我试图使用函数mutate是为了创建一个基于其他三个条件的变量。这些条件是使用case_when创建的，如下面的代码所示。
但是我有一些使用NA值的条件，这些似乎会导致mutate函数出错。请检查一下:
# About the variables being used:
unique(x1)
# [1] 1 0 NA
str(pemg$x1)
# num [1:1622989] 1 0 0 1 1 0 1 1 0 0 ...
unique(x2)
# [1] 16 66 38 11 8 6 14 17 53 59 10 31 50 19 48 42 44 21 54 55 56 18 57 61 13 43 7 4 15
# [30] 39 5 20 3 37 23 51 36 52 68 58 27 65 62 2 12 32 41 49 46 35 34 45 81 69 33 40 0 70
# [59] 9 47 63 29 25 22 64 24 60 30 67 26 71 72 28 1 75 80 87 77 73 78 76 79 74 83 92 102 85
# [88] 86 90 82 91 84 88 93 89 96 95 105 115 106 94 100 99 97 104 98 103 108 109 101 117 107 114 113 NA 112
# [117] 110 111
str(pemg$x2)
# num [1:1622989] 16 66 38 11 8 6 14 17 53 59 ...
unique(x3)
# [1] 6 3 4 5 0 8 2 1 11 9 10 7 NA 15
str(pemg$anoest)
# num [1:1622989] 6 3 4 5 3 0 5 8 4 2 ...
df <- mutate(df,
y = case_when(
x1 == 1 & x2 >= 7 & x3 == 0 ~ 1,
x1 == 1 & x2 >= 8 & x3 == 1 ~ 1,
x1 == 1 & x2 >= 10 & x3 == 3 ~ 1,
x1 == 1 & x2 >= 11 & x3 == 4 ~ 1,
x1 == 1 & x2 >= 12 & x3 == 5 ~ 1,
x1 == 1 & x2 >= 13 & x3 == 6 ~ 1,
x1 == 1 & x2 >= 14 & x3 == 7 ~ 1,
x1 == 1 & x2 >= 15 & x3 == 8 ~ 1,
x1 == 1 & x2 >= 16 & x3 == 9 ~ 1,
x1 == 1 & x2 >= 17 & x3 == 10 ~ 1,
x1 == 1 & x2 >= 18 & x3 == 11 ~ 1,
x1 == 1 & !is.na(x3) ~ 0,
x1 == 1 & x3 %in% 12:16 ~ 0,
x2 %in% 0:7 ~ NA,
x2 > 18 ~ NA,
x1 == 0 ~ NA,
is.na(x3) ~ NA))
# Error: Problem with `mutate()` input `defasado`.
# x must be a double vector, not a logical vector.
# i Input `defasado` is `case_when(...)`.
# Run `rlang::last_error()` to see where the error occurred.
last_error()
# <error/dplyr_error>
# Problem with `mutate()` input `y`.
# x must be a double vector, not a logical vector.
# i Input `y` is `case_when(...)`.
# Backtrace:
# 1. dplyr::mutate(...)
# 2. dplyr:::mutate.data.frame(...)
# 3. dplyr:::mutate_cols(.data, ...)
# Run `rlang::last_trace()` to see the full context.
last_trace()
# <error/dplyr_error>
# Problem with `mutate()` input `defasado`.
# x must be a double vector, not a logical vector.
# i Input `defasado` is `case_when(...)`.
# Backtrace:
# x
# 1. +-dplyr::mutate(...)
# 2. -dplyr:::mutate.data.frame(...)
# 3. -dplyr:::mutate_cols(.data, ...)
# <parent: error/rlang_error>
# must be a double vector, not a logical vector.
# Backtrace:
# x
# 1. +-mask$eval_all_mutate(dots[[i]])
# 2. -dplyr::case_when(...)
# 3. -dplyr:::replace_with(...)
# 4. -dplyr:::check_type(val, x, name)
# 5. -dplyr:::glubort(header, "must be {friendly_type_of(template)}, not {friendly_type_of(x)}.")
这里的问题是case_when的结果。if_else form dplyr比ifelse from base R更严格——所有结果值必须具有相同的类型。因为case_when是多个if_else的向量化，你必须告诉R输出应该是哪种NA类型:
library(dplyr)
# does not work
dplyr::tibble(d = c(6,2,4, NA, 5)) %>%
dplyr::mutate(v = case_when(d < 4 ~ 0,
is.na(d) ~ NA))
# works
dplyr::tibble(d = c(6,2,4, NA, 5)) %>%
dplyr::mutate(v = case_when(d < 4 ~ 0,
is.na(d) ~ NA_real_))
您需要确保您的NA's是正确的类。在您的情况下，将NA放在as.numeric()中的~之后。例如:
x2 %in% 0:7 ~ as.numeric(NA)
R有不同类型的NA。您正在使用的是逻辑类型，但是您需要双类型NA_real_，以便与其他条件的输出保持一致。欲了解更多信息，请参阅:https://stat.ethz.ch/R-manual/R-patched/library/base/html/NA.html
在base R中，我们可以构造一个逻辑向量，并根据该逻辑向量将列值赋给NA。与case_when不同，我们不需要真正指定NA的类型，因为它会自动转换。
df1$d[df1$d %in% 0:7] <- NA
对于简单的操作，可以在base R中以紧凑的方式完成