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如何利用dplyr获得多样本点的物种丰富度和丰富度

summarize mutate grouping dplyr r

2

Grubbmeister · 技术社区 · 6 年前

问题:

我有许多地点,每个地点有10个取样点。

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 ...
...

我想获得每个网站的丰富和丰富。丰富度是一个地点的物种总数,丰富度是一个地点所有物种的所有个体总数,如下所示:

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

7:34

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

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

如果我尝试在一个函数中同时执行这两个操作,则会出现以下错误

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

如何使此功能正常工作?

(注:此问题以前标记为此问题的副本: Manipulating seperated species quantity data into a species abundance matrix

全部的 跨柱物种(多柱)。此外,我认为这个问题的答案非常有用——像我这样的生态学家一直在计算丰富度和富足度,我相信他们会喜欢一个专门的问题。)

1 回复 | 直到 6 年前

1

2

akrun 6 年前

之后 summarise 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

也可以先做 group_by sum 然后 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 总和 具有 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))