变长df二次采样函数r
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【中文标题】变长df二次采样函数r【英文标题】:variable length df subsampling function r 【发布时间】:2020-07-04 21:12:02 【问题描述】:我需要编写一个函数,该函数涉及通过变量n
bins 对 df 进行子集。就像,如果n
是 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,我一直在手动为不同长度的 n 执行此操作,并且我知道必须有更好的方法来执行此操作。我想将它写入一个以n
作为输入的函数,但到目前为止我还不能让它工作。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,我只取了 2 个,但在上半年(1:23)随机一年,在下半年(24:46)随机一年。
# 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”样本年份的样本,那就太好了)。
所以,如果我想为三点而不是两点执行此操作,我会像这样重复该过程:
# 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,我想这样做很多次,最多采样 20 次(所以在 20 个 bin 中),所以这种手动方法是不可持续的。你能帮我写一个函数说“从n个箱子中挑选一个样本x次”吗?
顺便说一句,这是我用完整的 df 制作的情节:
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()
【问题讨论】:
【参考方案1】:这是一个使用循环的函数,更接近于你开始做的事情:
df <- data.frame(year = c(1:46),
sample = seq(from=10, to=30, length.out = 46) +
rnorm(46,mean=0,sd=2))
df[c(20,32), 2] <- NA
my_function <- function(n, sample_size, data = df)
plot_df <- data %>% mutate(group = 'full')
sample_matrix <- matrix(data = NA, nrow = sample_size, ncol = n)
first_row <- 1 # First subset has 1 as first row, no matter how many subsets
for (i in 1:n)
last_row <- round(first_row + nrow(df)/n - 1) # Determine last row of i-th subset
sample_matrix[, i] <- sample(first_row:last_row, sample_size, replace = T) # Store sample directly in matrix
first_row <- i + last_row # Determine first row for next i
group_name <- paste("group", i, sep = "_") # Column name for i-th group
plot_df[[group_name]] <- "all" # Column for i-th group
for (j in 1:sample_size)
# Creating a new data frame for new observations
new_obs <- df[sample_matrix[j,], ]
new_obs[["group"]] <- j
for (group_n in 1:n)
new_obs[[paste0("group_", group_n)]] <- sample_matrix[j, group_n]
plot_df <- rbind(plot_df, new_obs)
plot_df <<- plot_df
my_function(2, 200, data = df)
【讨论】:
【参考方案2】:如果我猜对了,以下函数会将您的 df 拆分为 n 个 bin,从每个 bin 中抽取 x 个样本,然后将结果放回 df 的 cols 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 (v0.3.0) 于 2020 年 3 月 24 日创建
【讨论】:
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