可以将此for循环向量化 [英] Can this for-loop be vectorized
问题描述
我使用for循环在R中实施了卡方检验，以便计算每个单元的检验统计量.但是，我想知道这是否可以优化.R中的chi = square是否可以用作我的代码?
I implemented a chi-square test in R using a for-loop in order to calculate the test statistic for every cell. However, I was wondering whether this can be optimized. And is the chi=square in R working as my code?
eval_preds <- function(df, observations, predictions, groups) {
# determine cells
cells <- unique(df[,groups])
my_sum = 0
# compute test statistic for every cell
for (i in 1:length(cells)) {
# get cell means
m_obs <- mean(df[df[,groups] == cells[i], observations])
m_pred <- mean(df[df[,groups] == cells[i], predictions])
# get cell variance
var_obs <- var(df[df[,groups] == cells[i], observations])
var_pred <- var(df[df[,groups] == cells[i], predictions])
# get cell's number of observations
N_obs <- length(df[df[,groups] == cells[i], observations])
N_pred <- length(df[df[,groups] == cells[i], predictions])
# sum up
my_sum = my_sum + (m_obs-m_pred) / ((var_obs/N_obs) + (var_pred/N_pred))
}
return(my_sum)
}
还有一个玩具示例:
# save this as df_head.csv
"RT","RTmodel_s","groups"
899,975.978308710825,"pl_sgDom"
1115,1095.61538562629,"sg_sgDom"
1077,1158.19266217845,"pl_sgDom"
850,996.410033287249,"sg_plDom"
854,894.862587823602,"pl_sgDom"
1720,1046.34200941684,"sg_sgDom"
# load dat into R
df <- read.csv('./df_head.csv')
my_chi <- eval_preds(df, 'RT', 'Pred', 'Group')
编辑
在for循环中调用eval_preds函数，在该循环中，基于自由参数t_parsing计算不同的预测.
The function eval_preds function is called in a for loop in which different predictions are computed based on a free parameter t_parsing.
p_grid = seq(0,1,0.1)
# tune p
for (i in 1:length(p_grid)) {
# set t_parsing
t_parsing = p_grid[i]
# compute model-time RTs
df$RTmodel <- ifelse(df$number == 'sing',
RT_lookup(df$sg_t_act, df$epsilon),
RT_decompose(df$sg_t_act, df$pl_t_act, df$epsilon))
# scale into real time
df$RTmodel_s <- scale_RTmodel(df$RT, df$RTmodel)
# compare model output to measured RT
# my_chi <- eval_preds(my_nouns, 'RT', 'RTmodel_s', 'groups')
my_chi <- eval_preds1(df, RT, RTmodel_s, groups) #function written by DaveArmstrong
print(paste(p_grid[i], ': ', my_chi))
}
RT 和 RTmodel_s 是数字变量， groups 是字符变量.
RT and RTmodel_s are numerical variables, groups is a character variable.
推荐答案
当然，您可以使用 dplyr 和 rlang :
eval_preds <- function(df, observations, predictions, groups) {
# determine cells
require(dplyr)
require(rlang)
groups <- enquo(groups)
observations <- enquo(observations)
predictions <- enquo(predictions)
out <- DF %>%
group_by(!!groups) %>%
summarise(m_obs = mean(!!observations),
m_pred = mean(!!predictions),
var_obs = var(!!observations),
var_pred = var(!!predictions),
N_obs = sum(!is.na(!!observations)),
N_pred = sum(!is.na(!!predictions))) %>%
mutate(my_sum = (m_obs - m_pred)/((var_obs/N_obs) + (var_pred/N_pred)))
sum(out$my_sum)
}
my_chi <- eval_preds(DF, RT, Pred, Group)
my_chi
# [1] 1.6
或者，更简化:
eval_preds <- function(df, observations, predictions, groups) {
# determine cells
require(dplyr)
require(rlang)
groups <- enquo(groups)
observations <- enquo(observations)
predictions <- enquo(predictions)
out <- DF %>%
group_by(!!groups) %>%
summarise(diff= mean(!!observations) - mean(!!predictions),
sum_v = (var(!!observations)/n()) + (var(!!predictions)/n())) %>%
mutate(my_sum = diff/sum_v)
sum(out$my_sum)
}
my_chi <- eval_preds(DF, RT, Pred, Group)
my_chi
# [1] 1.6
编辑-添加了基准
因此，我认为哪个更好的问题取决于数据集的大小.我还想将一个函数添加到混合中-一个使用基数R中的 by()的函数:
So, I think the question of which one is better depends on the size of the data set. I also want to throw one more function into the mix - one that uses by() from base R:
eval_preds <- function(df, observations, predictions, groups) {
# determine cells
cells <- unique(df[,groups])
my_sum = 0
# compute test statistic for every cell
for (i in 1:length(cells)) {
# get cell means
m_obs <- mean(df[df[,groups] == cells[i], observations])
m_pred <- mean(df[df[,groups] == cells[i], predictions])
