在某些观察结果之前先选择分组，然后将R中的var与NA对照分组即可

时间：2018-08-24 13:04:44

标签： r dplyr data.table

我的样品。 

 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, 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), .Label = c("x", 
"y"), class = "factor"), x1 = c(14L, 15L, 36L, 0L, 0L, 0L, 53L, 
10L, 39L, 27L, 67L, 25L, 19L, 49L, 53L, 64L, 61L, 12L, 75L, 34L, 
88L, 43L, 85L, 93L, 44L, 31L, 37L, 90L, 66L, 39L, 59L, 96L, 41L, 
23L, 20L, 26L, 69L, 28L, 35L, 96L, 87L, 82L, 70L, 68L, 26L, 12L, 
58L, 18L, 76L, 93L, 3L, 31L), group = structure(c(2L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
2L), .Label = c("female", "male"), class = "factor")), .Names = c("add", 
"x1", "group"), class = "data.frame", row.names = c(NA, -52L))

在此数据中有组变量（性别（男性和女性），我需要统计平均值，所有在女性之前的男性的平均百分比为25％。在女性之后，我不接触的男性。女性，我也不接触触摸。 这是根据添加列中的组x和y进行的分析。 如果对于男性而言，女性的价值比女性的价值高出x1> 25％（我们在女性之前计算得出），则该值必须用女性之前的男性平均值代替“女性类别，我们请勿触摸”。

AntoniosK的解决方案非常好

library(tidyverse)
library(data.table)

data %>%  
  group_by(add) %>%                                           # for each add do the below...
  mutate(group2 = rleid(group)) %>% 
  group_by(add, group, group2) %>%
  mutate(MEAN = mean(x1[group=="male" & group2==1]),               
         Q25 = quantile(x1[group=="male" & group2==1], 0.25)) %>%
  group_by(add) %>%                                            # for each add update x1 values....
  mutate(x1 = ifelse(group=="male" & group2==3 & x1 > unique(Q25[!is.na(Q25)]), unique(MEAN[!is.na(MEAN)]), x1)) %>%
  ungroup() %>%
  select(-group2) %>%
  data.frame()

但现在我想将x的0值替换为Na。

data$x1[data$x1 == 0] <- NA

在此之后，当我取消脚本编写时，出现错误

  

mutate_impl（.data，点）中的错误：评估错误：丢失   如果“ na.rm”为FALSE，则不允许使用值和NaN。

该怎么办，该脚本通过了NA，并且只能使用int值？

编辑

data=structure(list(add = c(11202L, 11202L, 11202L, 11202L, 11202L, 
                       11202L, 11202L, 11202L, 11202L, 11202L, 11202L, 11202L, 11202L, 
                       11202L, 11202L, 11202L, 11202L, 11202L, 11202L, 11202L, 11202L, 
                       11202L, 11202L, 11202L, 11202L, 11202L, 11202L, 11202L, 11202L, 
                       11202L, 11202L, 11202L, 11202L, 11202L, 11202L, 11202L, 11202L, 
                       11202L, 11202L, 11202L, 11202L, 11202L, 11202L, 11202L, 11202L, 
                       11202L, 11202L, 11202L, 11202L, 11202L, 11202L, 11202L), x1 = c(NA, 
                                                                                       2L, NA, NA, NA, NA, NA, NA, NA, NA, 1L, NA, 1L, 1L, NA, NA, NA, 
                                                                                       NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 1L, NA, NA, NA, NA, NA, 
                                                                                       NA, NA, NA, NA, NA, NA, NA, 3L, NA, NA, NA, NA, 1L, 1L, NA, NA, 
                                                                                       NA, NA, NA), 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")), .Names = c("add", "x1", "group"), class = "data.frame", row.names = c(NA, 
                                                                                                                                                                                                                                                                                           -52L))

library(tidyverse)
library(data.table)

data %>%  
  group_by(add) %>%                                          
  mutate(group2 = rleid(group)) %>% 
  group_by(add, group, group2) %>%
  mutate(MEAN = mean(x1[group=="male" & group2==1]),               
         Q25 = quantile(x1[group=="male" & group2==1], 0.25)) %>%
  group_by(add) %>%                                           
  mutate(x1 = ifelse(group=="male" & group2==3 & x1 > unique(Q25[!is.na(Q25)]), unique(MEAN[!is.na(MEAN)]), x1),
         x1 = ifelse(x1==0, NA, x1)) %>%  # new code added
  ungroup() %>%
  select(-group2) %>%
  data.frame()

