问题描述
这是data.frame，其中第二列是列表的列（请注意，还有一个NULL）。
我们如何将每个列表转换为常规元素，以便该列像其他任何字符类列一样？ （NULL可以是NA）
df <- structure(list(Year = c(2014L,2014L,2013L,2014L),Country = list(Country = "Canada",Country = "Germany",Country = "France",Country = "Mexico",Country = "Canada",NULL,Country = "United States of America",Country = "Germany")),class = "data.frame",row.names = c(NA,-20L))
注意
df %>% sapply(class)
Year Country
"integer" "list"
所需结果：
- 相同的数据，但
df %>% sapply(class)
Year Country
"integer" "character"
解决方法
我建议对您的margin数据使用函数的方法：
df输出：
myfun <- function(x)
{
if(is.null(x))
{y <- NA}
else
{
y <- x[[1]]
}
return(y)
}
#Apply
df$Newvar <- as.vector(do.call(rbind,lapply(df$Country,myfun)))
和一些检查：
Year Country Newvar
1 2014 Canada Canada
2 2014 Germany Germany
3 2014 France France
4 2014 Germany Germany
5 2014 Mexico Mexico
6 2014 Germany Germany
7 2014 Germany Germany
8 2014 Canada Canada
9 2014 NULL <NA>
10 2014 Germany Germany
11 2014 Mexico Mexico
12 2014 Canada Canada
13 2014 Mexico Mexico
14 2014 Germany Germany
15 2013 Canada Canada
16 2014 United States of America United States of America
17 2014 Canada Canada
18 2014 Mexico Mexico
19 2014 Canada Canada
20 2014 Germany Germany
str(df)
'data.frame': 20 obs. of 3 variables:
$ Year : int 2014 2014 2014 2014 2014 2014 2014 2014 2014 2014 ...
$ Country:List of 20
..$ Country: chr "Canada"
..$ Country: chr "Germany"
..$ Country: chr "France"
..$ Country: chr "Germany"
..$ Country: chr "Mexico"
..$ Country: chr "Germany"
..$ Country: chr "Germany"
..$ Country: chr "Canada"
..$ : NULL
..$ Country: chr "Germany"
..$ Country: chr "Mexico"
..$ Country: chr "Canada"
..$ Country: chr "Mexico"
..$ Country: chr "Germany"
..$ Country: chr "Canada"
..$ Country: chr "United States of America"
..$ Country: chr "Canada"
..$ Country: chr "Mexico"
..$ Country: chr "Canada"
..$ Country: chr "Germany"
$ Newvar : chr "Canada" "Germany" "France" "Germany" ...
现在不在列表中。
一个选项：
df$Country <- sapply(df$Country,function(x) if (length(x)) x else NA)
另一个：
df$Country[lengths(df$Country) == 0] <- list(NA)
df$Country <- as.vector(df$Country)
另一种使它与dplyr的mutate保持一致的方法。
df2 = df %>%
mutate(NewCountry = if_else(
sapply(df$Country,is.null),"MISSING",as.character(df$Country))
)
> sapply(df2,class)
Year Country NewCountry
"integer" "list" "character"