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 [英] How to obtain species richness and abundance for sites with multiple samples using dplyr
问题描述
问题:
我有多个站点，每个站点有10个采样点.
I have a number of sites, with 10 sampling points at each site.
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.丰富度是一个站点中物种的总数，而丰度是该站点中所有物种的所有个体的总数，例如:
I would like to obtain the richness and abundance of each site. Richness is the total number of species at a site, and abundance is the total number of all individuals of all species at a site, like this:
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).
I can get there with two functions (below). However, I would like both in one dplyr function. The range 7:34 refers to my species matrix (each row a site/sample, species as columns).
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
If I try to do both in one function, I get the following error
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.
The Richness part has to come after the summarise function, since it has to operate on summed and grouped data.
如何使此功能起作用?
(注意:先前已将其标记为该问题的重复项: 将分离的物种数量数据处理为物种丰度矩阵
(Note: This was previously marked as a duplicate of this question: 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.
It is a completely different question, however - that question is essentially about transposing a dataset and summing within a single species/column. This is about summing all species across columns (multiple columns). In addition, I actually think the answer to this question is very helpful - ecologists like me calculate richness and abundance all the time, and I'm sure they'll appreciate a dedicated question.)
推荐答案
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
也可以通过先执行group_by sum然后执行transmute
It can also be written by first doing the group_by sum and then transmute
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))
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