问题
I have a data frame similar to the one created by the code below. In this example, measurements of 5 variables are taken on are 30 individuals represented by ID. The individuals can be separated by any of three grouping variables: GroupVar1,GroupVar2,GroupVar3. For each of the grouping variables, I need to conduct an ANOVA for each of the 5 variables, and return the results of each (possibly onto a pdf or separate document?). How can I write a function, or use iteration, to handle this problem and minimize repetition in my code? What is the best way to extract and visualize the results if you have a large dataset (my real data set has several hundred individuals, and the grouping variables range in size from 6 to 30 groups)?
library(tidyverse)
GroupVar1 <- rep(c("FL", "GA", "SC", "NC", "VA", "GA"), each = 5)
GroupVar2 <- rep(c("alpha", "beta", "gamma"), each = 10)
GroupVar3 <- rep(c("Bravo", "Charlie", "Delta", "Echo"), times = c(7,8,10,5))
ID <- c("A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L", "M", "N", "O", "P", "Q", "R", "S", "T", "U", "V", "W", "X", "Y","Z", "a","b","c","d")
Var1 <- rnorm(30)
Var2 <- rnorm(30)
Var3 <- rnorm(30)
Var4 <- rnorm(30)
Var5 <- rnorm(30)
data <- tibble(GroupVar1,GroupVar2,GroupVar3,ID,Var1,Var2,Var3,Var4,Var5)
> dput(data[1:10,])
structure(list(Location = structure(c(21L, 21L, 21L, 21L, 21L,
21L, 21L, 21L, 21L, 21L), .Label = c("ALTE", "ASTR", "BREA",
"CAMN", "CFU", "COEN", "JENT", "NAT", "NEAU", "NOCO", "OOGG",
"OPMM", "PING", "PITC", "POMO", "REAN", "ROND", "RTD", "SANT",
"SMIT", "SUN", "TEAR", "WINC"), class = "factor"), PR = structure(c(16L,
16L, 16L, 16L, 16L, 16L, 16L, 16L, 16L, 16L), .Label = c("ALTE",
"ASTR", "CF", "CHOW", "JENT", "NAT", "NEAU", "NSE", "OOGG", "PALM",
"POMO", "REAN", "ROND", "RTD", "SS", "SUN", "WINC"), class = "factor"),
Est = structure(c(2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c("AS",
"CB", "CF", "CS", "OS", "PS", "SS", "WB"), class = "factor"),
State = structure(c(3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L
), .Label = c("FL", "GA", "MD", "NC", "SC", "VA"), class = "factor"),
Year = c(2017L, 2017L, 2017L, 2017L, 2017L, 2017L, 2017L,
2017L, 2017L, 2017L), ID = c(90L, 92L, 93L, 95L, 96L, 98L,
99L, 100L, 103L, 109L), Sex = structure(c(1L, 2L, 2L, 2L,
1L, 1L, 2L, 1L, 1L, 2L), .Label = c("F", "M"), class = "factor"),
DOB = c(-0.674706816, 2.10472846, 0.279952847, -0.26959379,
-1.243977657, 0.188828771, 0.026530709, 0.483363306, -0.63599302,
-0.979506001), Mg = c(-1.409815618, 1.180920604, 0.765102543,
1.828057339, -0.689841498, -0.604272366, 0.194867939, -1.015964127,
-0.520136693, 0.769042585), Mn7 = c(1.387385913, 0.320582444,
-0.490356598, -0.020540649, -0.594210249, -1.119170306, -0.225065868,
-1.892064456, -2.434101506, -0.816518662), Cu7 = c(-0.176599651,
0.100529267, 1.4967142, 0.094840221, 1.791653259, -0.191723817,
-1.526868086, -0.308696916, -2.046613977, -2.228513411),
Zn7 = c(-0.338454617, -0.235800727, -0.785876374, 0.114698826,
0.202960987, 0.432013987, 0.164099621, 0.609232311, 0.169329098,
-0.284402654), Sr7 = c(-0.010929071, -1.616835312, -0.208856,
-0.362538736, 1.662066318, -0.893155185, 0.699406559, -0.333176495,
-2.026364633, -1.324456127), Ba7 = c(-1.041126455, 0.551165907,
0.126849272, -1.069762666, -0.922501551, -1.36095076, 1.57800858,
-0.842518997, -1.017894235, 0.265895019)), row.names = c(NA,
10L), class = "data.frame")
回答1:
Edited answer based on updated question with dput data:
Assuming the columns representing the grouping variables are 1:5 and 7, and assuming the dependent numeric variables are in columns 8:14, this can be done using a double loop, with no other dependencies:
tests <- list()
Groups <- c(1:5, 7)
Variables <- 8:14
for(i in Groups)
{
Group <- as.factor(data[[i]])
for(j in Variables)
{
test_name <- paste0(names(data)[j], "_by_", names(data[i]))
Response <- data[[j]]
tests[[test_name]] <- anova(lm(Response ~ Group))
}
}
Now you can do what you like with all these tests using lapply, such as
lapply(tests, print)
I agree with @DaveGruenewald about multiple hypothesis testing though - in fact, this example gave a nice demonstration of why Buonferroni or Sidak's corrections are needed, since there were (as expected) a few "significant" p values among the random data simply due to the number of tests involved.
回答2:
Without knowing too much about the underlying data, my hunch is this may be improper use of an ANOVA. I would advise that you post to Cross Validated to confirm you aren't breaking any assumptions here.
Regardless, here is the code I would use to tackle the problem presented:
# We will use dplyr, tidyr, purrr, stats, and broom to accomplish this
# I am using tidyr v1.0.0. For older versions you will need to modify code for pivot_longer
results <- data %>%
# First pivot the data longer so each dependent variable is on its own row
pivot_longer(
cols = Var1:Var5,
names_to = "name",
values_to = "value"
) %>%
# Second, pivot longer again, so each row is now its unique grouping var
pivot_longer(
cols = GroupVar1:GroupVar3,
names_to = "group_name",
values_to = "group_value"
) %>%
# group by both group name and dependent variable
group_by(name, group_name) %>%
# nest the data, so each dataset is unique for each dependent and independent variable
nest() %>%
mutate(
# run an anova on each nested data frame
anova = map(data, ~aov(data = .x, value ~ group_value)), # may need to change aov() call here
# use broom to tidy the output
tidied_results = map(anova, broom::tidy)
)
# To easily access the ANOVA results, you can do something like the following:
results %>%
# select columns of interest
select(name, group_name, tidied_results) %>%
# unnest to access summary information of ANOVA
unnest(cols = c(tidied_results))
I think you'll also want to use some sort of multiple-comparison correction, such as Bonferroni Correction. Again, Cross Validated can lead you in the right direction here.
来源：https://stackoverflow.com/questions/59492324/how-to-conduct-an-anova-of-several-variables-taken-on-individuals-separated-by-m