英文:
Optimize data for t.test to avoid "data are essentially constant" error
问题
有关在R中使用t.test()时出现错误消息“数据基本上是恒定的”的情况，有几个StackOverflow帖子讨论了这个问题，这是因为组之间没有足够的差异（没有变化）来运行t.test()。（如果有其他原因，请纠正我）
我现在处于这种情况，我想通过改变我的数据以使数据的统计特性不发生 drastical 改变来解决这个问题，这样 t-检验结果仍然正确。我想知道如果我向数据中添加一些非常小的变化（例如将 0.301029995663981 更改为 0.301029995663990），或者还能做些什么？
例如，这是我的数据：
# 创建数据框data <- data.frame(Date = c("2021.08","2021.08","2021.09","2021.09","2021.09","2021.10","2021.10","2021.10","2021.11","2021.11","2021.11","2021.11","2021.11","2021.12","2021.12","2022.01","2022.01","2022.01","2022.01","2022.08","2022.08","2022.08","2022.08","2022.08","2022.09","2022.09","2022.10","2022.10","2022.10","2022.11","2022.11","2022.11","2022.11","2022.11","2022.12","2022.12","2022.12","2022.12","2023.01","2023.01","2023.01","2023.01"),Species = c("A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A"),Site = c("Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something"),Mean = c("0.301029995663981","1.07918124604762","0.698970004336019","1.23044892137827","1.53147891704226","1.41497334797082","1.7160033436348","0.698970004336019","1.39794000867204","1","0.301029995663981","0.301029995663981","0.477121254719662","0.301029995663981","0.301029995663981","0.301029995663981","0.477121254719662","0.301029995663981","0.301029995663981","0.845098040014257","0.301029995663981","0.301029995663981","0.477121254719662","0.698970004336019","1.23044892137827","1.41497334797082","1.95904139232109","1.5910646070265","1.53147891704226","1.14612803567824","1.57978359661681","1.34242268082221","0.778151250383644","0.301029995663981","0.301029995663981","0.477121254719662","0.301029995663981","1.20411998265592","0.845098040014257","1.17609125905568","1.20411998265592","0.698970004336019","0.301029995663981","0.698970004336019","0.698970004336019","0.903089986991944","1.14612803567824","0.301029995663981","0.602059991327962","0.301029995663981","0.845098040014257","0.698970004336019","0.698970004336019","0.301029995663981","0.698970004336019","0.301029995663981","0.301029995663981","0.301029995663981","0.301029995663981","0.301029995663981","0.602059991327962","0.301029995663981","0.845098040014257","1.92941892571429","1.27875360095283","0.698970004336019","1.38021124171161","1.20411998265592","1.38021124171161","1.14612803567824","1","1.07918124604762","1.17609125905568","0.845098040014257","0.698970004336019","0.778151250383644","0.301029995663981","0.845098040014257","1.64345267648619","1.46239799789896","1.34242268082221","1.34242268082221","0.778151250383644"))
然后，我设置了因子：
# 设置因子str(data)data$Date<-as.factor(data$Date)data$Site<-as.factor(data$Site)data$Species<-as.factor(data$Species)data$Mean<-as.numeric(data$Mean)str(data)
当我尝试使用t.test()时：
compare_means(Mean ~ Species, data = data, group.b = "Date", method = "t.test")
会出现以下错误：
错误 in `mutate()`:ℹ In argument: `p = purrr::map(...)`.Caused by error in `purrr::map()`:ℹ In index: 5.ℹ With name: Date.2021.12.Caused by error in `t.test.default()`:! data are essentially constantRun `rlang::last_trace()` to see where the error occurred.
