使用 while 循环存储 purrr:map_dfr 和 dplyr::group_split 的输出答案

【问题标题】：Store output from purrr:map_dfr and dplyr::group_split with while loop使用 while 循环存储 purrr:map_dfr 和 dplyr::group_split 的输出
【发布时间】：2019-08-17 02:19:29
【问题描述】：

我想使用map_dfr 和group_split 通过while 循环运行一组data.frame 并存储结果。

我可以为这样的一个小组做到这一点。

# df dput below
# this code finds the closet match for DIFF for Sample.x in Sample.y, then finds the next closest match, until 
df_f <- df %>% filter(grp == "AB" & VAR == "Var1")
HowMany <- length(unique(df_f$Sample.y))
i <- 1
MyList <- list()

while (i <= HowMany){
  res1 <- df_f %>%
    group_by(grp, VAR, Sample.x) %>%
    filter(DIFF == min(DIFF)) %>%
    ungroup() %>%
    mutate(Rank1 = dense_rank(DIFF))

  res2 <- res1 %>% group_by(grp, VAR) %>% filter(rank(Rank1, ties.method="first")==1)

  SY <- as.numeric(res2$Sample.y)
  SX <- as.numeric(res2$Sample.x)
  res3 <- df_f %>% filter(Sample.y != SY)
  res4 <- res3 %>% filter(Sample.x != SX)
  df_f <- res4

  MyList[[i]] <- res2

  i <- i + 1
}
df.result <- do.call("rbind", MyList)

但是，当尝试使用 while 循环创建函数以与 map_dfr 和 group_split 一起使用时，我无法和/或不确定如何存储输出。

MyResult <- df %>%
      dplyr::group_split(grp, VAR) %>%
      map_dfr(fun) # fun below

df.store <- data.frame() # attempt to store results

fun <- function(df){
  HowMany <- length(unique(df$Sample.y))
  i <- 1
  MyList_FF <- list()
  ThisDF <- df
  while (i <= HowMany){

    res1 <- ThisDF %>%
      group_by(grp, VAR, Sample.x) %>%
      filter(DIFF == min(DIFF)) %>%
      ungroup() %>%
      mutate(Rank1 = dense_rank(DIFF))
    res2 <- res1 %>% group_by(grp, VAR) %>% filter(rank(Rank1, ties.method="first")==1)
    # print(res2) # when printed to screen the desired output looks correct
    SY <- as.numeric(res2$Sample.y)
    SX <- as.numeric(res2$Sample.x)

    res3 <- ThisDF %>% filter(Sample.y != SY)
    res4 <- res3 %>% filter(Sample.x != SX)

    # df.store <- rbind(df.store, res4)
    # MyList_FF[[i]] <- res2
    ThisDF <- res4
    i <- i + 1
  }
}

我尝试使用rbind 或使用list 来存储输出，但我的尝试并不正确。如果我在屏幕上打印“res2”，我可以一次看到所需的输出一行。如何存储来自每个group_split 的fun 的输出？

# df dput
df <- structure(list(Location.x = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L, 2L), .Label = c("A", "C", "B"), class = "factor"), 
    Sample.x = c(6L, 6L, 10L, 10L, 9L, 9L, 6L, 6L, 10L, 10L, 
    9L, 9L, 6L, 6L, 6L, 10L, 10L, 10L, 9L, 9L, 9L, 6L, 6L, 6L, 
    10L, 10L, 10L, 9L, 9L, 9L, 1L, 1L, 1L, 9L, 9L, 9L, 1L, 1L, 
    1L, 9L, 9L, 9L), VAR = c("Var1", "Var1", "Var1", "Var1", 
    "Var1", "Var1", "Var2", "Var2", "Var2", "Var2", "Var2", "Var2", 
    "Var1", "Var1", "Var1", "Var1", "Var1", "Var1", "Var1", "Var1", 
    "Var1", "Var2", "Var2", "Var2", "Var2", "Var2", "Var2", "Var2", 
    "Var2", "Var2", "Var1", "Var1", "Var1", "Var1", "Var1", "Var1", 
    "Var2", "Var2", "Var2", "Var2", "Var2", "Var2"), value.x = c(56.48, 
    56.48, 57.03, 57.03, 55.04, 55.04, 6, 6, 10, 10, 9, 9, 56.48, 
    56.48, 56.48, 57.03, 57.03, 57.03, 55.04, 55.04, 55.04, 6, 
    6, 6, 10, 10, 10, 9, 9, 9, 55.62, 55.62, 55.62, 55.65, 55.65, 
    55.65, 1, 1, 1, 9, 9, 9), Location.y = structure(c(2L, 2L, 
    2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 
    3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 
    3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L), .Label = c("A", 
    "C", "B"), class = "factor"), Sample.y = c(1L, 9L, 1L, 9L, 
    1L, 9L, 1L, 9L, 1L, 9L, 1L, 9L, 3L, 7L, 9L, 3L, 7L, 9L, 3L, 
    7L, 9L, 3L, 7L, 9L, 3L, 7L, 9L, 3L, 7L, 9L, 3L, 7L, 9L, 3L, 
    7L, 9L, 3L, 7L, 9L, 3L, 7L, 9L), value.y = c(55.62, 55.65, 
    55.62, 55.65, 55.62, 55.65, 1, 9, 1, 9, 1, 9, 1.4, 111.6, 
    111.8, 1.4, 111.6, 111.8, 1.4, 111.6, 111.8, 10.2, 14.4, 
    20.9, 10.2, 14.4, 20.9, 10.2, 14.4, 20.9, 1.4, 111.6, 111.8, 
    1.4, 111.6, 111.8, 10.2, 14.4, 20.9, 10.2, 14.4, 20.9), DIFF = c(0.859999999999999, 
    0.829999999999998, 1.41, 1.38, 0.579999999999998, 0.609999999999999, 
    5, 3, 9, 1, 8, 0, 55.08, 55.12, 55.32, 55.63, 54.57, 54.77, 
    53.64, 56.56, 56.76, 4.2, 8.4, 14.9, 0.199999999999999, 4.4, 
    10.9, 1.2, 5.4, 11.9, 54.22, 55.98, 56.18, 54.25, 55.95, 
    56.15, 9.2, 13.4, 19.9, 1.2, 5.4, 11.9), grp = c("AC", "AC", 
    "AC", "AC", "AC", "AC", "AC", "AC", "AC", "AC", "AC", "AC", 
    "AB", "AB", "AB", "AB", "AB", "AB", "AB", "AB", "AB", "AB", 
    "AB", "AB", "AB", "AB", "AB", "AB", "AB", "AB", "CB", "CB", 
    "CB", "CB", "CB", "CB", "CB", "CB", "CB", "CB", "CB", "CB"
    )), row.names = c(NA, -42L), class = "data.frame")

