R 计算数值的平均值和字符串值的唯一连接

我有一个df，如下所示，现在我想要计算数值的平均值和字符串值的唯一连接

尝试以下代码

out = dcast(df, Date+Name+class~gender,fun.aggregate = mean,value.var='value')

输入数据帧如下所示

Date = c("8/20/2019","8/20/2019","8/20/2019","8/20/2019","8/20/2019","8/20/2019","8/20/2019","8/20/2019")
Name = c("ABC","ABC","CBC","CBC","XYLEM","XYLEM","XYLEM","XYLEM")
class = c("one","one","two","two","three","three","three","three")
gender = c("M","M","F","F","M","M","F","F")
value = c("1","2","top","topper","low","lower","1","3")

df = data.frame(Date,Name,class,gender,value)

输出数据框如下所示，共有五列

Date = c("8/20/2019","8/20/2019","8/20/2019")
Name = c("ABC","CBC","XYLEM")
class = c("one","two","three")
M=c("1.5","NA","low,lower")
F = c("NA","top,topper","2")

out = data.frame(Date,Name,class,M,F)

---

老实说，我很难理解为什么数据是这样格式化的，因为我们似乎在单个列中混合了数字值和字符串值。但这是可以做到的：

library(dplyr)
library(tidyr)

df.out = df %>%
  # Figure out which cells contain numbers and which contain strings, and store
  # them in separate columns.
  mutate(numeric.value = as.numeric(as.character(value)),
         string.value = ifelse(is.na(numeric.value), as.character(value), NA)) %>%
  # Group by date, name, class, and gender so we can summarize.
  group_by(Date, Name, class, gender) %>%
  # For each group, get the mean of the numeric values and the concatenation of
  # string values.
  summarize(numeric.mean = mean(numeric.value, na.rm = T),
            string.concat = gsub("^,|,,|,$", "", gsub("NA", "", paste0(string.value, collapse = ",")))) %>%
  # Fill in NAs where appropriate.
  mutate(numeric.mean = ifelse(is.nan(numeric.mean), NA, numeric.mean),
         string.concat = ifelse(string.concat == "", NA, string.concat)) %>%
  # The form of the desired output suggests that each group will have *either*
  # numeric values *or* string values, but not both.  So let's put the two
  # summary values back into a single column.
  mutate(summary.by.gender = coalesce(as.character(numeric.mean), string.concat)) %>%
  # Pivot so that each gender gets its own column.
  select(Date, Name, class, gender, summary.by.gender) %>%
  spread(gender, summary.by.gender)