_Group","Tertiary_Group")
Day <- c(1,2,3,1,2,3,1,2,3,1,2,3,1,2,3,1,2,3)
Type <- c("A","A","A","B","B","B","A","A","A","B","B","B","A","A","A","B","B","B")
Value <- c(7,3,10,3,9,4,0,9,3,10,1,6,3,4,10,2,3,1)
df <- as.data.frame(cbind(Group,Day,Type,Value))

Fun <- function(Group,Type, A, B, G){
    df <- Data[Data$Group== Group & Data$Type== Type, ]
    assign(paste(Group,Type,"_df",sep = ''), df, envir = parent.frame()) 
    df_holtwinters <- HoltWinters(ts(Data[Data$Group== Group & Data$Type== Type, ], 
                                  frequency = 365), alpha = A, beta = B, gamma = G)
    assign(paste(Group,Type,"_hw",sep = ''), df_holtwinters, envir = parent.frame()) 
}

您会注意到 Group 和 Type 是字符，而 A、B、G 是数字或 NULL。

如果我现在有一个由列表值组成的 data.frame，我怎样才能最好地循环上述函数（可能使用 mapply）以使用第一行中每一列的值......然后是第 2 行中的每一列等 - 创建几个数据框。

argGroup <- c("Primary_Group","Primary_Group","Secondary_Group","Secondary_Group","Tertiary_Group","Tertiary_Group")
argType <- c("A","B","A","B","A","B")
argA <- c(NA, NA, NA, NA, NA, NA)
argB <- c(0.05, 0.05, NA, NA, NA, NULL)
argG <- c(NA, NA, NA, NA, NA, NA)

argGroup[is.na(argGroup)] <- list(NULL)
argType[is.na(argType)] <- list(NULL)
argA[is.na(argA)] <- list(NULL)
argB[is.na(argB)] <- list(NULL)
argG[is.na(argG)] <- list(NULL)

Arguments <- cbind(argType, argType, argA, argB, argG)

理想情况下，我会生成以下 data.frames...

Primary_Group_A_df
Primary_Group_A_hw
Primary_Group_B_df
Primary_Group_B_hw
Secondary_Group_A_df
Secondary_Group_A_hw
Secondary_Group_B_df
Secondary_Group_B_hw
Tertiary_Group_A_df
Tertiary_Group_A_hw
Tertiary_Group_B_df
Tertiary_Group_B_hw

了解如何最好地（最自动化的方式）rbind 将所有 _df 和所有 _hw 放在一起也会很有帮助。

任何帮助都将是惊人的，非常感谢。非常感谢！

Answer 1

避免使用许多类似结构的对象淹没您的全局环境。考虑使用诸如列表之类的容器来保存许多数据帧。一种有用的方法是by 按一个或多个因素（例如 Group 和 Type）对数据帧进行子集化，以返回数据帧列表。此外，不要按行迭代，而是 merge 参数与每个子集的一次参数传递的数据。

具体来说，为 df 和 hw 列表调用两次by。但首先，按 Group 和 Type 合并 df 和 Arguments 数据框。一个挑战是NULL 不能存储在数据框中，因此请考虑保存"NULL" 字符串并分配临时变量以传递给HW 参数。不幸的是，这会将整个列转换为字符类型，您需要将其转换为 as.numeric 以获得非 NULL 值。

合并

Group <- c("Primary_Group","Primary_Group","Secondary_Group","Secondary_Group",
           "Tertiary_Group","Tertiary_Group")
Type <- c("A","B","A","B","A","B")
argA <- c("NULL", "NULL", "NULL", "NULL", "NULL", "NULL")
argB <- c(0.05, 0.05, "NULL", "NULL", "NULL", "NULL")
argG <- c("NULL", "NULL", "NULL", "NULL", "NULL", "NULL")

Arguments <- data.frame(Group, Type, argA, argB, argG, stringsAsFactors=FALSE)
df <- merge(df, Arguments, by=c("Group", "Type"))

数据框列表 （带有命名的df元素）

# ORDER FOR NAMING LATER
df <- with(df, df[order(Type, Group),])

# DATAFRAME LIST
df_list <- by(df, df[c("Group", "Type")], identity)
# RENAME LIST
df_list <- setNames(df_list, unique(paste0(df$Group, "_", df$Type, "_df")))