类似地，在ggplot中使用这个：
<details><summary>英文:</summary>There are several StackOverflow posts about situation where t.test() in R produce an error saying "data are essentially constant", this is due to that there is not enough difference between the groups (there is no variation) to run the t.test(). (Correct me if there is something else)I'm in this situation, and I would like to fix this buy altering my data the way the statistical features of the data don't change drastically, so the t-test stays correct. I was wondering what if I add some very little variation to the data (e.g. change 0.301029995663981 to 0.301029995663990), or what else can I do?For example, this is my data:# Create the data framedata <- data.frame(Date = c("2021.08","2021.08","2021.09","2021.09","2021.09","2021.10","2021.10","2021.10","2021.11","2021.11","2021.11","2021.11","2021.11","2021.12","2021.12","2022.01","2022.01","2022.01","2022.01","2022.08","2022.08","2022.08","2022.08","2022.08","2022.09","2022.09","2022.10","2022.10","2022.10","2022.11","2022.11","2022.11","2022.11","2022.11","2022.12","2022.12","2022.12","2022.12","2023.01","2023.01","2023.01","2023.01","2021.08","2021.08","2021.09","2021.09","2021.09","2021.10","2021.10","2021.10","2021.11","2021.11","2021.11","2021.11","2021.11","2021.12","2021.12","2022.01","2022.01","2022.01","2022.01","2022.08","2022.08","2022.08","2022.08","2022.08","2022.09","2022.09","2022.09","2022.09","2022.10","2022.10","2022.10","2022.10","2022.11","2022.11","2022.11","2022.11","2022.11","2022.12","2022.12","2022.12","2022.12","2023.01","2023.01","2023.01","2023.01"),Species = c("A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B"),Site = c("Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something"),Mean = c("0.301029995663981","1.07918124604762","0.698970004336019","1.23044892137827","1.53147891704226","1.41497334797082","1.7160033436348","0.698970004336019","1.39794000867204","1","0.301029995663981","0.301029995663981","0.477121254719662","0.301029995663981","0.301029995663981","0.301029995663981","0.477121254719662","0.301029995663981","0.301029995663981","0.845098040014257","0.301029995663981","0.301029995663981","0.477121254719662","0.698970004336019","1.23044892137827","1.41497334797082","1.95904139232109","1.5910646070265","1.53147891704226","1.14612803567824","1.57978359661681","1.34242268082221","0.778151250383644","0.301029995663981","0.301029995663981","0.477121254719662","0.301029995663981","1.20411998265592","0.845098040014257","1.17609125905568","1.20411998265592","0.698970004336019","0.301029995663981","0.698970004336019","0.698970004336019","0.903089986991944","1.14612803567824","0.301029995663981","0.602059991327962","0.301029995663981","0.845098040014257","0.698970004336019","0.698970004336019","0.301029995663981","0.698970004336019","0.301029995663981","0.301029995663981","0.301029995663981","0.477121254719662","0.301029995663981","0.301029995663981","0.301029995663981","0.301029995663981","0.301029995663981","0.602059991327962","0.301029995663981","0.845098040014257","1.92941892571429","1.27875360095283","0.698970004336019","1.38021124171161","1.20411998265592","1.38021124171161","1.14612803567824","1","1.07918124604762","1.17609125905568","0.845098040014257","0.698970004336019","0.778151250383644","0.301029995663981","0.845098040014257","1.64345267648619","1.46239799789896","1.34242268082221","1.34242268082221","0.778151250383644"))After, I set the factors:# Set factorsstr(data)data$Date<-as.factor(data$Date)data$Site<-as.factor(data$Site)data$Species<-as.factor(data$Species)data$Mean<-as.numeric(data$Mean)str(data)When I try t.test():compare_means(Mean ~ Species, data = data, group.b = "Date", method = "t.test")This is the error:Error in `mutate()`:ℹ In argument: `p = purrr::map(...)