使用tidyverse；在组内计算值变化前后的计数，为每个唯一的移位生成新变量。

Question

使用tidyverse；在组内计算值变化前后的计数，为每个唯一的移位生成新变量。

14

我正在寻找一种基于tidyverse的解决方案，该方案能够在数据表格tbl中针对每个组id内唯一值TF的出现次数进行计数。当TF发生变化时，我希望从该点向前和向后都进行计数。这些计数应存储在一个新变量PM##中，以便PM##包含每个唯一转换中的加减。

这个问题类似于我之前提出的一个问题，但是在这里，我特别地寻求使用tidyverse工具的解决方案。Uwe 在初始问题上使用了data.table提供了优雅的答案，可在此处查看。

如果此问题违反任何SO政策，请告诉我，我将很乐意重新打开我的初始问题或添加bounty-issue。

为了举例说明我的问题，我有如下数据：

# install.packages(c("tidyverse"), dependencies = TRUE)
library(tibble)

tbl <- tibble(id = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1,
                     1, 1, 1, 1, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7), 
              TF = c(NA, 0, NA, 0, 0, 1, 1, 1, NA, 0, 0, NA, 0, 0,
                     0, 1, 1, 1, NA, NA, 0, 0, 1, 0, 0, 1, 0, 1, 1, 1))
tbl
#> # A tibble: 30 x 2
#>       id    TF
#>    <dbl> <dbl>
#>  1     0    NA
#>  2     0     0
#>  3     0    NA
#>  4     0     0
#>  5     0     0
#>  6     0     1
#>  7     0     1
#>  8     0     1
#>  9     0    NA
#> 10     0     0
#> # ... with 20 more rows

这就是我试图获得的东西，

dfa <- tibble(id = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1,
                     1, 1, 1, 1, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7),
              TF = c(NA, 0, NA, 0, 0, 1, 1, 1, NA, 0, 0, NA, 0, 0,
                     0, 1, 1, 1, NA, NA, 0, 0, 1, 0, 0, 1, 0, 1, 1, 1),
              PM01 = c(NA, -3, NA, -2, -1, 1, 2, 3, NA, NA, NA, NA, -3, -2, -1,
                       1, 2, 3, NA, NA, -2, -1, 1, NA, NA, NA, NA, NA, NA, NA),
              PM02 = c(NA, NA, NA, NA, NA, -3, -2, -1, NA, 1, 2, NA, NA, NA, NA,
                       NA, NA, NA, NA, NA, NA, NA, -1, 1, 2, NA, NA, NA, NA, NA),
              PM03 = c(NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
                       NA, NA, NA, NA, NA, NA, NA, NA, -2, -1, 1, NA, NA, NA, NA),
              PM04 = c(NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
                       NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, -1, 1, NA, NA, NA),
              PM05 = c(NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
                       NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, -1, 1, 2, 3)
               )

dfa
#> # A tibble: 30 x 7
#>       id    TF  PM01  PM02  PM03  PM04  PM05
#>    <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
#>  1     0    NA    NA    NA    NA    NA    NA
#>  2     0     0    -3    NA    NA    NA    NA
#>  3     0    NA    NA    NA    NA    NA    NA
#>  4     0     0    -2    NA    NA    NA    NA
#>  5     0     0    -1    NA    NA    NA    NA
#>  6     0     1     1    -3    NA    NA    NA
#>  7     0     1     2    -2    NA    NA    NA
#>  8     0     1     3    -1    NA    NA    NA
#>  9     0    NA    NA    NA    NA    NA    NA
#> 10     0     0    NA     1    NA    NA    NA
#> # ... with 20 more rows

- Eric Fail

请问您所说的“可信和/或官方来源的答案”是什么意思？dplyr手册是否属于此类来源？ - m-dz

{btsdaf} - Eric Fail

{btsdaf} - m-dz

3个回答

3

使用优化后的data.table函数进行更新:

