r语言 - 以迭代方式将多个列中的一个合并到列表中的数据帧

Question

我在迭代地将单个数据帧中的列与列表中的其他数据帧组合时遇到了一个(可能很小的(问题。一些数据可以说明：

# load example data
library(vegan)
data(varechem)
data(varespec)
# generate predictor tables with overlapping rows and different amount of cols
varespec1 <- varespec[c(1:9), ]
varespec2 <- varespec[c(8:16), c(1:43)]
varespec3 <- varespec[c(14:24), c(1:41)]
# store predictor tables in list
subset_list <- list(varespec1 = varespec1, 
varespec2 = varespec2, 
varespec3 = varespec3)
# generate a table that holds ALL possible response variables as presence/absence
varechem_binary <- as.data.frame(apply(varechem, 2, cut, 
breaks = c(-Inf, 1.0, Inf), labels = c("Absent", "Present")))
row.names(varechem_binary) <- row.names(varechem)

上面的代码说明了我如何为分类任务准备数据。现在的想法是，列表中包含预测变量(varespec1，...(的data.frames应该用于预测响应表(varechem_binary(中的每一列，但一次只能预测一列。将响应表与每个预测变量表合并很容易：

# merge response table with each predictor table
merge_counter <- 0
merged_list <- list()
for(table in subset_list) {
merge_counter <- merge_counter + 1
current_name <- names(subset_list)[merge_counter]
tmp <- merge(table, varechem_binary, by = "row.names")
row.names(tmp) <- tmp$Row.names
tmp <- tmp[, -1]
merged_list[[current_name]] <- tmp
rm(tmp)
}

预期产出：

我现在(或代码的早期，如果这更有意义(正在寻找一种将每个预测变量表与响应表中的每一列和恰好一列组合在列表中的方法varechem。这基本上是：

# storing in data frames just for illustration, I would like to do this within the list
# subsets for the 3 predictor tables with the first response variable
aa <- merged_list[[1]][,-c(46:58)]  # column 1:44 are the predictor variables, then the different response variables start
bb <- merged_list[[2]][,-c(45:57)]  # column 1:43 are the predictor variables, then the different response variables start
cc <- merged_list[[3]][,-c(43:58)] # column 1:41 are the predictor variables, then the different response variables start
# subsets for the 3 predictor tables with the second response variable
dd <- merged_list[[1]][,-c(45, 47:58)]
ee <- merged_list[[2]][,-c(44, 46:57)]
ff <- merged_list[[3]][,-c(42, 44:58)]
# subsets for the 3 predictor tables with the third response variable
gg <- merged_list[[1]][,-c(45, 46, 48:58)]
...
# this is just to illustrate how the list could look like, I would like to keep all files in a list all the time
list_for_classification_runs <- list(aa, bb, cc, dd, ee, ff, gg, ...)

此结果列表将是随机森林分类调用的输入，其中响应变量将由来自varespec的所有其他预测变量进行分类，例如：

for (current_table in list_for_classification_runs) {
counter <- counter + 1 
# response_variable should be the one variable added to the predictor variables in the data frames 
RF_list[[counter]] <- ranger(response_variable ~ ., data = current_table)
}

Answer 1

根据格雷戈尔的评论，我想出了一个类似的方法。我没有将完整的varechem_binary与subset_list的所有元素合并，而是添加了另一个 for 循环并迭代了varechem_binary中的所有列。使用drop = FALSE将保留 row.names 和结构，因此合并有效：

merge_col_counter <- 0
column_counter <- 0
merged_column_list <- list()
for(table in subset_list) {
merge_col_counter <- merge_col_counter + 1
for (column in names(varechem_binary)) {
column_counter <- column_counter + 1
current_name <- paste(names(subset_list)[merge_col_counter], names(varechem_binary)[column_counter], sep = "_")
print(current_name)
tmp <- merge(table, varechem_binary[, column_counter, drop = FALSE], by = "row.names")
row.names(tmp) <- tmp$Row.names
tmp <- tmp[, -1]
merged_column_list[[current_name]] <- tmp
rm(tmp)
}
column_counter <- 0
}

可能有办法做得更干净或更有效，但它有效，所以我可以继续

Answer 2

使用应用函数的另一种解决方案：

lapply(subset_list, function(x) apply(varechem_binary, 2, function(var) merge(var, x, by= 'row.names')))

使用system.time(( 使用示例数据对这两种方法进行基准测试，此方法的速度(0.075 个用户时间(是使用 for 循环的解决方案(0.143 个用户时间(的两倍。