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处理数据集以考虑重复测量

tidyr dplyr statistics r

D500 · 技术社区 · 7 年前

鉴于:

df <- data.frame(
                  CompanyID=c("Drinkers","Drinkers","Drinkers","Drinkers","Drinkers","Drinkers","Drinkers","Drinkers"
                            ,"Drinkers","Drinkers", "Liquders","Liquders","Liquders","PelletCoffeeCo","PelletCoffeeCo"),
                  Email= c("john@coffee.com", "john@coffee.com","john@coffee.com","john@coffee.com", "john@coffee.com", 
                          "john@coffee.com", "john@coffee.com", "john@coffee.com", "john@coffee.com", "john@coffee.com",
                          "george@liquid.com","george@liquid.com","george@liquid.com","stacy@pelletcoffee.com",
                        "stacy@pelletcoffee.com"),
                  Day= c("1","2","3","4","5","6","7","8","9","10","1","2","3","1","2"),
                 var1= c(4,5,5,5,2,3,2,7,6,5,7,6,6,2,3))

df2 <- data.frame(CompanyID=c("Drinkers","Drinkers","Drinkers","Drinkers","Drinkers","Drinkers","Drinkers","Drinkers"
                            ,"Drinkers","Drinkers", "Liquders","Liquders","Liquders","Liquders","Liquders","Liquders",
                            "Liquders","Liquders","Liquders","Liquders", "PelletCoffeeCo","PelletCoffeeCo","PelletCoffeeCo",
                            "PelletCoffeeCo","PelletCoffeeCo","PelletCoffeeCo","PelletCoffeeCo","PelletCoffeeCo",
                            "PelletCoffeeCo","PelletCoffeeCo"),
                  Email= c("john@coffee.com", "john@coffee.com","john@coffee.com","john@coffee.com", "john@coffee.com", 
                             "john@coffee.com", "john@coffee.com", "john@coffee.com", "john@coffee.com", "john@coffee.com",
                           "george@liquid.com","george@liquid.com","george@liquid.com","george@liquid.com","george@liquid.com",
                           "george@liquid.com","george@liquid.com","george@liquid.com","george@liquid.com","george@liquid.com","stacy@pelletcoffee.com",
                           "stacy@pelletcoffee.com","stacy@pelletcoffee.com","stacy@pelletcoffee.com","stacy@pelletcoffee.com",
                           "stacy@pelletcoffee.com","stacy@pelletcoffee.com","stacy@pelletcoffee.com","stacy@pelletcoffee.com",
                           "stacy@pelletcoffee.com"),
                  Day= c("1","2","3","4","5","6","7","8","9","10","1","2","3","4","5","6","7","8","9","10",
                         "1","2","3","4","5","6","7","8","9","10"),
                  var1= c(4,5,5,5,2,3,2,7,6,5,7,6,6, NA,NA,NA,NA,NA,NA,NA, 2,3,NA,NA,NA,NA,NA,NA,NA,NA))

说明: 我有数据表明,我每天对人们进行一次为期10天的调查。在一个完美的世界里,我会从每个参与者那里得到10个回复,用day1:day10表示。然而,由于没有回应,一些参与者给出了3个回应,其他人,6个,其他人10个等等。我正在设置数据以运行增长模型,因此我需要“Day”列始终读取Day1-Day10,无论是否有这些回应的数据。我试图通过向没有全部10天数据的行中添加NA来证明这一点。

2 回复 | 直到 3 年前

www 7 年前

试试这个:

library(tidyr)

df %>% 
  complete(nesting(CompanyID,Email), Day = seq(min(Day), max(Day), 1L)) %>%
  data.frame()

输出:

        CompanyID                  Email Day var1
1        Drinkers        john@coffee.com   1    4
2        Drinkers        john@coffee.com   2    5
3        Drinkers        john@coffee.com   3    5
4        Drinkers        john@coffee.com   4    5
5        Drinkers        john@coffee.com   5    5
6        Drinkers        john@coffee.com   6    2
7        Drinkers        john@coffee.com   7    3
8        Drinkers        john@coffee.com   8    2
9        Drinkers        john@coffee.com   9    7
10       Drinkers        john@coffee.com  10    6
11       Liquders      george@liquid.com   1    7
12       Liquders      george@liquid.com   2   NA
13       Liquders      george@liquid.com   3    6
14       Liquders      george@liquid.com   4    6
15       Liquders      george@liquid.com   5   NA
16       Liquders      george@liquid.com   6   NA
17       Liquders      george@liquid.com   7   NA
18       Liquders      george@liquid.com   8   NA
19       Liquders      george@liquid.com   9   NA
20       Liquders      george@liquid.com  10   NA
21 PelletCoffeeCo stacy@pelletcoffee.com   1    2
22 PelletCoffeeCo stacy@pelletcoffee.com   2   NA
23 PelletCoffeeCo stacy@pelletcoffee.com   3    3
24 PelletCoffeeCo stacy@pelletcoffee.com   4   NA
25 PelletCoffeeCo stacy@pelletcoffee.com   5   NA
26 PelletCoffeeCo stacy@pelletcoffee.com   6   NA
27 PelletCoffeeCo stacy@pelletcoffee.com   7   NA
28 PelletCoffeeCo stacy@pelletcoffee.com   8   NA
29 PelletCoffeeCo stacy@pelletcoffee.com   9   NA
30 PelletCoffeeCo stacy@pelletcoffee.com  10   NA

编辑:

上述代码使用由该列中现有值的最小值和最大值(即分别为1和10)定义的一组完整的日值填充每组的日列值。填充这些日值的组可以根据需要重新定义,但我选择在这里将其定义为公司+电子邮件,并带有行“嵌套(CompanyID,Email)”。数据。frame()行正好用于将输出转换为数据。帧而不是tibble。如果是数据。帧输出是没有必要的,请随意更换或删除该行。

pyll 7 年前

首先,创建唯一公司ID的数据框架。接下来,创建所需日期的数据框。

将这些交叉连接在一起。

然后连接到原始数据集以填写表格。

comp <- data.frame(CompanyID = unique(df$CompanyID))
Day <- data.frame(Day = c("1","2","3","4","5","6","7","8","9","10"))

compDay <- merge(comp, Day, all = TRUE)

dfday <- merge(df, compDay, by = c("CompanyID", "Day"), all = TRUE)