可能应该回到旧问题上，但也许这会触发进一步的优化。

为了保持事情的流畅，我尝试使用 data.table 函数，并将执行时间减少到了 tidyverse 版本的两倍左右 - 瓶颈是 dcast() 函数，请参见下方的 profvis 截图:

dt_method <- function(dt_test) {
  tmp_dt <- dt_test[, rn := .I][!is.na(TF)][, rl := rleid(TF), by = id][
    , c("up", "dn") := .(seq_len(.N), -rev(seq_len(.N))), by = .(id, rl)][, ':='(
      rl_PM = sprintf("PM%02d", rl),
      United = paste(id, TF, rn, sep = '_')
    )]

  res_dt <- tmp_dt[, .(sprintf("PM%02d", seq_len(max(rl) - 1L)), seq_len(max(rl) - 1L)), by = .(id)] %>% 
    tmp_dt[., on = .(id), allow.cartesian = TRUE] %>%  
    .[rl == V2, PM := dn] %>%
    .[rl == V2 + 1L, PM := up] %>%
    dcast(., United ~ V1, value.var = "PM") %>%
    .[, c('id', 'TF', 'rn') := lapply(tstrsplit(United, '_'), as.numeric)] %>%
    .[dt_test, on = .(rn, id, TF)] %>% .[, -c('rn', 'United')]
  res_dt
}

需要使用管道来处理一些奇怪的错误，但我仍然认为即使针对 data.table ，它们也是允许的。

微基准测试结果：

Unit: milliseconds
                 expr      min       lq      mean    median        uq       max neval
   dt_method(dt_test) 868.1491 932.8076 1048.5077 1029.9609 1078.0735 1518.0327    10
 tidy_method(df_test) 478.6824 515.5639  557.9644  565.9422  585.3143  622.1093    10

并且使用固定列顺序的identical()：

identical(
  dt_method(dt_test)[order(id), lapply(.SD, as.numeric)] %>% setcolorder(c('id', 'TF', setdiff(names(.), c('id', 'TF')))) %>% as.data.frame(),
  as.data.frame(tidy_method(df_test))
)

profvis时间:

旧部分:

以Uwe的答案为基础：

(免责声明：我没有太多使用dplyr，这是我自己练习的一个例子，所以它肯定不是dplyr最优化的，例如查看dcast。)

library(data.table)
library(magrittr)
library(dplyr)
library(tibble)

df <- tibble(id = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 
                    1, 1, 1, 1,7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7),
             TF = c(NA, 0, NA, 0, 0, 1, 1, 1, NA, 0, 0, NA, 0, 0, 0,
                    1, 1, 1, NA, NA, 0, 0, 1, 0, 0, 1, 0, 1, 1, 1))

dfa <- tibble(id = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1,
                     1, 1, 1, 1, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7),
              TF = c(NA, 0, NA, 0, 0, 1, 1, 1, NA, 0, 0, NA, 0, 0,
                     0, 1, 1, 1, NA, NA, 0, 0, 1, 0, 0, 1, 0, 1, 1, 1),
              PM01 = c(NA, -3, NA, -2, -1, 1, 2, 3, NA, NA, NA, NA, -3, -2, -1,
                       1, 2, 3, NA, NA, -2, -1, 1, NA, NA, NA, NA, NA, NA, NA),
              PM02 = c(NA, NA, NA, NA, NA, -3, -2, -1, NA, 1, 2, NA, NA, NA, NA,
                       NA, NA, NA, NA, NA, NA, NA, -1, 1, 2, NA, NA, NA, NA, NA),
              PM03 = c(NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
                       NA, NA, NA, NA, NA, NA, NA, NA, -2, -1, 1, NA, NA, NA, NA),
              PM04 = c(NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
                       NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, -1, 1, NA, NA, NA),
              PM05 = c(NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
                       NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, -1, 1, 2, 3))

tmp_dt <- setDT(df)[, rn := .I][!is.na(TF)][, rl := rleid(TF), by = id][
  , c("up", "dn") := .(seq_len(.N), -rev(seq_len(.N))), by = .(id, rl)][]

res_dt <- tmp_dt[tmp_dt[, seq_len(max(rl) - 1L), by = .(id)], on = .(id), allow.cartesian = TRUE][
  rl == V1, PM := dn][rl == V1 + 1L, PM := up][
    , dcast(.SD, id + TF + rn ~ sprintf("PM%02d", V1), value.var = "PM")][
      df, on = .(rn, id, TF)][, -"rn"]
res_dt

all.equal(res_dt, as.data.table(dfa))