【问题讨论】：

如果您已经习惯使用map 函数，为什么不使用嵌套的maps 而不是while 循环？
我的map 舒适程度在我上面提供的示例中。我需要探索嵌套的maps。

标签： r dplyr purrr

【解决方案1】：

唯一缺少的是您的映射函数fun 没有返回值。正在计算和构建临时列表 MyList_FF 正确，您可以通过 print() 调用看到，但没有返回，它正在消失。

fun <- function(df) {
    HowMany <- length(unique(df$Sample.y))
    i <- 1
    MyList_FF <- list()
    df_f <- df
    while (i <= HowMany){
        res1 <- df_f %>%
            group_by(grp, VAR, Sample.x) %>%
            filter(DIFF == min(DIFF)) %>%
            ungroup() %>%
            mutate(Rank1 = dense_rank(DIFF))

        res2 <- res1 %>% group_by(grp, VAR) %>% filter(rank(Rank1, ties.method="first")==1)

        SY <- as.numeric(res2$Sample.y)
        SX <- as.numeric(res2$Sample.x)
        res3 <- df_f %>% filter(Sample.y != SY)
        res4 <- res3 %>% filter(Sample.x != SX)
        df_f <- res4

        MyList_FF[[i]] <- res2

        i <- i + 1
    }
    # this is the magic line
    do.call("rbind", MyList_FF)
    # this returns the list built inside of the function
}

神奇之处在于最后一行，类似于您在单个示例之后所做的，绑定了中间结果列表。在 R 中，仅当您尝试提前返回时才需要 return() 函数，因为默认情况下 R 函数将返回最后一个值。所以在这里我们不需要明确地说return(do.call("rbind", MyList_FF))，尽管如果你这样做不会有任何伤害。在非工作示例中，自分配 i 以来没有最后一个值，因此您没有得到任何对象，但也没有得到任何错误。

完整的工作示例：

MyResult <- df %>%
    dplyr::group_split(grp, VAR) %>%
    map_df(fun)

MyResult
# A tibble: 16 x 10
# Groups:   grp, VAR [1]
   Location.x Sample.x VAR   value.x Location.y Sample.y value.y  DIFF grp   Rank1
   <fct>         <int> <chr>   <dbl> <fct>         <int>   <dbl> <dbl> <chr> <int>
 1 A                 9 Var1     55.0 B                 3     1.4  53.6 AB        1
 2 A                10 Var1     57.0 B                 7   112.   54.6 AB        1
 3 A                 6 Var1     56.5 B                 9   112.   55.3 AB        1
 4 A                 9 Var1     55.0 B                 3     1.4  53.6 AB        1
 5 A                10 Var1     57.0 B                 7   112.   54.6 AB        1
 6 A                 6 Var1     56.5 B                 9   112.   55.3 AB        1
 7 A                 9 Var1     55.0 B                 3     1.4  53.6 AB        1
 8 A                10 Var1     57.0 B                 7   112.   54.6 AB        1
 9 A                 9 Var1     55.0 B                 3     1.4  53.6 AB        1
10 A                10 Var1     57.0 B                 7   112.   54.6 AB        1
11 A                 9 Var1     55.0 B                 3     1.4  53.6 AB        1
12 A                10 Var1     57.0 B                 7   112.   54.6 AB        1
13 A                 6 Var1     56.5 B                 9   112.   55.3 AB        1
14 A                 9 Var1     55.0 B                 3     1.4  53.6 AB        1
15 A                10 Var1     57.0 B                 7   112.   54.6 AB        1
16 A                 6 Var1     56.5 B                 9   112.   55.3 AB        1

请注意，如果您经常使用do.call("xbind", list)，您可能会喜欢dplyr::bind_rows(list) 和dplyr::bind_cols(list)。

【讨论】：