使用 R 中的中值替换值答案

【问题标题】：Replace values using the median in R使用 R 中的中值替换值
【发布时间】：2021-08-13 11:58:08
【问题描述】：

假设我的数据如下

df<-data.frame(name=c(rep("Aust", 20), rep("Fr", 20), rep("Spa", 20)),
       Threshold = c(rep(38.9, 20), rep(50.5, 20), rep(20, 20)),
       Fitted_Data= c(38.20784,35.52096, 37.05763, 36.19203,39.91685,38.19453,36.86204,38.51312,35.14895,35.41919,35.13218,35.46005,
                      37.48999,37.54950,38.68705,36.23085, 35.90234,38.50205,39.27153,38.03129, 48.19456,48.23224,51.25736,50.59195,
                      51.35283, 48.45300,50.81403,51.03964,50.97189,50.38674,50.59499,49.76958,49.93091,48.90412,51.19752,51.31885, 
                      50.54078,48.77288,48.11736,48.60201, 18.99013,21.63701,21.45867,21.96485,19.73159,18.76820,21.73579,18.68561,
                      21.62721,20.88826,21.66602,19.29559,21.39014,21.40296,20.17120,21.42481,19.05561,21.71352,19.36918,18.95138),
       Day= c("jueves", "martes",  "miércoles", "jueves", "viernes", "viernes", "domingo" , "lunes",  " martes", "miércoles", "domingo", "lunes"),
       Month = c(rep(c(1, 3, 1, 4, 1, 1, 7, 2, 3, 4, 6, 7, 12, 11, 1, 2, 10, 10, 7, 3),3))

)

我想使用Threshold 和Fitted_Data 创建一个新变量New_prices。如果Fitted_Data 中的观察值大于Fitted_data 中的观察值，那么我想估计同一天和同一天的中位数并替换它。我尝试过类似的方法。

df%>%group_by(name, Month, Day)%>%
mutate(New_prices = replace(New_prices, Fitted_Data>Threshold, median(Fitted_Data)))

但我有一个错误提示 Problem with mutate()inputNew_prices.。

例如，我对第一组 (Aus) 的期望输出将是：

name Threshold Fitted_Data       Day Month
Aust      38.9    38.20784    jueves     1
Aust      38.9    35.52096    martes     3
Aust      38.9    37.05763 miércoles     1
Aust      38.9    36.19203    jueves     4
Aust      38.9    39.05569   viernes     1  #Replaced with median(38.19453, 39.91685)
Aust      38.9    38.19453   viernes     1
Aust      38.9    36.86204   domingo     7
Aust      38.9    38.51312     lunes     2
Aust      38.9    35.14895    martes     3
Aust      38.9    35.41919 miércoles     4
Aust      38.9    35.13218   domingo     6
Aust      38.9    35.46005     lunes     7
Aust      38.9    37.48999    jueves    12
Aust      38.9    37.54950    martes    11
Aust      38.9    38.68705 miércoles     1
Aust      38.9    36.23085    jueves     2
Aust      38.9    35.90234   viernes    10
Aust      38.9    38.50205   viernes    10
Aust      38.9    38.06678   domingo     7 #replaced with median(39.27153,  36.86204)
Aust      38.9    38.03129     lunes     3

【问题讨论】：

标签： r replace tidyverse dplyr

【解决方案1】：

我不确定你到底想做什么；您提供的代码（带有replace）给出了错误，因为它提到了数据中不存在的变量。

但是使用 mutate 和 if_else 应该可以工作：

df %>%
  group_by(name, Month, Day) %>%
  mutate(Fitted_Data = if_else(Fitted_Data > Threshold, median(Fitted_Data), Fitted_Data))

当我在你的数据上运行它时给出你想要的结果：

# A tibble: 60 x 5
# Groups:   name, Month, Day [51]
   name  Threshold Fitted_Data Day         Month
   <fct>     <dbl>       <dbl> <fct>       <dbl>
 1 Aust       38.9        38.2 "jueves"        1
 2 Aust       38.9        35.5 "martes"        3
 3 Aust       38.9        37.1 "miércoles"     1
 4 Aust       38.9        36.2 "jueves"        4
 5 Aust       38.9        39.1 "viernes"       1
 6 Aust       38.9        38.2 "viernes"       1
 7 Aust       38.9        36.9 "domingo"       7
 8 Aust       38.9        38.5 "lunes"         2
 9 Aust       38.9        35.1 " martes"       3
10 Aust       38.9        35.4 "miércoles"     4
# … with 50 more rows

【讨论】：