全部,
我有一个dplyr sample_n()
问题。我在尝试使用weight选项时尝试更换,但似乎遇到了障碍。即,替换抽样总是使一个组过度抽样。在不进行替换的情况下采样时不会出现问题,但如果可以的话,我真的很想在进行替换时进行采样。
这是一个最小的工作示例,它使用来自apistrat
程序包的熟悉的apipop
和survey
数据。 R中的调查研究人员非常了解这些数据。在人口数据(apipop
)中,小学(stype == E
)约占所有学校的71.4%。中学(stype == M
)约占所有学校的12.2%,高中(stype == H
)约占所有学校的16.4%。 apistrat
有故意的失衡,其中小学占数据的50%,而中学和高中分别占200行样本的25%。
[我想做的是使用apistrat
功能对sample_n()
数据进行采样,并进行替换。但是,我似乎一直在对小学过度采样,而对初中和高中采样不足。这是R代码中的最小工作示例。请原谅我的cornball循环代码。我知道我需要在purrr
上有所进步,但我还没有到那儿。 :P
library(survey)
library(tidyverse)
apistrat %>% tbl_df() -> strat
apipop %>% tbl_df() -> pop
pop %>%
group_by(stype) %>%
summarize(prop = n()/6194) -> Census
Census
# p(E) = ~.714
# p(H) = ~.122
# p(M) = ~.164
strat %>%
left_join(., Census) -> strat
# Sampling with replacement seems to consistently oversample E and undersample H and M.
with_replace <- tibble()
set.seed(8675309) # Jenny, I got your number...
for (i in 1:1000) {
strat %>%
sample_n(100, replace=T, weight = prop) %>%
group_by(stype) %>%
summarize(i = i,
n = n(),
prop = n/100) -> hold_this
with_replace <- bind_rows(with_replace, hold_this)
}
# group_by means with 95% intervals
with_replace %>%
group_by(stype) %>%
summarize(meanprop = mean(prop),
lwr = quantile(prop, .025),
upr = quantile(prop, .975))
# ^ consistently oversampled E.
# meanprop of E = ~.835.
# meanprop of H = ~.070 and meanprop of M = ~.095
# 95% intervals don't include true probability for either E, H, or M.
# Sampling without replacement doesn't seem to have this same kind of sampling problem.
wo_replace <- tibble()
set.seed(8675309) # Jenny, I got your number...
for (i in 1:1000) {
strat %>%
sample_n(100, replace=F, weight = prop) %>%
group_by(stype) %>%
summarize(i = i,
n = n(),
prop = n/100) -> hold_this
wo_replace <- bind_rows(wo_replace, hold_this)
}
# group_by means with 95% intervals
wo_replace %>%
group_by(stype) %>%
summarize(meanprop = mean(prop),
lwr = quantile(prop, .025),
upr = quantile(prop, .975))
# ^ better in orbit of the true probability
# meanprob of E = ~.757. meanprob of H = ~.106. meanprob of M = ~.137
# 95% intervals include true probability as well.
我不确定这是否是dplyr
(v。0.8.3)问题。 95%的替换抽样间隔不包括真实的概率,每个样本(要达到峰值)始终在80年代中期的范围内进行小学抽样。在1,000个样本中(替换后)只有3个样本的小学少于100行样本的72%。那是一致的。我很好奇,如果有人在这里对正在发生的事情有任何见解,或者可能是我可能做错了什么,并且如果我误解了sample_n()
的功能。
提前感谢。
sample_n()
中的dplyr
功能是base::sample.int()
的包装。看base::sample.int()
-实际的功能是用C实现的。我们可以看到问题出在源头:
rows <- sample(nrow(strat), size = 100, replace=F, prob = strat$prop)
strat[rows, ] %>% count(stype)
# A tibble: 3 x 2
stype n
<fct> <int>
1 E 74
2 H 14
3 M 12
rows <- sample(nrow(strat), size = 100, replace=T, prob = strat$prop)
strat[rows, ] %>% count(stype)
# A tibble: 3 x 2
stype n
<fct> <int>
1 E 85
2 H 8
3 M 7
老实说,我不完全确定为什么会这样,但是如果您使概率总和为1并使它们在组内一致,那么它给出了预期的样本量:
library(tidyverse)
library(survey)
data(api)
apistrat %>% tbl_df() -> strat
apipop %>% tbl_df() -> pop
pop %>%
group_by(stype) %>%
summarize(prop = n()/6194) -> Census
strat %>%
left_join(., Census) -> strat
#> Joining, by = "stype"
set.seed(8675309) # Jenny, I got your number...
with_replace <- tibble()
for (i in 1:1000) {
strat %>%
group_by(stype) %>%
mutate(per_prob = sample(prop/n())) %>%
ungroup() %>%
sample_n(100, replace=T, weight = per_prob) %>%
group_by(stype) %>%
summarize(i = i,
n = n(),
prop = n/100) -> hold_this
with_replace <- bind_rows(with_replace, hold_this)
}
with_replace %>%
group_by(stype) %>%
summarize(meanprop = mean(prop),
lwr = quantile(prop, .025),
upr = quantile(prop, .975))
#> # A tibble: 3 x 4
#> stype meanprop lwr upr
#> <fct> <dbl> <dbl> <dbl>
#> 1 E 0.713 0.63 0.79
#> 2 H 0.123 0.06 0.19
#> 3 M 0.164 0.09 0.24
由reprex package(v0.3.0)在2020-04-17创建
我猜想这与p的向量中的实体没有被replace = TRUE
减少有关,但实际上我不知道幕后发生了什么。有C语言知识的人应该看看!