用于随机假设检验 r 的 lapply 而不是 for 循环 [英] lapply instead of for loop for randomised hypothesis testing r
问题描述
我有一个看起来像这样的 df:
I have a df that looks something like this like this:
set.seed(42)
ID <- sample(1:30, 100, rep=T)
Trait <- sample(0:1, 100, rep=T)
Year <- sample(1992:1999, 100, rep=T)
df <- cbind(ID, Trait, Year)
df <- as.data.frame(df)
ID 是个体生物体，性状是表型的存在/不存在，Year 是进行观察的年份.
Where ID is an individual organism, trait is a presence/absence of a phenotype and Year is the year an observation was made.
如果特征在个体之间是随机的，我想建模，就像这样
I would like to model if trait is random between individuals, something like this
library(MCMCglmm)
m <- MCMCglmm(Trait ~ ID, random = ~ Year, data = df, family = "categorical")
现在，想改组 Trait 列并运行 x 排列，以检查我观察到的均值和 CI 是否超出了随机预期的范围.我可以用 for 循环来做到这一点，但我宁愿使用 tidyverse 解决方案.我已经读过 lapply 是一个更好的选择(?)，但我正在努力寻找一个我可以遵循的足够具体的演练.
Now, would like to shuffle the Trait column and run x permutations, to check if my observed mean and CI fall outside of what's expected from random. I could do this with a for loop, but I'd rather use a tidyverse solution. I've read that lapply is a bette(?) alternative, but I am struggling to find a specific enough walk-through that I can follow.
我很感激这里提供的任何建议.
I'd appreciate any advice offered here.
干杯！
杰米
推荐答案
EDIT 10 月 10 日. 清理了代码，并在下面的每条评论中添加了代码，为您提供一个井井有条的 tibble\dataframe
EDIT October 10th. Cleaned up the code and per comment below added the code to give you back a nice organized tibble\dataframe
### decide how many shuffles you want and name them
### in an orderly fashion for the output
shuffles <- 1:10
names(shuffles) <- paste0("shuffle_", shuffles)
library(MCMCglmm)
library(dplyr)
library(tibble)
library(purrr)
ddd <- purrr::map(shuffles,
~ df %>%
mutate(Trait = sample(Trait)) %>%
MCMCglmm(fixed = Trait ~ ID,
random = ~ Year,
data = .,
family = "categorical",
verbose = FALSE)) %>%
purrr::map( ~ tibble::as_tibble(summary(.x)$solutions, rownames = "model_term")) %>%
dplyr::bind_rows(., .id = 'shuffle')
ddd
#> # A tibble: 20 x 7
#> shuffle model_term post.mean `l-95% CI` `u-95% CI` eff.samp pMCMC
#> <chr> <chr> <dbl> <dbl> <dbl> <dbl> <dbl>
#> 1 shuffle_1 (Intercept) 112. 6.39 233. 103. 0.016
#> 2 shuffle_1 ID -6.31 -13.5 -0.297 112. 0.014
#> 3 shuffle_2 (Intercept) 24.9 -72.5 133. 778. 0.526
#> 4 shuffle_2 ID -0.327 -6.33 5.33 849. 0.858
#> 5 shuffle_3 (Intercept) 4.39 -77.3 87.4 161. 0.876
#> 6 shuffle_3 ID 1.04 -3.84 5.99 121. 0.662
#> 7 shuffle_4 (Intercept) 7.71 -79.0 107. 418. 0.902
#> 8 shuffle_4 ID 0.899 -4.40 6.57 408. 0.694
#> 9 shuffle_5 (Intercept) 30.4 -62.4 144. 732. 0.51
#> 10 shuffle_5 ID -0.644 -6.61 4.94 970. 0.866
#> 11 shuffle_6 (Intercept) -45.5 -148. 42.7 208. 0.302
#> 12 shuffle_6 ID 4.73 -0.211 11.6 89.1 0.058
#> 13 shuffle_7 (Intercept) -16.2 -133. 85.9 108. 0.696
#> 14 shuffle_7 ID 2.47 -2.42 10.3 47.8 0.304
#> 15 shuffle_8 (Intercept) 0.568 0.549 0.581 6.60 0.001
#> 16 shuffle_8 ID -0.0185 -0.0197 -0.0168 2.96 0.001
#> 17 shuffle_9 (Intercept) -6.95 -112. 92.2 452. 0.886
#> 18 shuffle_9 ID 2.07 -3.30 8.95 370. 0.476
#> 19 shuffle_10 (Intercept) 43.8 -57.0 159. 775. 0.396
#> 20 shuffle_10 ID -1.36 -7.44 5.08 901. 0.62
您的原始数据
set.seed(42)
ID <- sample(1:30, 100, rep=T)
Trait <- sample(0:1, 100, rep=T)
Year <- sample(1992:1999, 100, rep=T)
df <- cbind(ID, Trait, Year)
df <- as.data.frame(df)
这篇关于用于随机假设检验 r 的 lapply 而不是 for 循环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！
