使用lapply在R中出现问题，该函数在全局环境中使用常量。

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英文:

Problem using lapply in R with a function that uses constants in the global environment

问题

我正在尝试应用一个函数，该函数从全局环境中获取一个向量（BASELINE_CLASSIFICATION_THRESHOLDS）中的一些输入，并使用lapply将其应用于数据框。实质上，它将数字转换为级别（轻微、中等、严重、极端）：

  1. BASELINE_CLASSIFICATION_THRESHOLDS <- c(0, 3.5, 6.5, 10.0000001)
  3. value_to_classification <- function(x){
  5.   if((x >= BASELINE_CLASSIFICATION_THRESHOLDS[1]) && (x < BASELINE_CLASSIFICATION_THRESHOLDS[2])){
  6.     classification <- "轻微"
  7.   } else if((x >= BASELINE_CLASSIFICATION_THRESHOLDS[2]) && (x < BASELINE_CLASSIFICATION_THRESHOLDS[3])){
  8.     classification <- "中等"
  9.   } else if((x >= BASELINE_CLASSIFICATION_THRESHOLDS[3]) && (x < round(BASELINE_CLASSIFICATION_THRESHOLDS[4]))){
  10.     classification <- "严重"
  11.   } else {
  12.     classification <- "极端"
  13.   }
  14.   return(classification)
  15. }
  17. df <- data.frame(x = runif(10, min = 0, max = 10),
  18.                  y = runif(10, min = 0, max = 10),
  19.                  z = runif(10, min = 0, max = 10))

但是，当我尝试将value_to_classification应用于x列时，我遇到了一个错误：

  1. lapply(df["x"], value_to_classification)
  2. $x
  3. [1] "轻微"
  5. 警告信息：
  6. 1: 在 (x >= BASELINE_CLASSIFICATION_THRESHOLDS[1]) && (x < BASELINE_CLASSIFICATION_THRESHOLDS[2]) 中：
  7.   'length(x) = 10 > 1' 在强制类型转换为 'logical(1)' 时
  8. 2: 在 (x >= BASELINE_CLASSIFICATION_THRESHOLDS[1]) && (x < BASELINE_CLASSIFICATION_THRESHOLDS[2]) 中：
  9.   'length(x) = 10 > 1' 在强制类型转换为 'logical(1)' 时

另一方面，如果我写成：

  1. lapply(df[["x"]], value_to_classification)

它可以工作。最终我想做的是类似于：

  1. df[c("x1", "x2")] <-  lapply(df[c("x", "y")], value_to_classification)

一些搜索似乎表明我的语法是正确的，但我显然做错了什么。我做错了什么，该如何修复？

诚挚地感谢您提前的帮助。

Thomas Philips

英文:

I'm trying to apply a function that takes some inputs from the global environment held in a vector (BASELINE_CLASSIFICATION_THRESHOLDS) to a dataframe using lapply. In essence it transforms numbers to levels (Mild, Moderate, Severe, Extreme):

  1. BASELINE_CLASSIFICATION_THRESHOLDS  &lt;- c(0, 3.5, 6.5, 10.0000001)
  3. value_to_classification &lt;- function(x){
  5.   if((x &gt;= BASELINE_CLASSIFICATION_THRESHOLDS[1]) &amp;&amp; (x &lt; BASELINE_CLASSIFICATION_THRESHOLDS[2])){
  6.     classification &lt;- &quot;Mild&quot;
  7.   } else if((x &gt;= BASELINE_CLASSIFICATION_THRESHOLDS[2]) &amp;&amp; (x &lt; BASELINE_CLASSIFICATION_THRESHOLDS[3])){
  8.     classification &lt;- &quot;Moderate&quot;
  9.   } else if((x &gt;= BASELINE_CLASSIFICATION_THRESHOLDS[3]) &amp;&amp; (x &lt; round(BASELINE_CLASSIFICATION_THRESHOLDS[4]))){
  10.     classification &lt;- &quot;Severe&quot;
  11.   } else {
  12.     classification &lt;- &quot;Extreme&quot;
  13.   }
  14.   return(classification)
  15. }
  17. df &lt;- data.frame(x = runif(10, min = 0, max = 10),
  18.                  y = runif(10, min = 0, max = 10),
  19.                  z = runif(10, min = 0, max = 10))

