按组汇总 wtd.quantile答案

【问题标题】：summarise wtd.quantile by group按组汇总 wtd.quantile
【发布时间】：2019-02-24 09:28:02
【问题描述】：

我想为具有许多重复日期的数据框使用 Hmisc::wtd.quantile 创建一个新的 df。我按日期分组，使用summarize() 聚合日期，并尝试在每个日期使用wtd.quantile()（带权重）。这也是一个相当大的数据集。下面是一些示例代码：

# sample data
# grouping_var = dt_time
require(Hmisc)
require(plyr)
require(dplyr)
df <- data.frame(type = sample(letters[1:2], 10e6, replace = TRUE), 
             score = sample(500:899, 10e6, replace = TRUE),
             dt_time = sample(seq(as.Date('2010/01/01'), 
                                  as.Date('2018/01/01'), 
                                  by="day"), 10e6, replace = TRUE),
             weight = sample(1.0:2.0, 10e6, replace = TRUE))
# my attempt:
ptiles <- df %>%
group_by(dt_time) %>%
plyr::ddply(~dt_time, dplyr::summarize,
            ptile10 = Hmisc::wtd.quantile(., .$score, weights = .$weight, 
probs = .1, na.rm = TRUE),
            ptile50 = Hmisc::wtd.quantile(., .$score, weights = .$weight, 
probs = .5, na.rm = TRUE),
            ptile90 = Hmisc::wtd.quantile(., .$score, weights = .$weight, 
probs = .9, na.rm = TRUE))

# desired df,
# where each new variable would be created using the
# wtd.quantile function:
desired_ptiles <- data.frame(dt_time = seq(as.Date('2010/01/01'),
                                       as.Date('2010/01/06'),
                                       by = "day"),
                         # only 6 because lol 10e6
                         ptile10 = sample(500:899, 6, replace = TRUE),
                         ptile50 = sample(500:899, 6, replace = TRUE),
                         ptile90 = sample(500:899, 6, replace = TRUE))

到目前为止，我的努力导致了这个错误：

Error in summarise_impl(.data, dots) :
Evaluation error: 'arg' must be NULL or a character vector.

当使用公式表示法时：

ptiles <- df %>%
    group_by(dt_time) %>%
plyr::ddply(~dt_time, dplyr::summarize,
ptile10 = Hmisc::wtd.quantile(., .$score, weights = .$weight,
                                     probs = .1, na.rm = TRUE),
ptile50 = Hmisc::wtd.quantile(., .$score, weights = .$weight, 
                                        probs = .5, na.rm = TRUE),
          ptile90 = Hmisc::wtd.quantile(., .$score, weights = .$weight,
          probs = .9, na.rm = TRUE))
# error message:
Error in summarise_impl(.data, dots) : 
  Evaluation error: 'arg' must be NULL or a character vector.

我是否以错误的方式处理这个问题？我见过使用split() 的方法，但这似乎很烦人。是否有data.table 方法允许以这种方式总结wtd.quantile()？

谢谢！

【问题讨论】：

标签： r dplyr plyr hmisc

【解决方案1】：

使用group_by 时不需要ddply，因为数据已经被分组分割。另外，分组后不需要在summarize内定义数据。

这行得通：

ptiles <- df %>%
  group_by(dt_time) %>%
  summarize(ptile10 = wtd.quantile(score, weights = weight, 
                                            probs = .1, na.rm = TRUE),
              ptile50 = wtd.quantile(score, weights = weight, 
                                            probs = .5, na.rm = TRUE),
              ptile90 = wtd.quantile(score, weights = weight, 
                                            probs = .9, na.rm = TRUE))

> ptiles
# A tibble: 2,923 x 4
      dt_time ptile10 ptile50 ptile90
       <date>   <dbl>   <dbl>   <dbl>
 1 2010-01-01   539.0     697   859.0
 2 2010-01-02   538.0     704   861.7
 3 2010-01-03   541.0     706   862.0
 4 2010-01-04   541.0     702   859.0
 5 2010-01-05   540.0     706   860.0
 6 2010-01-06   537.0     695   859.0
 7 2010-01-07   539.0     696   859.0
 8 2010-01-08   536.0     700   857.0
 9 2010-01-09   538.0     694   861.0
10 2010-01-10   538.4     701   859.0
# ... with 2,913 more rows

【讨论】：

【解决方案2】：

这是一个替代解决方案，它使用一个函数（基于@LAP 的解决方案）检查您提供的一组分位数（即c(.1, .5, .9)）并自动创建相应的列：

require(Hmisc)
require(plyr)
require(tidyverse)

# for reproducibility
set.seed(5)

# example dataset
df <- data.frame(type = sample(letters[1:2], 50000, replace = TRUE), 
                 score = sample(500:899, 50000, replace = TRUE),
                 dt_time = sample(seq(as.Date('2010/01/01'), 
                                      as.Date('2018/01/01'), 
                                      by="day"), 50000, replace = TRUE),
                 weight = sample(1.0:2.0, 50000, replace = TRUE))

# function to get quantiles per date
f = function(x) {
    df %>%
      group_by(dt_time) %>%
      dplyr::summarize(value = wtd.quantile(score, weights = weight, probs = x, na.rm = TRUE)) %>%
      mutate(ptile = x) }

# input quantiles of interest
q = c(.1, .5, .9)

# apply function to each quantile of interest
# and reshape
map_df(q, f) %>%
  spread(ptile, value, sep = "_")

# # A tibble: 2,923 x 4
#   dt_time    ptile_0.1 ptile_0.5 ptile_0.9
#   <date>         <dbl>     <dbl>     <dbl>
# 1 2010-01-01      543       670.      857 
# 2 2010-01-02      549       800       868.
# 3 2010-01-03      529.      705       829.
# 4 2010-01-04      543       690       850.
# 5 2010-01-05      528.      642.      870.
# 6 2010-01-06      534.      649       810.
# 7 2010-01-07      512.      668       801.
# 8 2010-01-08      554       664       869.
# 9 2010-01-09      533.      650       765 
#10 2010-01-10      535.      696       843.
# # ... with 2,913 more rows

【讨论】：