尽可能使用整洁的语法，类似于tidyverse：

tmp_dplyr <- df %>%
  # create row id column (required for final join to get NA rows back in)
  mutate(rn = row_number()) %>%
  # ignore NA rows 
  filter(complete.cases(.)) %>%
  # number streaks of unique values within each group
  group_by(id) %>%
  mutate(rl = rleid(TF)) %>%
  # create ascending and descending counts for each streak
  # this is done once to avoid repeatedly creation of counts for each PM 
  # (slight performance gain)
  group_by(id, rl) %>%
  mutate(
    up = seq_len(n()),
    dn = -rev(seq_len(n()))
  )

res_dplyr <- tmp_dplyr %>%
  ## Replicating tmp[tmp[, seq_len(max(rl) - 1L), by = .(id)], on = .(id), allow.cartesian = TRUE]
  group_by(id) %>%
  ## Part below can for sure be optimized for code length, it's just too early now...
  transmute(rl = max(rl)) %>% # Cannot transmute id directly
  unique() %>%
  ungroup() %>%
  slice(rep(1:n(), times = rl - 1L)) %>%
  group_by(id) %>%
  transmute(V1 = seq_len(max(rl) - 1L)) %>%
  ungroup() %>%
  right_join(tmp_dplyr, by = 'id') %>%
  ## End or replicating tmp[tmp[, seq_len(max(rl) - 1L), by = .(id)], on = .(id), allow.cartesian = TRUE]
  ## Copy descending counts to rows before the switch and ascending counts to rows after the switch
  mutate(
    PM = ifelse(rl == V1, dn, NA),
    PM = ifelse(rl == V1 + 1L, up, PM)
  ) %>%
  ## This is very not tidyverse-sque, but I don't get the gather/spread ...
  dcast(id + TF + rn ~ sprintf("PM%02d", V1), value.var = "PM") %>%
  full_join(df, by = c('rn', 'id', 'TF')) %>%
  select(-rn)

all.equal( ## Using data.table all.equal
  res_dplyr[do.call(order, res_dplyr),] %>% as.data.table(),
  res_dt[do.call(order, res_dt),]
)

- m-dz

1

我之前有一个不使用 data.table 的答案，但是它没有使用 dplyr。这是我使用 dplyr 的尝试：

        #Remove the NAs 
dfr <-  df %>% filter(!is.na(TF)) %>% 
  # group by id
  group_by(id) %>% 
  # Calculate the rle on TF for each group
  do(., mrle = rle(.$TF)) %>% mutate(Total=sum(mrle$lengths)) %>%
  # Trasform the rle result in a data.frame counting the values after and before changes
  do( {
  t<- .$mrle
  #for each length generate the columns
  res <- as.data.frame(lapply(seq_along(t$lengths[-length(t$lengths)]), function(i) {

      #before change counts
      n1 <- t$lengths[i]
      #position  the counts
      if(i==1) {
        before <- 0
      } else {
        before <- sum(t$lengths[1:i-1])
      }

      #after change conts
      n2 <- t$lengths[i+1]

      if(i == (length(t$lengths)-1))
        after  <- 0
      else
        after <- .$Total - before - n1 - n2

      # assemble the column
      c(rep(NA,before),-n1:-1,1:n2, rep(NA,after))

    } ))

  colnames(res) <- paste0("PM", 1:ncol(res))
  #preserve the id
  cbind(id=.$id,res)

 })

#Join with the original data.frame
res <-  df %>% mutate(rn = row_number()) %>% filter(!is.na(TF)) %>% bind_cols(dfr) %>% right_join( df %>% mutate(rn = row_number()) ) %>% select(-rn, -id1)

#Verify
mapply(all.equal, dfa,res)
#  id   TF PM01 PM02 PM03 PM04 PM05 
#TRUE TRUE TRUE TRUE TRUE TRUE TRUE