But when I try to lapply value_to_classification to a column of x, I get an error:

  1. lapply(df[&quot;x&quot;], value_to_classification)
  2. $x
  3. [1] &quot;Mild&quot;
  5. Warning messages:
  6. 1: In (x &gt;= BASELINE_CLASSIFICATION_THRESHOLDS[1]) &amp;&amp; (x &lt; BASELINE_CLASSIFICATION_THRESHOLDS[2]) :
  7.   &#39;length(x) = 10 &gt; 1&#39; in coercion to &#39;logical(1)&#39;
  8. 2: In (x &gt;= BASELINE_CLASSIFICATION_THRESHOLDS[1]) &amp;&amp; (x &lt; BASELINE_CLASSIFICATION_THRESHOLDS[2]) :
  9.   &#39;length(x) = 10 &gt; 1&#39; in coercion to &#39;logical(1)&#39;

On the other hand, if I write

  1. lapply(df[[&quot;x&quot;]], value_to_classification)

it works. What I eventually want to do is to write something like

  1. df[c(&quot;x1&quot;, &quot;x2&quot;)] &lt;-  lapply(df[c(&quot;x&quot;, &quot;y&quot;)], value_to_classification)

Some searching seems to suggest that my syntax is OK, but I'm clearly getting something wrong. What am I doing wrong, and how can I fix this?

Sincerely and with many thanks in advance

Thomas Philips

答案1

得分: 1

问题是value_to_classification不适用于向量。您可以运行value_to_classification(c(1,2,3))，它将只返回一个值（而不是3个）。

一种解决方法是将函数向量化：

  1. vectorized_value_to_classification <- Vectorize(value_to_classification)
  3. df[c("x1", "x2")] <-  lapply(df[c("x", "y")], vectorized_value_to_classification)
  5. df
  6.           x        y         z       x1       x2
  7. 1  3.233599 5.612147 2.9525939     轻度     中度
  8. 2  5.453014 3.659298 8.1642952     中度     中度
  9. 3  7.104259 6.333049 7.1706136     重度     中度
  10. 4  4.199447 3.277607 8.9458447     中度     轻度
  11. 5  9.352140 7.135801 2.6721405     重度     重度
  12. 6  7.682951 1.358830 4.2102313     重度     轻度
  13. 7  6.551999 9.986188 1.9995422     重度     重度
  14. 8  7.436272 9.260056 0.1093833     重度     重度
  15. 9  5.163593 7.689474 0.2999034     中度     重度
  16. 10 7.500994 4.599129 8.5266752     重度     中度

(Note: I've translated the variable names in the code as well.)

英文:

The issue is value_to_classification does not work for a vector. You can run value_to_classification(c(1,2,3)) and it would only return one value (instead of 3).

One solution is to vectorize the function:

  1. vectorized_value_to_classification &lt;- Vectorize(value_to_classification)
  3. df[c(&quot;x1&quot;, &quot;x2&quot;)] &lt;-  lapply(df[c(&quot;x&quot;, &quot;y&quot;)], vectorized_value_to_classification)
  5. df
  6.           x        y         z       x1       x2
  7. 1  3.233599 5.612147 2.9525939     Mild Moderate
  8. 2  5.453014 3.659298 8.1642952 Moderate Moderate
  9. 3  7.104259 6.333049 7.1706136   Severe Moderate
  10. 4  4.199447 3.277607 8.9458447 Moderate     Mild
  11. 5  9.352140 7.135801 2.6721405   Severe   Severe
  12. 6  7.682951 1.358830 4.2102313   Severe     Mild
  13. 7  6.551999 9.986188 1.9995422   Severe   Severe
  14. 8  7.436272 9.260056 0.1093833   Severe   Severe
  15. 9  5.163593 7.689474 0.2999034 Moderate   Severe
  16. 10 7.500994 4.599129 8.5266752   Severe Moderate

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  • 本文由 发表于 2023年6月15日 06:01:45
  • 转载请务必保留本文链接：https://go.coder-hub.com/76477861.html
  • dataframe
  • lapply
  • r
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