Edit2

代码结果

add x1  group   MEAN    Q25
x   14.00000    male    23.72727    5.0
x   15.00000    male    23.72727    5.0
x   36.00000    male    23.72727    5.0
x   0.00000 male    23.72727    5.0
x   0.00000 male    23.72727    5.0
x   0.00000 male    23.72727    5.0
x   53.00000    male    23.72727    5.0
x   10.00000    male    23.72727    5.0
x   39.00000    male    23.72727    5.0
x   27.00000    male    23.72727    5.0
x   67.00000    male    23.72727    5.0
x   25.00000    female  NaN NA
x   19.00000    female  NaN NA
x   49.00000    female  NaN NA
x   53.00000    female  NaN NA
x   64.00000    female  NaN NA
x   61.00000    female  NaN NA
x   12.00000    female  NaN NA
x   23.72727    male    NaN NA
x   23.72727    male    NaN NA
x   23.72727    male    NaN NA
x   23.72727    male    NaN NA
x   23.72727    male    NaN NA
x   23.72727    male    NaN NA
x   23.72727    male    NaN NA
x   23.72727    male    NaN NA

之后

add x1     group
x   94.90   male

女性之后的前4位男性的总和= 94.90

1 个答案:

答案 0 :(得分：1)

我添加了一段代码来解决您的问题，并简要说明了错误。

更新代码 

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, 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), .Label = c("x", 
"y"), class = "factor"), x1 = c(14L, 15L, 36L, 0L, 0L, 0L, 53L, 
10L, 39L, 27L, 67L, 25L, 19L, 49L, 53L, 64L, 61L, 12L, 75L, 34L, 
88L, 43L, 85L, 93L, 44L, 31L, 37L, 90L, 66L, 39L, 59L, 96L, 41L, 
23L, 20L, 26L, 69L, 28L, 35L, 96L, 87L, 82L, 70L, 68L, 26L, 12L, 
58L, 18L, 76L, 93L, 3L, 31L), group = structure(c(2L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
2L), .Label = c("female", "male"), class = "factor")), .Names = c("add", 
"x1", "group"), class = "data.frame", row.names = c(NA, -52L))

library(tidyverse)
library(data.table)

data %>%  
  group_by(add) %>%                                          
  mutate(group2 = rleid(group)) %>% 
  group_by(add, group, group2) %>%
  mutate(MEAN = mean(x1[group=="male" & group2==1]),               
         Q25 = quantile(x1[group=="male" & group2==1], 0.25)) %>%
  group_by(add) %>%                                           
  mutate(x1 = ifelse(group=="male" & group2==3 & x1 > unique(Q25[!is.na(Q25)]), unique(MEAN[!is.na(MEAN)]), x1),
         x1 = ifelse(x1==0, NA, x1)) %>%  # new code added
  ungroup() %>%
  select(-group2) %>%
  data.frame()

错误说明

您必须运行代码的上半部分，最后只更新x1列。之所以会出现此错误，是因为NA值破坏了您需要执行的mean和quantile计算。

另一种方法是在开始时更新x1，然后使用na.rm=T进行计算。

对于您提到的新案例，从NA的{​​{1}}值开始，请尝试以下操作：

x1

对于您提到的新案例（编辑2），首先将之前代码的输出保存为data %>% group_by(add) %>% mutate(group2 = rleid(group)) %>% group_by(add, 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) %>% mutate(x1 = ifelse(group=="male" & group2==3 & x1 > unique(Q25[!is.na(Q25)]), unique(MEAN[!is.na(MEAN)]), x1))%>% ungroup() %>% select(-group2) %>% data.frame() ：

data2

然后运行：

data2 = data %>% ...
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