`.Caused by error in `purrr::map()`:ℹ In index: 5.ℹ With name: Date.2021.12.Caused by error in `t.test.default()`:! data are essentially constantRun `rlang::last_trace()` to see where the error occurred.Similarly, when I use this in ggplot:ggplot(data, aes(x = Date, y = Mean, fill=Species)) +geom_boxplot()+stat_compare_means(data=data,method="t.test", label = "p.signif") +theme(axis.text.x = element_text(angle = 90, vjust = 0.5, hjust=1))Warning message:Computation failed in `stat_compare_means()`Caused by error in `mutate()`:ℹ In argument: `p = purrr::map(...)`.Caused by error in `purrr::map()`:ℹ In index: 5.ℹ With name: x.5.Caused by error in `t.test.default()`:! data are essentially constantWhat is the best solution, which keeps the data still usable in t-test?</details># 答案1**得分**: 1针对每个日期-物种组合找到`Mean`的标准差，然后筛选掉任何标准差为0的日期，这样就可以实现。您甚至可以将筛选后的数据传递给`compare_means()`函数：``` rlibrary(dplyr)library(ggpubr)data <- data.frame(Date = c("2021.08","2021.08","2021.09","2021.09","2021.09","2021.10","2021.10","2021.10","2021.11","2021.11","2021.11","2021.11","2021.11","2021.12","2021.12","2022.01","2022.01","2022.01","2022.01","2022.08","2022.08","2022.08","2022.08","2022.08","2022.09","2022.09","2022.10","2022.10","2022.10","2022.11","2022.11","2022.11","2022.11","2022.11","2022.12","2022.12","2022.12","2022.12","2023.01","2023.01","2023.01","2023.01"),Species = c("A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B"),Site = c("Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something"),Mean = c("0.301029995663981","1.07918124604762","0.698970004336019","1.23044892137827","1.53147891704226","1.41497334797082","1.7160033436348","0.698970004336019","1.39794000867204","1","0.301029995663981","0.301029995663981","0.477121254719662","0.301029995663981","0.301029995663981","0.301029995663981","0.477121254719662","0.301029995663981","0.301029995663981","0.845098040014257","0.301029995663981","0.301029995663981","0.477121254719662","0.698970004336019","1.23044892137827","1.41497334797082","1.95904139232109","1.5910646070265","1.53147891704226","1.14612803567824","1.57978359661681","1.34242268082221","0.778151250383644","0.301029995663981","0.301029995663981","0.477121254719662","0.301029995663981","1.20411998265592","0.845098040014257","1.17609125905568","1.20411998265592","0.698970004336019","0.301029995663981","0.698970004336019","0.698970004336019","0.903089986991944","1.14612803567824","0.301029995663981","0.602059991327962","0.301029995663981","0.845098040014257","0.698970004336019","0.698970004336019","0.301029995663981","0.698970004336019","0.301029995663981","0.301029995663981","0.301029995663981","0.477121254719662","0.301029995663981","0.301029995663981","0.301029995663981","0.301029995663981","0.301029995663981","0.602059991327962","0.301029995663981","0.845098040014257","1.92941892571429","1.27875360095283","0.698970004336019","1.38021124171161","1.20411998265592","1.38021124171161","1.14612803567824","1","1.07918124604762","1.17609125905568","0.845098040014257","0.698970004336019","0.778151250383644","0.301029995663981","0.845098040014257","1.64345267648619","1.46239799789896","1.34242268082221","1.34242268082221","0.778151250383644"))data$Date <- as.factor(data$Date)data$Site <- as.factor(data$Site)data$Species <- as.factor(data$Species)data$Mean <- as.numeric(data$Mean)data %>%group_by(Date, Species) %>%mutate(s = sd(Mean)) %>%group_by(Date) %>%filter(!any(s == 0)) %>%compare_means(Mean ~ Species, data = ., group.b = "Date", method = "t.test")#> # A tibble: 11 × 9#> Date .y. group1 group2 p p.adj p.format p.signif method#> <fct> <chr> <chr> <chr> <dbl> <dbl> <chr> <chr> <chr>#> 1 2021.08 Mean A B 0.718 1 0.718 ns T-test#> 2 2021.09 Mean A B 0.451 1 0.451 ns T-test#> 3 2021.10 Mean A B 0.0889 0.89 0.089 ns T-test#> 4 2021.11 Mean A B 0.850 1 0.850 ns T-test#> 5 2022.01 