问题

I've translated the code for you. Here's the translated code:

我试图找出三个时期内物种丰富度的变化率（即 abundance.period2 / abundance.period1 * 100）。

我的数据框看起来像这样：

1 时期 物种 丰富度
2 1 A 10
3 2 A 2
4 3 A 4
5 1 B 5
6 2 B 20
7 3 B 0
8 1 C 1

我需要使用循环，因为我有太多的物种要处理：

```R
value.final <- data.frame() # 空数据框用于存储循环结果

reps <- length(unique(df$species)) # 每个物种的重复次数

for(i in 1:reps) {
  # 按物种子集
  subs <- df[df$species == i, ]
  subs$species <- factor(subs$species)
  
  # 时期1
  per.1 <- subs[subs$period == "1",]
  
  # 时期2
  per.2 <- subs[subs$period == "2",]
  
  # 时期3
  per.3 <- subs[subs$period == "3",]
  
  # 值计算
  per2.per1 <- per.2$abundance / per.1$abundance * 100
  per3.per2 <- per.3$abundance / per.2$abundance * 100
  
  # 保存到数据框
  values <- data.frame(species = rep(subs$species, length.out = 2), 
             period = c("per2.per1", "per3.per2"),
             value = c(per2.per1, per3.per2))
  
  # 绑定以保存
  value.final <- rbind(values, value.final)
}

我不断收到以下错误：
Error in data.frame(species = rep(subs$species, length.out = 2), period = c("per2.per1", : arguments imply differing number of rows: 2, 0

然而，当我尝试逐个物种地进行操作而不使用循环时，一切正常：

subs <- df[df$species == "A", ]
subs$species <- factor(subs$species)

per.1 <- subs[subs$period == "1",]
per.2 <- subs[subs$period == "2",]
per.3 <- subs[subs$period == "3",]

per2.per1 <- per.2$abundance / per.1$abundance * 100
per3.per2 <- per.3$abundance / per.2$abundance * 100

data.frame(species = rep(subs$species, length.out = 2), 
             period = c("per2.per1", "per3.per2"),
             value = c(per2.per1, per3.per2))

我不知道如何修复这个错误，因为我不明白为什么会出现这个错误。非常感谢任何帮助。

Please note that I've translated the code, but the issue you're facing is related to the R programming logic, and it would require debugging in the original code. If you have any questions or need further assistance with understanding the error, feel free to ask.

<details>
<summary>英文:</summary>

I am trying to find the rate of change in species abundance over 3 periods (i.e. abundance.period2 / abundance.period1 * 100).

My dataframe looks like this: 

1 Period species Abundance
2 1 A 10
3 2 A 2
4 3 A 4
5 1 B 5
6 2 B 20
7 3 B 0
8 1 C 1

I need to use a loop because I have too many species to do each individually:

value.final <- data.frame() # Empty data frame to fill with loop results

reps <- length(unique(df$species)) # repetition per species

for(i in 1:reps) {

Subset by species

subs <- df[df$species == i, ]
subs$species <- factor(subs$species)

Period 1

per.1 <- subs[subs$period == "1",]

Period 2

per.2 <- subs[subs$period == "2",]

Period 3

per.3 <- subs[subs$period == "3",]

Value calculation

per2.per1 <- per2$abundance/per1$abundance * 100
per3.per2 <- per3$abundance/per2$abundance * 100

Save in dataframe

values <- data.frame(species = rep(subs$species, length.out = 2),
period = c("per2.per1", "per3.per2"),
value = c(per2.per1, per3.per2))

Bind to save

value.final <- rbind(values, value.final)
}

I keep receiving the folloing error:
**Error in data.frame(species = rep(subs$species, length.out = 2), period = c("per2.per1",  : arguments imply differing number of rows: 2, 0**

However, when I try doing it species by species without using the loop, it works fine:

subs <- df[df$species == "A", ]
subs$species <- factor(subs$species)

per.1 <- subs[subs$period == "1",]
per.2 <- subs[subs$period == "2",]
per.3 <- subs[subs$period == "3",]

per2.per1 <- per2$abundance/per1$abundance * 100
per3.per2 <- per3$abundance/per2$abundance * 100

data.frame(species = rep(subs$species, length.out = 2),
period = c("per2.per1", "per3.per2"),
value = c(per2.per1, per3.per2))

I don't know how to fix the error because I don't understand why I am receiving it. Any help is greatly appreciated.

</details>


# 答案1
**得分**: 1

这是一种可能的方法来解决这个问题...

```R
# 创建你提供的数据框
df <- data.frame(Period = c(1, 2, 3, 1, 2, 3, 1), 
                 Species = c(rep("A", 3), rep("B", 3), rep("C", 1)),
                 Abundance = c(10, 2, 4, 5, 20, 0, 1))

# 按照物种拆分成列表元素
specList <- split(df, df$Species)

# 使用sapply返回差异的列表
result <- sapply(specList, 
                 function(x) {
                   100*(exp(diff(log(x$Abundance))) - 1)
                 })

# 将结果重新组合成数据框并打印出来
print(as.data.frame(result[sapply(result, length) == 2]))

我假设你想计算每个期间之间的百分增长（类似于：100*(a1/a0 - 1) ），我使用了diff()函数以及log()和exp()来完成除法。你可以使用你感觉最舒服的方法。

希望这能帮助到你。

# build the data.frame you presented
df <- data.frame(Period = c(1, 2, 3, 1, 2, 3, 1), 
                 Species = c(rep("A", 3), rep("B", 3), rep("C", 1)),
                 Abundance = c(10, 2, 4, 5, 20, 0, 1))

# split it by Species into list elements 
specList <- split(df, df$Species)

# use sapply to return a list of the differences
result <- sapply(specList, 
                 function(x) {
                   100*(exp(diff(log(x$Abundance))) - 1)
                 })

# reform the results into a data.frame and print it
print(as.data.frame(result[sapply(result, length) == 2]))

答案2

得分: 1

以下是代码部分的翻译：

library(tidyverse)

df <- tibble(Period = c(1, 2, 3, 1, 2, 3, 1), 
             Species = c(rep("A", 3), rep("B", 3), rep("C", 1)),
             Abundance = c(10, 2, 4, 5, 20, 0, 1))

df %>%
  group_by(Species) %>%
  reframe(per2.per1 = (Abundance[Period == 2]/ Abundance[Period == 1])*100,
          per3.per2 = (Abundance[Period == 3]/ Abundance[Period == 2])*100) %>%
  pivot_longer(-Species, names_to = "period")
#> # A tibble: 4 x 3
#>   Species period    value
#>   <chr>   <chr>     <dbl>
#> 1 A       per2.per1    20
#> 2 A       per3.per2   200
#> 3 B       per2.per1   400
#> 4 B       per3.per2     0

希望这有帮助。

library(tidyverse)

df <- tibble(Period = c(1, 2, 3, 1, 2, 3, 1), 
             Species = c(rep("A", 3), rep("B", 3), rep("C", 1)),
             Abundance = c(10, 2, 4, 5, 20, 0, 1))

df |>
  group_by(Species)|>
  reframe(per2.per1 = (Abundance[Period == 2]/ Abundance[Period == 1])*100,
          per3.per2 = (Abundance[Period == 3]/ Abundance[Period == 2])*100) |>
  pivot_longer(-Species, names_to = "period")
#> # A tibble: 4 x 3
#>   Species period    value
#>   <chr>   <chr>     <dbl>
#> 1 A       per2.per1    20
#> 2 A       per3.per2   200
#> 3 B       per2.per1   400
#> 4 B       per3.per2     0