在R中使用mutate函数时出现错误消息 [英] error message when using mutate function in r

问题描述

收到以下错误消息：
mutate_impl（.data，点）中的错误：
评估错误：缺少参数 no，没有默认值。

receiving the following error message: Error in mutate_impl(.data, dots) : Evaluation error: argument "no" is missing, with no default.

 mutate(x,perfLev= ifelse(SS< 1438, "Below Basic",
                   ifelse(SS>= 1439 & SS <= 1499, "Basic",
                   ifelse(SS >= 1500 & SS <= 1545, "Proficient",
                   ifelse(SS >= 1546, "Advanced")))))

推荐答案

使用Make212和Renu，这是修复它的一种选择：

Using comments by Make212 and Renu, here's one option for fixing it:

library(dplyr)
mutate(x,
       perfLev = case_when(
         SS <  1438              ~ "Below Basic",
         SS >= 1439 & SS <= 1499 ~ "Basic",
         SS >= 1500 & SS <= 1545 ~ "Proficient",
         SS >= 1546              ~ "Advanced",
         TRUE                    ~ "huh?"
       ) )

我添加了一个默认值（ TRUE ），通常很好（显式代码）。请注意，如果您不包含 TRUE ，则它将获得 NA 值，以防万一， 。如果满足以下任何条件，我就会在这里看到它发生：

I added a "default" (TRUE), which is generally good (explicit code). Note that if you do not include the TRUE, then it would get an NA value, in case that's what you want. I can see it happening here if any of the following are true:

• is.na（SS）


• SS> = 1438& SS ＜ 1439


• SS> 1499年SS ＜ 1500


• SS> 1545年SS ＜ 1546

• is.na(SS)
• SS >= 1438 & SS < 1439
• SS > 1499 & SS < 1500
• SS > 1545 & SS < 1546

如果 NA 是可以接受的，并且您保证具有 SS 的完整性。

You may not need it if NA is acceptable and you are guaranteed of SS's integrality.

此代码相当于对您的代码进行了轻微修复：

This code is equivalent to a slight fix to your code:

mutate(x,
       perfLev = 
         ifelse(SS < 1438, "Below Basic",
                ifelse(SS >= 1439 & SS <= 1499, "Basic",
                       ifelse(SS >= 1500 & SS <= 1545, "Proficient",
                              ifelse(SS >= 1546, "Advanced", "huh?"))))
       )

仅用于样式/清晰度的缩进。

Indentation for style/clarity only.

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