如果在R中使用超过1个组，则na.rm函数不起作用

时间：2018-09-08 08:33:16

标签： r dplyr data.table lapply

在这篇文章中 select group before certain observations separated by grouping var in R with NA control，在使用一组add na.rm=T时有效。 但是新数据，其中三组

data=structure(list(add = 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, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = "x", class = "factor"), 
    x1 = c(0L, 2L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 1L, 0L, 1L, 
    1L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 1L, 
    0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 3L, 0L, 0L, 
    0L, 0L, 1L, 1L, 0L, 0L, 0L, 0L, 0L), add1 = c(514L, 514L, 
    514L, 514L, 514L, 514L, 514L, 514L, 514L, 514L, 514L, 514L, 
    514L, 514L, 514L, 514L, 514L, 514L, 514L, 514L, 514L, 514L, 
    514L, 514L, 514L, 514L, 514L, 514L, 514L, 514L, 514L, 514L, 
    514L, 514L, 514L, 514L, 514L, 514L, 514L, 514L, 514L, 514L, 
    514L, 514L, 514L, 514L, 514L, 514L, 514L, 514L, 514L, 514L
    ), group = structure(c(2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
    2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
    2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 2L, 2L, 
    2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c("female", 
    "male"), class = "factor"), add2 = c(2018L, 2018L, 2018L, 
    2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 
    2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 
    2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 
    2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 
    2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 
    2018L, 2018L, 2018L, 2018L)), .Names = c("add", "x1", "add1", 
"group", "add2"), class = "data.frame", row.names = c(NA, -52L
))

所以当我运行代码

library(tidyverse)
library( data.table)
data %>%  
  group_by(add,add1,add2) %>%                                          
  mutate(group2 = rleid(group)) %>% 
  group_by(add,add1,add2, group, group2) %>%
  mutate(MEAN = mean(x1[group=="male" & group2==1], na.rm = T),      ## extra code here ##    
         Q25 = quantile(x1[group=="male" & group2==1], 0.25, na.rm = T)) %>%  ## extra code here ##
  group_by(add,add1,add2) %>%                                           
  mutate(x1 = ifelse(group=="male" & group2==3 & x1 > unique(Q25[!is.na(Q25)]), unique(MEAN[!is.na(MEAN)]), x1))%>%
  ungroup() %>%
  select(-group2) %>%
  data.frame()

我遇到错误

Error in mutate_impl(.data, dots) : 
  Column `x1` must be length 24 (the group size) or one, not 0

PS。我只是提供了一个示例来给出数据结构，原因是有1000个组。我找不到群组 哪里有错误

如何解决此错误。

1 个答案:

答案 0 :(得分：2)

如果我理解正确，则该错误是由第一部分（x1）中所有 all NA是group == 1L的第一男性群体引起的。 / p>

恕我直言，一种更干净的方法是按照建议的here首先计算所有组的统计信息，并按照建议的here使用非等参连接来更新第二个男性组中的受影响行。 

library( data.table)
grp_stats <- setDT(data)[, group2 :=rleid(group), by = .(add, add1, add2)][
  group2 == 1L & group == "male", 
  .(group2 = 3L, mean = mean(x1, na.rm = TRUE), q25 = quantile(x1, 0.25, na.rm = TRUE)), 
  by = .(add, add1, add2)] 
grp_stats 

   add add1 add2 group2 mean  q25
1:   x  514 2018      3  1.5 1.25
2:   y  515 2018      3  NaN   NA
3:   z  516 2018      3  2.0 2.00

可以清楚地识别出产生错误统计信息的组。由OP决定从数据集中删除受影响的组。

但是，对于以后的联接，我们可以将它们保留在其中，因为它们不会产生任何影响。

具有常数值group2的列3已添加到组统计信息中，以简化后续的update in a non-equi join：

data[, x1 := as.numeric(x1)][
  grp_stats, on = .(group2, add, add1, add2, x1 > q25), x1 := mean][]
data

    add  x1 add1  group add2 group2
 1:   x 1.0  514   male 2018      1
 2:   x 2.0  514   male 2018      1
 3:   x  NA  514 female 2018      2
 4:   x  NA  514 female 2018      2
 5:   x 1.5  514   male 2018      3
 6:   x 1.0  514   male 2018      3
 7:   y  NA  515   male 2018      1
 8:   y  NA  515   male 2018      1
 9:   y  NA  515 female 2018      2
10:   y  NA  515 female 2018      2
11:   y 7.0  515   male 2018      3
12:   y 1.0  515   male 2018      3
13:   z 2.0  516   male 2018      1
14:   z  NA  516   male 2018      1
15:   z  NA  516 female 2018      2
16:   z  NA  516 female 2018      2
17:   z 2.0  516   male 2018      3
18:   z 1.0  516   male 2018      3

请注意，第5行和第17行已更新，而未触及产生错误统计信息的第二组行。

x1被强制键入numeric，然后再加入以匹配mean()返回的结果的类型。

样本数据

这里是由三组组成的样本数据。在第二个组中，第一个公节的所有x1值均为NA。

data <- data.table::fread("
add x1 add1  group add2
x    1  514   male 2018
x    2  514   male 2018
x   NA  514 female 2018
x   NA  514 female 2018
x    7  514   male 2018
x    1  514   male 2018
y   NA  515   male 2018
y   NA  515   male 2018
y   NA  515 female 2018
y   NA  515 female 2018
y    7  515   male 2018
y    1  515   male 2018
z    2  516   male 2018
z   NA  516   male 2018
z   NA  516 female 2018
z   NA  516 female 2018
z    7  516   male 2018
z    1  516   male 2018
")

验证错误消息是由不适用的第一个男性组引起的

将上述示例数据集传递到OP的代码中后，我们可以重现错误消息：

library(dplyr)
data %>% 
  group_by(add,add1,add2) %>%                                          
  mutate(group2 = rleid(group)) %>% 
  group_by(add,add1,add2, group, group2) %>%
  mutate(MEAN = mean(x1[group=="male" & group2==1], na.rm = T),      ## extra code here ##    
         Q25 = quantile(x1[group=="male" & group2==1], 0.25, na.rm = T)) %>%  ## extra code here ##
  group_by(add,add1,add2) %>%                                           
  mutate(x1 = ifelse(group=="male" & group2==3 & x1 > unique(Q25[!is.na(Q25)]), unique(MEAN[!is.na(MEAN)]), x1))%>%
  ungroup() %>%
  select(-group2) %>%
  data.frame()
  

mutate_impl（.data，点）中的错误：
    x1列的长度必须为6（组大小）或1，而不是0

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