# REFERENCE ELEMENTS
df_list$Primary_Group_A_df
df_list$Secondary_Group_A_df
df_list$Tertiary_Group_A_df
...

硬件列表 （带有命名的硬件元素）

# HW LIST
hw_list <- by(df, df[c("Group", "Type")], function(sub) {
  # CONDITIONALLY ASSIGN TEMP VARIABLES 
  # (BEING SUBSETS: max(arg*)==min(arg*)==mean(arg*)==median(arg*))
  if(!is.na(max(sub$argA)) & max(sub$argA) == "NULL") { tmpA <- NULL } 
  else { tmpA <- max(as.numeric(sub$argA)) }

  if(!is.na(max(sub$argB)) & max(sub$argB) == "NULL") { tmpB <- NULL } 
  else { tmpB <- max(as.numeric(sub$argB)) }

  if(!is.na(max(sub$argG)) & max(sub$argG) == "NULL") { tmpG <- NULL } 
  else { tmpG <- max(as.numeric(sub$argG)) }

  # PASS ARGS ONCE PER SUBSET 
  return(HoltWinters(ts(sub, frequency = 365), alpha=tmpA, beta=tmpB, gamma=tmpG))
})

# RENAME LIST
hw_list <- setNames(hw_list, unique(paste0(df$Group, "_", df$Type, "_hw")))

# REFERENCE ELEMENTS
hw_list$Primary_Group_A_hw
hw_list$Secondary_Group_A_hw
hw_list$Tertiary_Group_A_hw
...

输出 （使用 3 表示 HW 的频率以与发布的数据对齐）

> hw_list$Primary_Group_A_hw
Holt-Winters exponential smoothing with trend and additive seasonal component.

Call:
HoltWinters(x = ts(sub[c("Group", "Day", "Type", "Value")], frequency = 3),     alpha = tmpA, beta = tmpB, gamma = tmpG)

Smoothing parameters:
 alpha: 0.2169231
 beta : 0.05
 gamma: 0.1

Coefficients:
          [,1]
a   2.89129621
b   0.08783715
s1  0.54815382
s2 -0.12485260
s3  0.21087038

> hw_list$Secondary_Group_A_hw
Holt-Winters exponential smoothing with trend and additive seasonal component.

Call:
HoltWinters(x = ts(sub[c("Group", "Day", "Type", "Value")], frequency = 3),     alpha = tmpA, beta = tmpB, gamma = tmpG)

Smoothing parameters:
 alpha: 0.752124
 beta : 0
 gamma: 0

Coefficients:
            [,1]
a   3.691664e+00
b   3.333333e-01
s1  3.333333e-01
s2 -1.480388e-16
s3 -3.333333e-01

> hw_list$Tertiary_Group_A_hw
Holt-Winters exponential smoothing with trend and additive seasonal component.

Call:
HoltWinters(x = ts(sub[c("Group", "Day", "Type", "Value")], frequency = 3),     alpha = tmpA, beta = tmpB, gamma = tmpG)

Smoothing parameters:
 alpha: 0.3145406
 beta : 0
 gamma: 0

Coefficients:
            [,1]
a   3.022946e+00
b  -3.333333e-01
s1 -3.333333e-01
s2 -1.480388e-16
s3  3.333333e-01

Answer 2

使用as.data.frame(cbind(...)) 会丢失类型信息， 直接使用data.frame：

Data <- data.frame(
  Group = rep(c("Primary_Group", "Secondary_Group", "Tertiary_Group"), each = 6L),
  Day = rep(1L:3L, 6L),
  Type = rep(rep(c("A", "B"), each = 3L), 3L),
  Value = c(7,3,10,3,9,4,0,9,3,10,1,6,3,4,10,2,3,1)
)

之后，我认为您可以执行以下操作：

split_data <- split(Data, as.list(Data[, c("Group", "Type")]))
dfs <- do.call(rbind, split_data)

dfs_hw <- lapply(split_data, function(sub_data) {
  Map(argA, argB, argG, f = function(A, B, G) {
    HoltWinters(ts(sub_data, frequency = 365), alpha = A, beta = B, gamma = G)
  })
})

dfs_hw <- do.call(rbind, unlist(dfs_hw, recursive = FALSE))

但我收到来自HoltWinters 的错误， 所以我不能肯定地说。 另外，我认为dfs 又是Data，只是重新排序。