我有一個具有以下基本結構的巨大資料框：
data <- data.frame(species = factor(c(rep("species1", 4), rep("species2", 4), rep("species3", 4))),
trap = c(rep(c("A","B","C","D"), 3)),
count=c(6,3,7,9,5,3,6,6,5,8,1,3))
data
我希望同時對每個單獨物種的四個陷阱之間的物種計數資料進行卡方檢驗，但不是在它們之間。對于每個單獨的物種，可以使用以下代碼來解決它，但由于我的原始資料框很大，這對我來說不是一個合適的解決方案。
chi_species1 <- xtabs(count~trap, data,
subset = species=="species1")
chi_species1
chisq.test(chi_species1)
謝謝你的幫助！！
uj5u.com熱心網友回復：
根據
df <- data.frame(species = factor(c(rep("species1", 4), rep("species2", 4), rep("species3", 4))),
trap = c(rep(c("A","B","C","D"), 3)),
count=c(6,3,7,9,5,3,6,6,5,8,1,3))
df
#> species trap count
#> 1 species1 A 6
#> 2 species1 B 3
#> 3 species1 C 7
#> 4 species1 D 9
#> 5 species2 A 5
#> 6 species2 B 3
#> 7 species2 C 6
#> 8 species2 D 6
#> 9 species3 A 5
#> 10 species3 B 8
#> 11 species3 C 1
#> 12 species3 D 3
species <- unique(df$species)
chi_species <- lapply(species, function(x) xtabs(count~trap, df,
subset = species== x))
chi_species <- setNames(chi_species, species)
lapply(chi_species, chisq.test)
#> $species1
#>
#> Chi-squared test for given probabilities
#>
#> data: X[[i]]
#> X-squared = 3, df = 3, p-value = 0.3916
#>
#>
#> $species2
#>
#> Chi-squared test for given probabilities
#>
#> data: X[[i]]
#> X-squared = 1.2, df = 3, p-value = 0.753
#>
#>
#> $species3
#>
#> Chi-squared test for given probabilities
#>
#> data: X[[i]]
#> X-squared = 6.2941, df = 3, p-value = 0.09815
由reprex 包于 2022-04-25 創建(v2.0.1)
tidyverse
df %>%
group_by(species, trap) %>%
summarise(count = sum(count)) %>%
summarise(pvalue= chisq.test(count)$p.value)
# A tibble: 3 × 2
species pvalue
<fct> <dbl>
1 species1 0.392
2 species2 0.753
3 species3 0.0981
uj5u.com熱心網友回復：
你想要這樣的東西：
library(dplyr)
data %>%
group_by(species) %>%
summarise(pvalue= chisq.test(count, trap)$p.value)
輸出：
# A tibble: 3 × 2
species pvalue
<fct> <dbl>
1 species1 0.213
2 species2 0.238
3 species3 0.213
轉載請註明出處，本文鏈接：https://www.uj5u.com/gongcheng/464888.html