我需要编写一个函数,涉及用变量n
bins 子集 df。例如,如果 n 为 2,则在两个 bin 中对 df 进行多次二次采样(从前半部分开始,然后从后半部分开始)。如果 n 为 3,则在 3 个 bin 中进行子采样(第一个 1/3、第二个 1/3、第三个 1/3)。到目前为止,我一直在手动对不同长度的 n 执行此操作,并且我知道一定有更好的方法来执行此操作。我想将其写入一个以 n
作为输入的函数,但到目前为止我还无法使其工作。代码如下。
# create df
df <- data.frame(year = c(1:46),
sample = seq(from=10,to=30,length.out = 46) + rnorm(46,mean=0,sd=2) )
# real df has some NAs, so we'll add some here
df[c(20,32),2] <- NA
这个df是46年的采样。我想假装不是 46 个样本,而是只取了 2 个样本,但在上半年随机一年 (1:23),在下半年随机一年 (24:46)。
# to subset in 2 groups, say, 200 times
# I'll make a df of elements to sample
samplelist <- data.frame(firstsample = sample(1:(nrow(df)/2),200,replace = T), # first sample in first half of vector
secondsample = sample((nrow(df)/2):nrow(df),200, replace = T) )# second sample in second half of vector
samplelist <- as.matrix(samplelist)
# start a df to add to
plot_df <- df %>% mutate(first='all',
second = 'all',
group='full')
# fill the df using coords from expand.grid
for(i in 1:nrow(samplelist)){
plot_df <<- rbind(plot_df,
df[samplelist[i,] , ] %>%
mutate(
first = samplelist[i,1],
second = samplelist[i,2],
group = i
))
print(i)
}
(如果我们可以让它跳过“NA”样本年的样本,那就太好了)。
所以,如果我想获得三个点而不是两个点,我会像这样重复这个过程:
# to subset in 3 groups 200 times
# I'll make a df of elements to sample
samplelist <- data.frame(firstsample = sample(1:(nrow(df)/3),200,replace = T), # first sample in first 1/3
secondsample = sample(round(nrow(df)/3):round(nrow(df)*(2/3)),200, replace = T), # second sample in second 1/3
thirdsample = sample(round(nrow(df)*(2/3)):nrow(df), 200, replace=T) # third sample in last 1/3
)
samplelist <- as.matrix(samplelist)
# start a df to add to
plot_df <- df %>% mutate(first='all',
second = 'all',
third = 'all',
group='full')
# fill the df using coords from expand.grid
for(i in 1:nrow(samplelist)){
plot_df <<- rbind(plot_df,
df[samplelist[i,] , ] %>%
mutate(
first = samplelist[i,1],
second = samplelist[i,2],
third = samplelist[i,3],
group = i
))
print(i)
}
但是,我想多次执行此操作,采样最多约 20 次(因此在 20 个容器中),因此这种手动方法不可持续。你能帮我写一个函数来表示“从 n 个箱子中选取一个样本 x 次”吗?
顺便说一句,这是我用完整的 df 绘制的图:
plot_df %>%
ggplot(aes(x=year,y=sample)) +
geom_point(color="grey40") +
stat_smooth(geom="line",
method = "lm",
alpha=.3,
aes(color=group,
group=group),
se=F,
show.legend = F) +
geom_line(color="grey40") +
geom_smooth(data = plot_df %>% filter(group %in% c("full")),
method = "lm",
alpha=.7,
color="black",
size=2,
#se=F,
# fill="grey40
show.legend = F
) +
theme_classic()
最佳答案
如果我没理解错的话,下面的函数将你的 df 分成 n 个容器,从每个容器中抽取 x 个样本,然后将结果放回到 df 的列中:
library(tidyverse)
set.seed(42)
df <- data.frame(year = c(1:46),
sample = seq(from=10,to=30,length.out = 46) + rnorm(46,mean=0,sd=2) )
get_df_sample <- function(df, n, x) {
df %>%
# bin df in n bins of (approx.) equal length
mutate(bin = ggplot2::cut_number(seq_len(nrow(.)), n, labels = seq_len(n))) %>%
# split by bin
split(.$bin) %>%
# sample x times from each bin
map(~ .x[sample(seq_len(nrow(.x)), x, replace = TRUE),]) %>%
# keep only column "sample"
map(~ select(.x, sample)) %>%
# Rename: Add number of df-bin from which sample is drawn
imap(~ rename(.x, !!sym(paste0("sample_", .y)) := sample)) %>%
# bind
bind_cols() %>%
# Add group = rownames
rownames_to_column(var = "group")
}
get_df_sample(df, 3, 200) %>%
head()
#> sample_1 sample_2 sample_3 group
#> 1 12.58631 18.27561 24.74263 1
#> 2 19.46218 24.24423 23.44881 2
#> 3 12.92179 18.47367 27.40558 3
#> 4 15.22020 18.47367 26.29243 4
#> 5 12.58631 24.24423 24.43108 5
#> 6 19.46218 23.36464 27.40558 6
由reprex package于2020年3月24日创建(v0.3.0)
关于r - 变长 df 二次采样函数 r,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/60829944/