# get cell variance
var_obs <- var(df[df[,groups] == cells[i], observations])
var_pred <- var(df[df[,groups] == cells[i], predictions])
# get cell's number of observations
N_obs <- length(df[df[,groups] == cells[i], observations])
N_pred <- length(df[df[,groups] == cells[i], predictions])
# sum up
my_sum = my_sum + (m_obs-m_pred) / ((var_obs/N_obs) + (var_pred/N_pred))
}
return(my_sum)
}
eval_preds1 <- function(df, observations, predictions, groups) {
# determine cells
require(dplyr)
require(rlang)
groups <- enquo(groups)
observations <- enquo(observations)
predictions <- enquo(predictions)
out <- df %>%
group_by(!!groups) %>%
summarise(m_obs = mean(!!observations),
m_pred = mean(!!predictions),
var_obs = var(!!observations),
var_pred = var(!!predictions),
N_obs = sum(!is.na(!!observations)),
N_pred = sum(!is.na(!!predictions))) %>%
ungroup %>%
mutate(my_sum = (m_obs - m_pred)/((var_obs/N_obs) + (var_pred/N_pred)))
sum(out$my_sum)
}
eval_preds2 <- function(df, observations, predictions, groups) {
# determine cells
require(dplyr)
require(rlang)
groups <- enquo(groups)
observations <- enquo(observations)
predictions <- enquo(predictions)
out <- df %>%
group_by(!!groups) %>%
summarise(diff= mean(!!observations) - mean(!!predictions),
sum_v = (var(!!observations)/n()) + (var(!!predictions)/n())) %>%
ungroup %>%
mutate(my_sum = diff/sum_v)
sum(out$my_sum)
}
eval_preds3<- function(df, observations, predictions, groups) {
# determine cells
m <- by(df[,c(observations, predictions)], list(df[[groups]]), function(x)diff(-colMeans(x)))
v <- by(df[,c(observations, predictions)], list(df[[groups]]), function(x)sum(apply(x, 2, var)/nrow(x)))
sum(m/v)
}
因此， eval_preds()是原始代码， eval_preds1()是第一组 dplyr 代码和 eval_preds2()， eval_preds3()是带有 by()的基数R.在原始数据集上，这是 microbenchmark()的输出.
So, eval_preds() is the original, eval_preds1() is the first set of dplyr code and eval_preds2(), eval_preds3() is the base R with by(). On the original data set, here is the output from microbenchmark().
microbenchmark(eval_preds(DF, 'RT', 'Pred', 'Group'),
eval_preds1(DF, RT, Pred, Group),
eval_preds2(DF, RT, Pred, Group),
eval_preds4(DF, 'RT', 'Pred', 'Group'), times=100)
# Unit: microseconds
# expr min lq mean median uq max neval cld
# eval_preds(DF, "RT", "Pred", "Group") 236.513 279.4920 324.9760 295.190 321.0125 774.213 100 a
# eval_preds1(DF, RT, Pred, Group) 5236.850 5747.5095 6503.0251 6089.343 6937.8670 12950.677 100 d
# eval_preds2(DF, RT, Pred, Group) 4871.812 5372.2365 6070.7297 5697.686 6548.8935 14577.786 100 c
# eval_preds4(DF, "RT", "Pred", "Group") 651.013 739.9405 839.1706 773.610 923.9870 1582.218 100 b
在这种情况下，原始代码是最快的.但是，如果我们制作一个更大的数据集-这是一个具有1000个组且每个组10个观测值的数据集.
In this case, the original code is the fastest. However, if we made a bigger dataset - here's one with 1000 groups and 10 observations per group.
DF2 <- data.frame(
Group = rep(1:1000, each=10),
RT = rpois(10000, 3),
Pred = rpois(10000, 4)
)
microbenchmark()的输出在这里讲述了一个完全不同的故事.
The output from microbenchmark() here tells a quite different story.
microbenchmark(eval_preds(DF2, 'RT', 'Pred', 'Group'),
eval_preds1(DF2, RT, Pred, Group),
eval_preds2(DF2, RT, Pred, Group),
eval_preds3(DF2, 'RT', 'Pred', 'Group'), times=25)
# Unit: milliseconds
# expr min lq mean median uq max neval cld
# eval_preds(DF2, "RT", "Pred", "Group") 245.67280 267.96445 353.6489 324.29416 419.4193 565.4494 25 c
# eval_preds1(DF2, RT, Pred, Group) 74.56522 88.15003 102.6583 92.24103 106.6766 211.2368 25 a
# eval_preds2(DF2, RT, Pred, Group) 79.00919 89.03754 125.8202 94.71703 114.5176 606.8830 25 a
# eval_preds3(DF2, "RT", "Pred", "Group") 193.94042 240.35447 272.0004 254.85557 316.5098 420.5394 25 b
两个基于 dplyr()的函数都比其基本R竞争对手要快得多.因此，问题是最佳".方法取决于要解决的问题的大小.
Both dplyr() based functions are much faster than their base R competitors. So, the question of which is the "optimal" method depends on the size of the problem to be solved.
这篇关于可以将此for循环向量化的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
