Question

问题：

我有多个站点，每个站点有10个采样点。

Site Time Sample Species1 Species2 Species3 etc
Home    A      1        1        0        4 ...
Home    A      2        0        0        2 ...
Work    A      1        0        1        1 ...
Work    A      2        1        0        1 ...
Home    B      1        1        0        4 ...
Home    B      2        0        0        2 ...
Work    B      1        0        1        1 ...
Work    B      2        1        0        1 ...
...

我想获得每个站点的丰富性和丰富性。丰富度是站点中物种的总数，而丰度是站点中所有物种的所有个体的总数，例如：

Site Time Richness Abundance
Home    A        2         7
Work    A        3         4
Home    B        2         7
Work    B        3         4

我可以通过以下两个功能到达那里。但是，我希望两者都在一个dplyr函数中。范围7:34是指我的物种矩阵（每行一个站点/一个样本，物种为一列）。

df1 <- df %>% mutate(Abundance = rowSums(.[,4:30])) %>%
group_by(Site,Time) %>%   
    summarise_all(sum)

df1$Richness <- apply(df1[,4:30]>0, 1, sum)

如果我尝试同时执行一项功能，则会收到以下错误消息

df1 <- df  %>% mutate(Abundance = rowSums(.[,4:30]) ) %>%
   group_by(Site, Time) %>%   
   summarise_all(sum) %>% 
   mutate(Richness = apply(.[,4:30]>0, 1, sum))

Error in mutate_impl(.data, dots) : 
  Column `Richness` must be length 5 (the group size) or one, not 19

“丰富度”部分必须位于汇总功能之后，因为它必须对汇总和分组的数据进行运算。

如何使此功能起作用？

（注意：先前已将此问题标记为该问题的副本： Manipulating seperated species quantity data into a species abundance matrix

但是，这是一个完全不同的问题-这个问题本质上是关于转置数据集并在单个物种/列中求和。这是关于跨列（多列）对 all 种进行求和。另外，我实际上认为该问题的答案非常有帮助-像我这样的生态学家一直都在计算丰富度和丰富度，我相信他们会喜欢这个专门的问题。

Answer 1

在summarise之后，我们需要ungroup

library(tidyverse)
df %>% 
  mutate(Abundance = rowSums(.[4:ncol(.)])) %>% 
  group_by(Site, Time) %>% 
  summarise_all(sum) %>%
  ungroup %>% 
  mutate(Richness = apply(.[4:(ncol(.)-1)] > 0, 1, sum)) %>%
  #or
  #mutate(Richness = rowSums(.[4:(ncol(.)-1)] > 0)) %>%
  select(Site, Time, Abundance, Richness)
# A tibble: 4 x 4
#  Site  Time  Abundance Richness
#  <chr> <chr>     <dbl>    <int>
#1 Home  A             7        2
#2 Home  B             7        2
#3 Work  A             4        3
#4 Work  B             4        3

也可以先写group_by sum然后写transmute

来写

df %>% 
  group_by(Site, Time) %>%
  summarise_at(vars(matches("Species")), sum)  %>% 
  ungroup %>%
  transmute(Site, Time, Abundance = rowSums(.[3:ncol(.)]), 
                        Richness =  rowSums(.[3:ncol(.)] > 0))

或者另一个选择是sum和map

df %>% 
   group_by(Site, Time) %>%
   summarise_at(vars(matches("Species")), sum) %>% 
   group_by(Time, add = TRUE) %>%
   nest %>% 
   mutate(data = map(data, ~ 
                 tibble(Richness = sum(.x > 0), 
                        Abundance = sum(.x)))) %>% 
   unnest

数据

df <- structure(list(Site = c("Home", "Home", "Work", "Work", "Home", 
"Home", "Work", "Work"), Time = c("A", "A", "A", "A", "B", "B", 
"B", "B"), Sample = c(1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L), Species1 = c(1L, 
0L, 0L, 1L, 1L, 0L, 0L, 1L), Species2 = c(0L, 0L, 1L, 0L, 0L, 
0L, 1L, 0L), Species3 = c(4L, 2L, 1L, 1L, 4L, 2L, 1L, 1L)), 
class = "data.frame", row.names = c(NA, 
 -8L))

如何使用dplyr获取具有多个样本的站点的物种丰富度和丰度

1 个答案:

数据