如何在dplyr+;呼噜声
我有一个问题,在复制了培训和测试集的数据后,我在Rstudio中显示了分配给用户的大量内存,但没有在R会话中使用。我创建了一个小示例来重现我的情况:) 这段代码基于我给出的不同公式、算法和参数集运行一系列模型。这是一个函数,但我为reprex创建了一个简单的脚本如何在dplyr+;呼噜声,r,dplyr,tidyverse,purrr,R,Dplyr,Tidyverse,Purrr,我有一个问题,在复制了培训和测试集的数据后,我在Rstudio中显示了分配给用户的大量内存,但没有在R会话中使用。我创建了一个小示例来重现我的情况:) 这段代码基于我给出的不同公式、算法和参数集运行一系列模型。这是一个函数,但我为reprex创建了一个简单的脚本 library(dplyr) library(purrr) library(modelr) library(tidyr) library(pryr) # set my inputs data <- mtcars formulas
library(dplyr)
library(purrr)
library(modelr)
library(tidyr)
library(pryr)
# set my inputs
data <- mtcars
formulas <- c(test1 = mpg ~ cyl + wt + hp,
test2 = mpg ~ cyl + wt)
params = list()
methods <- "lm"
n <- 20 # num of cv splits
mult <- 10 # number of times I want to replicate some of the data
frac <- .25 # how much I want to cut down other data (fractional)
### the next few chunks get the unique combos of the inputs.
if (length(params) != 0) {
cross_params <- params %>%
map(cross) %>%
map_df(enframe, name = "param_set", .id = "method") %>%
list
} else cross_params <- NULL
methods_df <- tibble(method = methods) %>%
list %>%
append(cross_params) %>%
reduce(left_join, by = "method") %>%
split(1:nrow(.))
# wrangle formulas into a split dataframe
formulas_df <- tibble(formula = formulas,
name = names(formulas)) %>%
split(.$name)
# split out the data into n random train-test combos
cv_data <- data %>%
crossv_kfold(n) %>% # rsample?
mutate_at(vars(train:test), ~map(.x, as_tibble))
# sample out if needed
cv_data_samp <- cv_data %>%
mutate(train = modify(train,
~ .x %>%
split(.$gear == 4) %>%
# take a sample of the non-vo data
modify_at("FALSE", sample_frac, frac) %>%
# multiply out the vo-on data
modify_at("TRUE", function(.df) {
map_df(seq_along(1:mult), ~ .df)
}) %>%
bind_rows))
# get all unique combos of formula and method
model_combos <- list(cv = list(cv_data_samp),
form = formulas_df,
meth = methods_df) %>%
cross %>%
map_df(~ bind_cols(nest(.x$cv), .x$form, .x$meth)) %>%
unnest(data, .preserve = matches("formula|param|value")) %>%
{if ("value" %in% names(.)) . else mutate(., value = list(NULL))}
# run the models
model_combos %>%
# put all arguments into a single params column
mutate(params = pmap(list(formula = formula, data = train), list)) %>%
mutate(params = map2(params, value, ~ append(.x, .y))) %>%
mutate(params = modify(params, discard, is.null)) %>%
# run the models
mutate(model = invoke_map(method, params))
mem_change(rm(data, cv_data, cv_data_samp))
mem_used()
库(dplyr)
图书馆(purrr)
库(建模器)
图书馆(tidyr)
图书馆(普赖尔)
#设置我的输入
数据%
分割(.$gear==4)%>%
#对非vo数据进行采样
修改(错误),样本分数,分数)%>%
#乘以数据上的vo
在函数(.df)的“真”处修改{
地图测向(沿(1:mult)的顺序,~.df)
}) %>%
绑定(行)
#获取公式和方法的所有唯一组合
型号组合%
交叉%>%
map_df(~bind_cols(nest(.x$cv),.x$form,.x$meth))%>%
unnest(数据,.preserve=matches(“公式|参数|值”))%>%
{if(“值”%in%names(.)。else mutate(,value=list(NULL))}
#运行模型
型号组合%>%
#将所有参数放入单个参数列中
突变(参数=pmap(列表(公式=公式,数据=序列),列表))%>%
mutate(params=map2(params,value,~append(.x,.y)))%>%
mutate(params=modify(params,discard,is.null))%>%
#运行模型
变异(模型=调用映射(方法,参数))
mem_变更(rm(数据、cv_数据、cv_数据、samp))
mem_used()
现在,在我这样做之后,我的mem_used
显示为77.3mb,但我看到分配给我的R用户的内存(160Mb)大约是原来的两倍。当我的数据为3 Gb时,这真的会爆炸,这是我的真实情况。我最终使用了100Gb并占用了整个服务器:(
发生了什么以及如何优化
感谢任何帮助!!!我明白了!问题是我正在将我的
modeler
resample
系列对象转换为tibble
s,这会爆炸内存,即使我随后对它们进行了采样。解决方案是编写处理resample
对象的方法,这样我就不会er必须将重采样
对象转换为TIBLE
。这些看起来像:
# this function just samples the indexes instead of the data
sample_frac.resample <- function(data, frac) {
data$idx <- sample(data$idx, frac * length(data$idx))
data
}
# this function replicates the indexes. I should probably call it something else.
augment.resample <- function(data, n) {
data$idx <- unlist(map(seq_along(1:n), ~ data$idx))
data
}
# This function does simple splitting (logical only) of resample obejcts
split.resample <- function(data, .p) {
pos <- list(data = data$data, idx = which(.p, 1:nrow(data$data)))
neg <- list(data = data$data, idx = which(!.p, 1:nrow(data$data)))
class(pos) <- "resample"
class(neg) <- "resample"
list("TRUE" = pos,
"FALSE" = neg)
}
# This function takes the equivalent of a `bind_rows` for resample objects.
# Since bind rows does not call `useMethod` I had to call it something else
bind <- function(data) {
out <- list(data = data[[1]]$data, idx = unlist(map(data, pluck, "idx")))
class(out) <- "resample"
out
}
#此函数仅对索引而不是数据进行采样
sample\u frac.resample我发现了这个问题!问题是我正在将我的modeler
resample
对象系列转换为tibble
s,这会爆炸内存,即使我随后对它们进行了采样。解决方案是编写处理resample
对象的方法,这样我就永远不会o将重采样
对象转换为tibble
。它们看起来像:
# this function just samples the indexes instead of the data
sample_frac.resample <- function(data, frac) {
data$idx <- sample(data$idx, frac * length(data$idx))
data
}
# this function replicates the indexes. I should probably call it something else.
augment.resample <- function(data, n) {
data$idx <- unlist(map(seq_along(1:n), ~ data$idx))
data
}
# This function does simple splitting (logical only) of resample obejcts
split.resample <- function(data, .p) {
pos <- list(data = data$data, idx = which(.p, 1:nrow(data$data)))
neg <- list(data = data$data, idx = which(!.p, 1:nrow(data$data)))
class(pos) <- "resample"
class(neg) <- "resample"
list("TRUE" = pos,
"FALSE" = neg)
}
# This function takes the equivalent of a `bind_rows` for resample objects.
# Since bind rows does not call `useMethod` I had to call it something else
bind <- function(data) {
out <- list(data = data[[1]]$data, idx = unlist(map(data, pluck, "idx")))
class(out) <- "resample"
out
}
#此函数仅对索引而不是数据进行采样
样品压裂重取样