R 用空白(空字符串)替换缺少的值(NA)
我有一个带有NA行的数据帧:R 用空白(空字符串)替换缺少的值(NA),r,na,R,Na,我有一个带有NA行的数据帧: df = data.frame(c("classA", NA, "classB"), t(data.frame(rep("A", 5), rep(NA, 5), rep("B", 5)))) rownames(df) <- c(1,2,3) colnames(df) <- c("class", paste("Year", 1:5, sep = "")) > df class Year1 Year2 Year3 Year4 Year5
df = data.frame(c("classA", NA, "classB"), t(data.frame(rep("A", 5), rep(NA, 5), rep("B", 5))))
rownames(df) <- c(1,2,3)
colnames(df) <- c("class", paste("Year", 1:5, sep = ""))
> df
class Year1 Year2 Year3 Year4 Year5
1 classA A A A A A
2 <NA> <NA> <NA> <NA> <NA> <NA>
3 classB B B B B B
df=data.frame(c(“classA”,NA,“classB”),t(data.frame(rep(“A”,5),rep(NA,5),rep(“B”,5)))
行名(df)另一种选择:
df <- sapply(df, as.character) # since your values are `factor`
df[is.na(df)] <- 0
这个答案更像是一个延伸的评论
你想做的不是我想做的好事。比如说,R不是Excel,所以做这样的事情只是为了在数据中创建可视的分离,这会让你以后感到头疼
如果您真的只关心视觉输出,我可以提供两个建议:
当您希望以该视觉分隔查看数据时,请使用na.print
参数来print
print(df, na.print = "")
# class Year1 Year2 Year3 Year4 Year5
# 1 classA A A A A A
# 2
# 3 classB B B B B B
要意识到,即使是上述建议也不是最好的建议。通过将数据.frame
转换为列表
,实现视觉和内容的分离:
split(df, df$class)
# $classA
# class Year1 Year2 Year3 Year4 Year5
# 1 classA A A A A A
#
# $classB
# class Year1 Year2 Year3 Year4 Year5
# 3 classB B B B B B
你的第一次尝试对我来说很好。那它不起作用呢?我仍然在数据帧中看到,代码似乎没有影响任何与因素有关的事情(当然!)。。。试试str(df)
(我对我的答案一语道破!)顺便说一句,不要只说“它没用”。您忽略了在运行该代码时肯定会收到的六条(!)警告消息。警告信息应该非常具有启发性,你不认为吗?
周围的括号表示它们不是字符串。更多信息。我认为“”是空格,而“”是空白。我说得对吗?
> df <- sapply(df, as.character)
> df[is.na(df)] <- " "
> df
class Year1 Year2 Year3 Year4 Year5
[1,] "classA" "A" "A" "A" "A" "A"
[2,] " " " " " " " " " " " "
[3,] "classB" "B" "B" "B" "B" "B"
> as.data.frame(df)
class Year1 Year2 Year3 Year4 Year5
1 classA A A A A A
2
3 classB B B B B B
print(df, na.print = "")
# class Year1 Year2 Year3 Year4 Year5
# 1 classA A A A A A
# 2
# 3 classB B B B B B
split(df, df$class)
# $classA
# class Year1 Year2 Year3 Year4 Year5
# 1 classA A A A A A
#
# $classB
# class Year1 Year2 Year3 Year4 Year5
# 3 classB B B B B B