- Marcelo

{btsdaf} - Eric Fail

{btsdaf} - Marcelo

网页内容由stack overflow 提供, 点击上面的

可以查看英文原文，
原文链接

- Psidom · Accepted Answer

这里是另一种使用dplyr、tidyr和zoo（用于其na.locf函数）包的整洁方法：

首先，与所有其他建议的方法（包括data.table方法）一样，不要在TF列中删除NAs然后再连接回来，我编写了一个帮助方法，在这里按块向前计数并忽略NAs;

forward_count <- function(v) {
    valid <- !is.na(v)
    valid_v <- v[valid]
    chunk_size = head(rle(valid_v)$lengths, -1)
    idx <- cumsum(chunk_size) + 1
    ones <- rep(1, length(valid_v))
    ones[idx] <- 1 - chunk_size
    v[valid] <- cumsum(ones)
    v
}

并且它按照更改后的计数所需的方式运行：

v <- sample(c(NA, 0, 1), 15, replace = T)
v
# [1] NA NA NA  0  1 NA  1 NA  1  1  0  1  0  0  0
forward_count(v)
# [1] NA NA NA  1  1 NA  2 NA  3  4  1  1  1  2  3

在变更之前的计数可以通过使用完全相同的函数两次反转向量来实现：

-rev(forward_count(rev(v)))
# [1] NA NA NA -1 -4 NA -3 NA -2 -1 -1 -1 -3 -2 -1

现在使用dplyr包定义标题，将向前列计数为fd，将向后列计数为bd：

library(dplyr); library(tidyr); library(zoo);

tidy_method <- function(df) {
    df %>% 
        group_by(id) %>% 
        mutate(
            rle_id = cumsum(diff(na.locf(c(0, TF))) != 0),   # chunk id for constant TF
            PM_fd = if_else(                 # PM count after change headers
                rle_id == head(rle_id, 1), 
                "head", sprintf('PM%02d', rle_id)
            ), 
            PM_bd = if_else(                 # shift the header up as before change headers
                rle_id == tail(rle_id, 1), 
                "tail", sprintf('PM%02d', rle_id+1)
            ), 
            fd = forward_count(TF),             # after change count
            bd = -rev(forward_count(rev(TF))),  # before change count
            rn = seq_along(id)) %>%             # row number
        gather(key, value, PM_fd, PM_bd) %>%    # align headers with the count
        mutate(count_ = if_else(key == "PM_fd", fd, bd)) %>%
        select(-key) %>% spread(value, count_) %>%    # reshaper PM column as headers
        select(id, TF, rn, matches('PM')) %>%  # drop no longer needed columns
        arrange(id, rn) %>% select(-rn)
}

与data.table方法相比，时间上的差异：

将data.table方法定义为：

dt_method <- function(df) {
    tmp_dt <- setDT(df)[, rn := .I][!is.na(TF)][, rl := rleid(TF), by = id][
        , c("up", "dn") := .(seq_len(.N), -rev(seq_len(.N))), by = .(id, rl)][]

    res_dt <- tmp_dt[tmp_dt[, seq_len(max(rl) - 1L), by = .(id)], on = .(id), allow.cartesian = TRUE][
        rl == V1, PM := dn][rl == V1 + 1L, PM := up][
            , dcast(.SD, id + TF + rn ~ sprintf("PM%02d", V1), value.var = "PM")][
                df, on = .(rn, id, TF)][, -"rn"]
    res_dt
}

数据：通过将样本数据框重复200次来生成中等大小的数据集：

df_test <- bind_rows(rep(list(df), 200))

microbenchmark::microbenchmark(dt_method(df_test), tidy_method(df_test), times = 10)
#Unit: milliseconds
#                 expr       min        lq      mean    median        uq       max neval
#   dt_method(df_test) 2321.5852 2439.8393 2490.8583 2456.1118 2557.4423 2834.2399    10
# tidy_method(df_test)  402.3624  412.2838  437.0801  414.5655  418.6564  540.9667    10

按 id 对数据表进行排序，并将所有列的数据类型转换为数值型；使用 data.table 和 tidyverse 的结果是相同的：

identical(
    as.data.frame(dt_method(df_test)[order(id), lapply(.SD, as.numeric)]), 
    as.data.frame(tidy_method(df_test))
)
# [1] TRUE