替换空白而不是NA

Question

我有两个变量a和b

a         b
vessel    hot
parts     
nest      NA
best      true
neat      smooth

我想用b

替换a中的空白

la$b[i1] <- ifelse(la$b[i1] == "",la$a[i1],la$b[i1])

但它不起作用

Answer 1

我们可以使用data.table。将'data.frame'转换为'data.table'（setDT(df1)），在'i'（b==''）中指定条件，并指定对应于TRUE值的'a'值'我'到'b'。它应该很快，因为我们正在分配。

library(data.able)
setDT(df1)[b=='', b:= a]
df1
#       a      b
#1: vessel    hot
#2:  parts  parts
#3:   nest     NA
#4:   best   true
#5:   neat smooth

或者我们可以base R

i1 <- df1$b=='' & !is.na(df1$b)
df1$b[i1] <- df1$a[i1]

数据

df1 <- structure(list(a = c("vessel", "parts", "nest", "best", "neat"
), b = c("hot", "", NA, "true", "smooth")), .Names = c("a", "b"
), class = "data.frame", row.names = c(NA, -5L))

Answer 2

而不是

# la$b[i1] <- ifelse(la$b[i1] == "",la$a[i1],la$b[i1])
# what is i1? it doesn't seem to have any obvious function here

......应该是：

la$b <- ifelse(la$b == "", la$a, la$b)

假设您要将b中的空格替换为a并且适用于所有空白

它有效：

df <- structure(list(a = c("vessel", "parts", "nest", "best", "neat"
), b = c("hot", "parts", NA, "true", "smooth")), .Names = c("a", 
"b"), row.names = c(NA, -5L), class = "data.frame")

df$b <- ifelse(df$b=="", df$a, df$b)
# or, with `with`: df$b <- with(df, ifelse(b=="",a,b))
#  > df
#       a      b
# 1 vessel    hot
# 2  parts  parts
# 3   nest   <NA>
# 4   best   true
# 5   neat smooth