Question

我正在处理一个数据集，其中每个参与者（ID）均被评估1、2或3次。这是一个纵向研究。不幸的是，当第一位分析师对数据集进行编码时，她/他没有分配有关该数据集的任何信息。

由于所有参与者都有年龄信息（以月为单位），因此很容易确定何时进行第一次评估，何时进行第二次评估，依此类推。在第一次评估中，参与者比第二次评估年轻，依此类推。

我使用了tidyverse工具来解决这个问题，并且一切正常。但是，我真的知道（想象...）还有许多其他（更多）优雅的解决方案，我来到这个论坛是为了解决这个问题。有人可以让我思考如何使这段代码更短，更清晰吗？

这是复制代码的伪造数据：

ds <- data.frame(id = seq(1:6),
                 months = round(rnorm(18, mean=12, sd=2),0),
                 x1 = sample(0:2), 
                 x2 = sample(0:2),
                 x3 = sample(0:2),
                 x4 = sample(0:2))

#add how many times each child was acessed
ds <- ds %>% group_by(id) %>% mutate(how_many = n())
#Add position
ds %>% group_by(id) %>% 
  mutate(first = min(months), 
         max = max(months), 
         med = median(months)) -> ds

#add label to the third evaluation (the second will be missing)
ds %>% 
  mutate(group = case_when((how_many == 3) & (months %in% first) ~ "First evaluation",
                           (how_many == 3) & (months %in% max) ~ "Third evaluation",
                           TRUE ~ group)) -> ds
#add label to the second evaluation for all children evaluated two times 
ds %>% mutate_at(vars(group), funs(if_else(is.na(.),"Second Evaluation",.))) -> ds

这是我的原始代码：

temp <- dataset %>% select(idind, arm, infant_sex,infant_age_months)
#add how many times each child was acessed
temp <- temp %>% group_by(idind) %>% mutate(how_many = n())
#Add position
temp %>% group_by(idind) %>% 
  mutate(first = min(infant_age_months), 
         max = max(infant_age_months), 
         med = median(infant_age_months)) -> temp

#add label to the first evaluation
temp %>% 
  mutate(group = case_when(how_many == 1 ~ "First evaluation")) -> temp

#add label to the second evaluation (and keep all previous results)
temp %>% 
  mutate(group = case_when((how_many == 2) & (infant_age_months %in% first) ~ "First evaluation",
                           (how_many == 2) & (infant_age_months %in% max) ~ "Second evaluation",
                           TRUE ~ group)) -> temp

#add label to the third evaluation (the second will be missing)
temp %>% 
  mutate(group = case_when((how_many == 3) & (infant_age_months %in% first) ~ "First evaluation",
                           (how_many == 3) & (infant_age_months %in% max) ~ "Third evaluation",
                           TRUE ~ group)) -> temp
#add label to the second evaluation for all children evaluated two times 
temp %>% mutate_at(vars(group), funs(if_else(is.na(.),"Second Evaluation",.))) -> temp

请记住，在询问之前我使用了搜索框，并且我真的想象其他人在编写程序时会想到相同的问题。非常感谢

Answer 1

你去了。我用rank()给出了处理顺序。

ds <- data.frame(id = seq(1:6),
             months = round(rnorm(18, mean=12, sd=2),0),
             x1 = sample(0:2), 
             x2 = sample(0:2),
             x3 = sample(0:2),
             x4 = sample(0:2))

ds2 = ds %>% group_by(id) %>% mutate(rank = rank(months,ties.method="first"))
labels = c("First", "Second","Third")
ds2$labels = labels[ds2$rank]

Answer 2

或者只是按年龄排列并使用1:n()而不是n()，它会创建一个序列：

ds <- ds %>% group_by(id) %>% arrange(months) %>% mutate(how_many = 1:n())
ds %>% arrange(id, months)

# A tibble: 18 x 7
# Groups:   id [6]
      id months    x1    x2    x3    x4 how_many
   <int>  <dbl> <int> <int> <int> <int>    <int>
 1     1     10     1     2     0     1        1
 2     1     11     1     2     0     1        2
 3     1     12     1     2     0     1        3
 4     2     11     0     1     2     2        1
 5     2     14     0     1     2     2        2
 6     2     14     0     1     2     2        3

然后，您可以根据需要使用factor附加标签。

ds$label <- factor(ds$how_many, level = 1:3, label = c("First", "Second","Third"))
head(ds)
# A tibble: 18 x 8
# Groups:   id [6]
      id months    x1    x2    x3    x4 how_many label 
   <int>  <dbl> <int> <int> <int> <int>    <int> <fct> 
 1     1     10     1     2     0     1        1 First 
 2     1     11     1     2     0     1        2 Second
 3     1     12     1     2     0     1        3 Third 
 4     2     11     0     1     2     2        1 First 
 5     2     14     0     1     2     2        2 Second
 6     2     14     0     1     2     2        3 Third

改善dplyr解决方案-根据其他信息通过条件排序（位置）创建变量

2 个答案: