问题

我有多列类似名称的数据，但属于不同的类别。我需要用0替换所有的NA，但保留列的类别，因为稍后会为不同的字符串分配其他数值。

这是示例数据：

                 tricep_wt = c(2,NA,3))```

这是我尝试将NA更改为0的方法：

mutate(qdf, across(contains("wt"), ~case_when(is.numeric(.x) ~ tidyr::replace_na(., 0),
is.character(.x) ~ tidyr::replace_na(., "0"))))

我收到错误：

Error in mutate():
i In argument: across(...).
Caused by error in across():
! Can't compute column bicep_wt.
Caused by error in case_when():
! Failed to evaluate the right-hand side of formula 1.
Caused by error in vec_assign():
! Can't convert replace <double> to match type of data <character>.

使用相同的方法也会出现错误：

mutate(qdf, across(contains("wt"), ~case_when(is.numeric(.x) ~ tidyr::replace_na(.x, 0),
is.character(.x) ~ tidyr::replace_na(.x, "0"))))

<details>
<summary>英文:</summary>

I have data with multiple columns of similar names, but of different classes. I need to replace all `NA` with `0` but retain the columns class, as different character strings will be assigned other numeric values later.

This is example data:

qdf = data.frame(bicep_wt = c("black band", "5", NA),
tricep_wt = c(2,NA,3))

and this is my attempt to change the NA to 0:

mutate(qdf, across(contains("wt"), ~case_when(is.numeric(.x) ~ tidyr::replace_na(., 0),
is.character(.x) ~ tidyr::replace_na(., "0"))))

I get error:

Error in mutate():
i In argument: across(...).
Caused by error in across():
! Can't compute column bicep_wt.
Caused by error in case_when():
! Failed to evaluate the right-hand side of formula 1.
Caused by error in vec_assign():
! Can't convert replace <double> to match type of data <character>.

same error with:

mutate(qdf, across(contains("wt"), ~case_when(is.numeric(.x) ~ tidyr::replace_na(.x, 0),
is.character(.x) ~ tidyr::replace_na(.x, "0"))))

</details>


# 答案1
**得分**: 2

以下是翻译好的部分：

"While being 'class-safe' is usually a good thing, functions that are not class-safe can be used advantageously."

"虽然通常来说，具有“类安全性”通常是一件好事，但那些不具备类安全性的函数也可以有优势的用途。"

By "class-safe", I mean that you are guaranteed of the class returns from an expression. For instance,

"所谓“类安全性”，指的是你可以从表达式中得到类别安全的返回值。例如，"

The first call is ambiguous, since the `yes=` is class numeric and the `no=` is class character. A class-safe function should complain about this, as in

"第一个调用是模糊的，因为`yes=`的类别是数值，而`no=`的类别是字符。一个具备类安全性的函数应该会抱怨这一点，如下所示："

Note that `ifelse` and `replace` are _not_ class-safe, but in this case it is acceptable based on your requirements.

"请注意，`ifelse` 和 `replace` _不具备_ 类安全性，但在这种情况下，根据您的需求，这是可以接受的。"

<details>
<summary>英文:</summary>

While being &quot;class-safe&quot; is usually a good thing, functions that are not class-safe can be used advantageously.

```r
mutate(qdf, across(contains(&quot;wt&quot;), ~ replace(.x, is.na(.x), 0)))
#     bicep_wt tricep_wt
# 1 black band         2
# 2          5         0
# 3          0         3

By "class-safe", I mean that you are guaranteed of the class returns from an expression. For instance,

ifelse(c(T, T), 1, "1")
# [1] 1 1
ifelse(c(T, F), 1, "1")
# [1] "1" "1"

The first call is ambiguous, since the yes= is class numeric and the no= is class character. A class-safe function should complain about this, as in

dplyr::if_else(c(T, T), 1, "1")
# Error in dplyr::if_else(c(T, T), 1, "1") : 
#   Can't combine `true` <double> and `false` <character>.
data.table::fifelse(c(T, T), 1, "1")
# Error in data.table::fifelse(c(T, T), 1, "1") : 
#   'yes' is of type double but 'no' is of type character. Please make sure that both arguments have the same type.

Note that ifelse and replace are not class-safe, but in this case it is acceptable based on your requirements.

答案2

得分: 1

你可以将条件传递给 replace_na 函数的内部：

mutate(qdf, across(contains("wt"), ~ replace_na(.x, `if`(is.numeric(.x), 0, "0"))))

结果：

    bicep_wt tricep_wt
1 black band         2
2          5         0
3          0         3

注：它可以使用 ifelse() 或 `if`()。

mutate(qdf, across(contains("wt"), ~ replace_na(.x, `if`(is.numeric(.x), 0, "0"))))

    bicep_wt tricep_wt
1 black band         2
2          5         0
3          0         3

答案3

得分: 0

mutate(
qdf,
across(
contains("wt"),
(.x) { # could use ~ instead
if (is.numeric(.x)) tidyr::replace_na(.x, 0)
else if (is.character(.x)) tidyr::replace_na(.x, "0")
}
)
)

bicep_wt tricep_wt

1 black band 2

2 5 0

3 0 3

mutate(
  qdf, 
  across(
    contains("wt"), 
    \(.x) { # could use ~ instead
      if      (is.numeric(.x)  ) tidyr::replace_na(.x, 0)
      else if (is.character(.x)) tidyr::replace_na(.x, "0")
    }
  )
)


#     bicep_wt tricep_wt
# 1 black band         2
# 2          5         0
# 3          0         3

答案4

得分: 0

1. 用循环的方式：

for (col in colnames(qdf)) {
  qdf[is.na(qdf[, col]), col] <- 0
}

2. 使用函数和 lapply()：

naToZero <- function(v) {
  v[is.na(v)] <- 0
  return(v)
}
qdf <- data.frame(lapply(X=qdf, FUN=function(u) naToZero(u)))
qdf
    bicep_wt tricep_wt
1 black band         2
2          5         0
3          0         3

for (col in colnames(qdf)) {
  qdf[is.na(qdf[, col]), col] <- 0
}

naToZero <- function(v) {
  v[is.na(v)] <- 0
  return(v)
}
qdf <- data.frame(lapply(X=qdf, FUN=function(u) naToZero(u)))
qdf
    bicep_wt tricep_wt
1 black band         2
2          5         0
3          0         3