英文:
Does the order of the statement in ifelse() matter in R
问题
这是我的数据集：
a <- c(1, NA, 1, 1, 1, NA, NA, 1, NA, 1, 1)b <- c(NA, 1, NA, NA, 1, 1, 1, 1, 1, NA, NA)df <- data.frame(a, b)
我想要创建一个新变量，即：
当a为1，b为1时，产生"both"
当a为NA，b为1时，产生"B"
当a为1，b为NA时，产生"A"
我不明白，为什么只有最后一段代码给了我正确的结果：
df$c <- ifelse(df$a == 1 & df$b == 1, "Both",ifelse(df$a == 1 & is.na(df$b), "A",ifelse(df$b == 1 & is.na(df$a), "B", NA)))df$d <- ifelse(df$a == 1 & is.na(df$b), "A",ifelse(df$a == 1 & df$b == 1, "Both",ifelse(df$b == 1 & is.na(df$a), "B", NA)))df$e <- ifelse(df$a == 1 & is.na(df$b), "A",ifelse(df$b == 1 & is.na(df$a), "B",ifelse(df$b == 1 & df$a == 1, "Both", NA)))df
希望这些代码有帮助。
英文:
This is my dataset:
a <- c(1, NA, 1, 1, 1, NA, NA, 1, NA, 1, 1)b <- c(NA, 1, NA, NA,1, 1, 1, 1,1, NA, NA)df <- data.frame(a,b)
I want to create a new variable,that is
when a is 1, b is 1, produce "both"
when a is NA, B is 1, produce "B"
when a is 1, b is NA, produce "A"
I don't understand, why only the last chuck of codes gave me the correct results
df$c <- ifelse(df$a ==1 & df$b == 1, "Both",ifelse(df$a == 1 & is.na(df$b), "A",ifelse(df$b == 1 & is.na(df$a), "B", NA)))df$d <- ifelse(df$a ==1 & is.na(df$b), "A",ifelse(df$a == 1 & df$b==1, "Both",ifelse(df$b == 1 & is.na(df$a), "B", NA)))df$e <- ifelse(df$a ==1 & is.na(df$b), "A",ifelse(df$b ==1 & is.na(df$a), "B",ifelse(df$b == 1 & df$a == 1, "Both", NA)))df
| a | b | c | d | e |
|---|---|---|---|---|
| 1 | NA | NA | A | A |
| NA | 1 | NA | NA | B |
| 1 | NA | NA | A | A |
| 1 | NA | NA | A | A |
| 1 | 1 | Both | Both | Both |
| NA | 1 | NA | NA | B |
| NA | 1 | NA | NA | B |
| 1 | 1 | Both | Both | Both |
| NA | 1 | NA | NA | B |
| 1 | NA | NA | A | A |
| 1 | NA | NA | A | A |
答案1
得分: 2
Yes, the order matters. The evaluation from left to right will stop as soon as the result is known.
是的，顺序很重要。从左到右的评估将在结果已知时停止。
ifelse returns NA in case the evaluation gives NA.
`ifelse` 在评估为 `NA` 时返回 `NA`。
So using ifelse it could look like:
因此，使用 `ifelse` 可能如下所示：
An alternative could be to use switch.
另一种选择是使用 `switch`。
英文:
Yes, the order matters. The evaluation from left to right will stop as soon as the result is known.
TRUE & NA#[1] NATRUE | NA#[1] TRUEFALSE & NA#[1] FALSEFALSE | NA#[1] NA
ifelse returns NA in case the evaluation gives NA.
ifelse(TRUE, 1, 0)#[1] 1ifelse(FALSE, 1, 0)#[1] 0ifelse(NA, 1, 0)#[1] NA
So using ifelse it could look like:
with(df, ifelse(!is.na(a) & a == 1 & !is.na(b) & b == 1, "Both",ifelse(!is.na(a) & a == 1 & is.na(b), "A",ifelse(!is.na(b) & b == 1 & is.na(a), "B", NA))))# [1] "A" "B" "A" "A" "Both" "B" "B" "Both" "B" "A"#[11] "A"
An alternative could be to use switch.
sapply(paste(df$a, df$b), switch,"1 1" = "Both","1 NA" = "A","NA 1" = "B",NA)# 1 NA NA 1 1 NA 1 NA 1 1 NA 1 NA 1 1 1 NA 1 1 NA 1 NA# "A" "B" "A" "A" "Both" "B" "B" "Both" "B" "A" "A"
答案2
得分: 1
Sure, here's the translated code:
由于您的 df$a 和 df$b 中都有 NA，如果进行值比较，您将得到 NA。此外，在 ifelse 中不会评估 NA，因此它将保持为 NA。
例如：
df$a == 1 & df$b == 1[1] NA NA NA NA TRUE NA NA TRUE NA NA NA
使用 dplyr::case_when() 可以解决这个问题，而且代码更可读，而不是嵌套一堆 ifelse。
library(dplyr)df %>%mutate(c = case_when(a == 1 & b == 1 ~ "Both",a == 1 & is.na(b) ~ "A",b == 1 & is.na(a) ~ "B",TRUE ~ NA_character_))
a b c
1 1 NA A
2 NA 1 B
3 1 NA A
4 1 NA A
5 1 1 Both
6 NA 1 B
7 NA 1 B
8 1 1 Both
9 NA 1 B
10 1 NA A
11 1 NA A
英文:
Since you have NA in your df$a and df$b, you would get NA if you do value comparison. Moreover, NA is not evaluated in your ifelse and would therefore stays NA.
For example:
df$a == 1 & df$b == 1[1] NA NA NA NA TRUE NA NA TRUE NA NA NA
Using dplyr::case_when() would solve the problem and the code is more readable without nesting a bunch of ifelse.
library(dplyr)df |> mutate(c = case_when(a == 1 & b == 1 ~ "Both",a == 1 & is.na(b) ~ "A",b == 1 & is.na(a) ~ "B",TRUE ~ NA_character_))a b c1 1 NA A2 NA 1 B3 1 NA A4 1 NA A5 1 1 Both6 NA 1 B7 NA 1 B8 1 1 Both9 NA 1 B10 1 NA A11 1 NA A
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