将 R for 循环的结果存储到对象中并使用应用函数绘制图形的问题

没有您的数据，我们无法重现该行为。但这里有一些示例代码可以做到这一点。我使用purrr::map代替循环或apply家庭，但您可以替换map为lapply并获得相同的结果。

library(tidyverse)
library(cowplot) # to plot a list of plots

# create some fake data
# make a vector for table size
sz <- 21:60

# function to make data frames
make_tbl <- function(size) {
  a <- sample(x = 1:50, size = size, replace = TRUE)
  b <- sample(x = LETTERS[1:3], size = size, replace = TRUE)
  return(tibble(a,b))
}

# a list of tables
list_of_tbls <- map(sz, make_tbl )

# function to plot
make_plot <- function(tbl) {
  ggplot(data=tbl) + geom_boxplot(aes(x=b, y=a, fill=b))
}

# make plot for all tables
list_of_plots <- map(list_of_tbls, make_plot)

# plot (all 40 if on a big screen)
cowplot::plot_grid(plotlist = list_of_plots[1:8], nrow=2)

另一种方法是按行绑定表并使用分面绘图。当然，这些表应该具有相同的列。这里的参数.id将创建一个tbl跟踪表的列，因此按 40 个表进行分面很简单。

# alternative to bind the tables if they have the same columns
bound_tbls <- bind_rows(list_of_tbls, .id = "tbl")

# then plot with facet
ggplot(bound_tbls) + geom_boxplot(aes(x=b, y=a, fill=b)) + facet_wrap("tbl", ncol=8)

编辑：使用 OP 的代码和数据。稍微修改绘图功能以匹配虚拟数据。注意最后的使用cowplot::plot_grid。但是，如果您只运行all_graphs[[graph_number]].

# library(xlsx)
library(ggplot2)
library(tidyverse)
library(readxl)
library(ggplot2)
library(reshape2)

d1 <-
  data.frame(
    options = c("Strongly Agree", "Agree", "Disagree", "N/A", NA),
    foo2016 =
      c(1, 4, 5, 6, NA),
    foo2017 = c(10, 7, 8, 9, NA)
  )
d2 <-
  data.frame(
    options = c("options", "Strongly Agree", "Agree", "Disagree", "N/A", NA),
    foo2016 = c(11, 4, 3, 2, 1, NA),
    foo2017 = c(12, 6, 5, 4, 5, NA)
  )

mytables_in_a_list <- list(d1, d2)

# where x = my tables in a list, n = table index in the list
foo_graph <- function(x, n) {
  tbl1 <- x[[n]]

  if (tbl1[1, 1] != "Strongly Agree") {
    tbl1 <- tbl1[-1, ]
  }

  #rename column
  # edited to match input data that doesn't have 2018
  names(tbl1) <- c("Options", "2016", "2017")

  # remove rofoo with NAs
  tbl1 <- tbl1 %>% drop_na()

  # edited to match input data that doesn't have 2018
  cols.num <- c("2016", "2017")

  tbl1[cols.num] <- sapply(tbl1[cols.num], as.numeric)

  sapply(tbl1, class)

  # alternative to removing rofoo with NAs
  # na.omit(tbl)
  mdf <-
    melt(
      tbl1,
      value.name = "value",
      variable.name = "year",
      id.vars = "Options"
    )

  foo_graph <-
    ggplot(data = mdf, aes(
      x = year,
      y = value,
      group = Options,
      colour = Options
    )) +
    geom_line() +
    geom_point(size = 4,
               shape = 21,
               fill = "white")

  foo_graph
}

all_graphs <-
  lapply(1:length(x), function(i)
    foo_graph(x = mytables_in_a_list, n = i))

# plot all of them
library(cowplot)
pp <- plot_grid(plotlist = all_graphs,
          align = "hv",
          axis = "ltbr")

# to save:
# ggsave(pp, filename = "all_plots.pdf", width=10, height=5)

要查看绘制在一起的图，只需调用pp对象：

pp

拥有 之后all_graphs，您应该能够通过调用来查看各个图：

all_graphs[[1]]

如果您只是调用，all_graphs您将只能在显示窗口中看到最后一个图，因为每个图都被显示并替换为以下内容。在 Rstudio 中，您可以在显示窗格中向后浏览以查看列表中的先前绘图。

> all_graphs
[[1]]

[[2]]

**编辑 2：使用刻面代替cowplot. 对于 40 个表，这应该会更好。不过，问题是是否有办法从 40 个表中的每一个中总结/提取有趣的信息并制作一个汇总图。而不是绘制 40 次调查的原始结果。

library(tidyverse)

d1 <-
  data.frame(
    options = c("Strongly Agree", "Agree", "Disagree", "N/A", NA),
    foo2016 =
      c(1, 4, 5, 6, NA),
    foo2017 = c(10, 7, 8, 9, NA)
  )
d2 <-
  data.frame(
    options = c("options", "Strongly Agree", "Agree", "Disagree", "N/A", NA),
    foo2016 = c(11, 4, 3, 2, 1, NA),
    foo2017 = c(12, 6, 5, 4, 5, NA)
  )

mytables_in_a_list <- list(d1, d2)

# combine into a single table
mytables_df <- bind_rows(mytables_in_a_list, .id="table")

# a single chain instead of function. 
# You could make this a function, but not necessary

mytables_df %>%
  drop_na() %>%
  rename("Options" = options,
         "2016" = foo2016,
         "2017" = foo2017) %>%
  filter(Options %in% c("Strongly Agree", "Agree", "Disagree", "N/A")) %>%
# make sure the options are ordered appropriatelly
  mutate(Options = factor(Options, levels = c(
    "Strongly Agree", "Agree", "Disagree", "N/A"
  ))) %>%
# using `gather` instead of `melt`, but its the same operation
  gather("Year", "Value", -table, -Options) %>% 
  ggplot(data = ., aes(x=Year, y=Value, group=Options, color=Options)) +
  geom_line() +
  geom_point() +
  facet_wrap("table", ncol=2) +
  theme(legend.position = "top")

使这个情节：