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：
df <- data.frame(
L = c(13,15,19,NA,NA,32,35,NA,NA,18,15),
W = c(NA,NA,50, NA,NA,NA,80,NA,NA,30,NA)
)
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。
lwr <- function(data, length, weight, a, b) {
if(!is.na(data$length) && is.na(data$weight)) {
data$weight = 10^(log(a) + b*log(data$length))
} else {
data$weight
}
}
在这里使用函数：
lwr(data=df, length = L, weight = W, a=0.003, b=3.2)
但它似乎不起作用。
如果你能帮助我，我将不胜感激。非常感谢您的时间。
英文:
I have Length and Weight values in a data frame. However some of them are missing. The data frame is like:
df <- data.frame(
L = c(13,15,19,NA,NA,32,35,NA,NA,18,15),
W = c(NA,NA,50, NA,NA,NA,80,NA,NA,30,NA)
)
I need a function which will work when length is not NA and weight is NA. it will calculate the weight for length, and else it will do nothing.
lwr <- function(data, length, weight, a, b) {
if(!is.na(data$length) && is.na(data$weight)) {
data$weight = 10^(log(a) + b*log(data$length))
} else {
data$weight
}
}
Here we go
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...。
lwr <- function(length, weight, a, b) {
ifelse(is.na(weight), 10^(log(a) + b*log(length)), weight)
}
df %>%
transform(W2 = lwr(L, W, a=0.003, b=3.2))
# 等价于：
# df$W2 <- lwr(df$L, df$W, a=0.003, b=3.2)
# L W W2
# 1 13 NA 250.4350
# 2 15 NA 718.8159
# 3 19 50 50.0000
# 4 NA NA NA
# 5 NA NA NA
# 6 32 NA 191078.5331
# 7 35 80 80.0000
# 8 NA NA NA
# 9 NA NA NA
# 10 18 30 30.0000
# 11 15 NA 718.8159
英文:
You should use the vectorized ifelse(), instead of if()...else....
lwr <- function(length, weight, a, b) {
ifelse(is.na(weight), 10^(log(a) + b*log(length)), weight)
}
df |>
transform(W2 = lwr(L, W, a=0.003, b=3.2))
# equivalent:
# df$W2 <- lwr(df$L, df$W, a=0.003, b=3.2)
# L W W2
# 1 13 NA 250.4350
# 2 15 NA 718.8159
# 3 19 50 50.0000
# 4 NA NA NA
# 5 NA NA NA
# 6 32 NA 191078.5331
# 7 35 80 80.0000
# 8 NA NA NA
# 9 NA NA NA
# 10 18 30 30.0000
# 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。
f <- \(x, a, b) 10^(log(a) + b*log(x))
naw <- is.na(df$W)
df$W[naw] <- f(df$L[naw], .003, 3.2)
# L W
# 1 13 250.4350
# 2 15 718.8159
# 3 19 50.0000
# 4 NA NA
# 5 NA NA
# 6 32 191078.5331
# 7 35 80.0000
# 8 NA NA
# 9 NA NA
# 10 18 30.0000
# 11 15 718.8159
英文:
You probably could do that easier.
f <- \(x, a, b) 10^(log(a) + b*log(x))
naw <- is.na(df$W)
df$W[naw] <- f(df$L[naw], .003, 3.2)
# L W
# 1 13 250.4350
# 2 15 718.8159
# 3 19 50.0000
# 4 NA NA
# 5 NA NA
# 6 32 191078.5331
# 7 35 80.0000
# 8 NA NA
# 9 NA NA
# 10 18 30.0000
# 11 15 718.8159
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库，让每个人都能够通过互相帮助和分享经验来进步。
评论