变长df二次采样函数r

Question

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()

Answer 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)

Answer 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}