R中一个函数中使用了”If”、”else”和”is.na()”。

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英文:

The usage of "If", "else", "is.na()" in one function of R

问题

我有一个数据框中的Length和Weight数值。然而，其中一些数值是缺失的。数据框如下：

  1. df <- data.frame(
  2.   L = c(13,15,19,NA,NA,32,35,NA,NA,18,15),
  3.   W = c(NA,NA,50, NA,NA,NA,80,NA,NA,30,NA)
  4. )

我需要一个函数，当长度不是NA而重量是NA时可以计算长度对应的重量，否则不执行任何操作。

  1. lwr <- function(data, length, weight, a, b) {
  2.   if(!is.na(data$length) && is.na(data$weight)) { 
  3.     data$weight = 10^(log(a) + b*log(data$length))
  4.   } else {
  5.     data$weight
  6.   }
  7. }

在这里使用函数：

  1. lwr(data=df, length = L, weight = W, a=0.003, b=3.2)

但它似乎不起作用。

如果你能帮助我，我将不胜感激。非常感谢您的时间。

英文:

I have Length and Weight values in a data frame. However some of them are missing. The data frame is like:

  1. df &lt;- data.frame(
  2.   L = c(13,15,19,NA,NA,32,35,NA,NA,18,15),
  3.   W = c(NA,NA,50, NA,NA,NA,80,NA,NA,30,NA)
  4. )

I need a function which will work when length is not NA and weight is NA. it will calculate the weight for length, and else it will do nothing.

  1. lwr &lt;- function(data, length, weight, a, b) {
  2.   if(!is.na(data$length) &amp;&amp; is.na(data$weight)) { 
  3.     data$weight = 10^(log(a) + b*log(data$length))
  4.   } else {
  5.     data$weight
  6.   }
  7. }

Here we go

  1. lwr(data=df, length = L, weight = W, a=0.003, b=3.2)

but it does not work.

If you help me, I would be appreciated. Thank you very much for your time.

答案1

得分: 1

你应该使用向量化的 ifelse()，而不是 if()...else...。

  1. lwr <- function(length, weight, a, b) {
  2.   ifelse(is.na(weight), 10^(log(a) + b*log(length)), weight)
  3. }
  5. df %>%
  6.   transform(W2 = lwr(L, W, a=0.003, b=3.2))
  8. # 等价于：
  9. # df$W2 <- lwr(df$L, df$W, a=0.003, b=3.2)
  1. #     L  W          W2
  2. # 1  13 NA    250.4350
  3. # 2  15 NA    718.8159
  4. # 3  19 50     50.0000
  5. # 4  NA NA          NA
  6. # 5  NA NA          NA
  7. # 6  32 NA 191078.5331
  8. # 7  35 80     80.0000
  9. # 8  NA NA          NA
  10. # 9  NA NA          NA
  11. # 10 18 30     30.0000
  12. # 11 15 NA    718.8159
英文:

You should use the vectorized ifelse(), instead of if()...else....

  1. lwr &lt;- function(length, weight, a, b) {
  2.   ifelse(is.na(weight), 10^(log(a) + b*log(length)), weight)
  3. }
  5. df |&gt;
  6.   transform(W2 = lwr(L, W, a=0.003, b=3.2))
  8. # equivalent:
  9. # df$W2 &lt;- lwr(df$L, df$W, a=0.003, b=3.2)
  1. #     L  W          W2
  2. # 1  13 NA    250.4350
  3. # 2  15 NA    718.8159
  4. # 3  19 50     50.0000
  5. # 4  NA NA          NA
  6. # 5  NA NA          NA
  7. # 6  32 NA 191078.5331
  8. # 7  35 80     80.0000
  9. # 8  NA NA          NA
  10. # 9  NA NA          NA
  11. # 10 18 30     30.0000
  12. # 11 15 NA    718.8159

答案2

得分: 1

你可能可以更轻松地做到这一点。

  1. f <- \(x, a, b) 10^(log(a) + b*log(x))
  3. naw <- is.na(df$W)
  5. df$W[naw] <- f(df$L[naw], .003, 3.2)
  6. #     L           W
  7. # 1  13    250.4350
  8. # 2  15    718.8159
  9. # 3  19     50.0000
  10. # 4  NA          NA
  11. # 5  NA          NA
  12. # 6  32 191078.5331
  13. # 7  35     80.0000
  14. # 8  NA          NA
  15. # 9  NA          NA
  16. # 10 18     30.0000
  17. # 11 15    718.8159
英文:

You probably could do that easier.

  1. f &lt;- \(x, a, b) 10^(log(a) + b*log(x))
  3. naw &lt;- is.na(df$W)
  5. df$W[naw] &lt;- f(df$L[naw], .003, 3.2)
  6. #     L           W
  7. # 1  13    250.4350
  8. # 2  15    718.8159
  9. # 3  19     50.0000
  10. # 4  NA          NA
  11. # 5  NA          NA
  12. # 6  32 191078.5331
  13. # 7  35     80.0000
  14. # 8  NA          NA
  15. # 9  NA          NA
  16. # 10 18     30.0000
  17. # 11 15    718.8159

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  • 本文由 发表于 2023年2月18日 16:00:44
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  • function
  • if-statement
  • r
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