Mean A B<details><summary>英文:</summary>Finding the sd of `Mean` for each Date-Species combination and then filtering out any Dates where any sd is 0 will do the trick. You could even just pipe the filtered data to `compare_means()`:``` rlibrary(dplyr)library(ggpubr)data <- data.frame(Date = c("2021.08","2021.08","2021.09","2021.09","2021.09","2021.10","2021.10","2021.10","2021.11","2021.11","2021.11","2021.11","2021.11","2021.12","2021.12","2022.01","2022.01","2022.01","2022.01","2022.08","2022.08","2022.08","2022.08","2022.08","2022.09","2022.09","2022.10","2022.10","2022.10","2022.11","2022.11","2022.11","2022.11","2022.11","2022.12","2022.12","2022.12","2022.12","2023.01","2023.01","2023.01","2023.01","2021.08","2021.08","2021.09","2021.09","2021.09","2021.10","2021.10","2021.10","2021.11","2021.11","2021.11","2021.11","2021.11","2021.12","2021.12","2022.01","2022.01","2022.01","2022.01","2022.08","2022.08","2022.08","2022.08","2022.08","2022.09","2022.09","2022.09","2022.09","2022.10","2022.10","2022.10","2022.10","2022.11","2022.11","2022.11","2022.11","2022.11","2022.12","2022.12","2022.12","2022.12","2023.01","2023.01","2023.01","2023.01"),Species = c("A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B"),Site = c("Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something","Something"),Mean = c("0.301029995663981","1.07918124604762","0.698970004336019","1.23044892137827","1.53147891704226","1.41497334797082","1.7160033436348","0.698970004336019","1.39794000867204","1","0.301029995663981","0.301029995663981","0.477121254719662","0.301029995663981","0.301029995663981","0.301029995663981","0.477121254719662","0.301029995663981","0.301029995663981","0.845098040014257","0.301029995663981","0.301029995663981","0.477121254719662","0.698970004336019","1.23044892137827","1.41497334797082","1.95904139232109","1.5910646070265","1.53147891704226","1.14612803567824","1.57978359661681","1.34242268082221","0.778151250383644","0.301029995663981","0.301029995663981","0.477121254719662","0.301029995663981","1.20411998265592","0.845098040014257","1.17609125905568","1.20411998265592","0.698970004336019","0.301029995663981","0.698970004336019","0.698970004336019","0.903089986991944","1.14612803567824","0.301029995663981","0.602059991327962","0.301029995663981","0.845098040014257","0.698970004336019","0.698970004336019","0.301029995663981","0.698970004336019","0.301029995663981","0.301029995663981","0.301029995663981","0.477121254719662","0.301029995663981","0.301029995663981","0.301029995663981","0.301029995663981","0.301029995663981","0.602059991327962","0.301029995663981","0.845098040014257","1.92941892571429","1.27875360095283","0.698970004336019","1.38021124171161","1.20411998265592","1.38021124171161","1.14612803567824","1","1.07918124604762","1.17609125905568","0.845098040014257","0.698970004336019","0.778151250383644","0.301029995663981","0.845098040014257","1.64345267648619","1.46239799789896","1.34242268082221","1.34242268082221","0.778151250383644"))
data$Date<-as.factor(data$Date)data$Site<-as.factor(data$Site)data$Species<-as.factor(data$Species)data$Mean<-as.numeric(data$Mean)data %>%group_by(Date, Species) %>%mutate(s = sd(Mean)) %>%group_by(Date) %>%filter(!any(s == 0)) %>%compare_means(Mean ~ Species, data = ., group.b = "Date", method = "t.test")#> # A tibble: 11 × 9#> Date .y. group1 group2 p p.adj p.format p.signif method#> <fct> <chr> <chr> <chr> <dbl> <dbl> <chr> <chr> <chr>#> 1 2021.08 Mean A B 0.718 1 0.718 ns T-test#> 2 2021.09 Mean A B 0.451 1 0.451 ns T-test#> 3 2021.10 Mean A B 0.0889 0.89 0.089 ns T-test#> 4 2021.11 Mean A B 0.850 1 0.850 ns T-test#> 5 2022.01 Mean A B 1 1 1.000 ns T-test#> 6 2022.08 Mean A B 0.234 1 0.234 ns T-test#> 7 2022.09 Mean A B 0.670 1 0.670 ns T-test#> 8 2022.10 Mean A B 0.0707 0.78 0.071 ns T-test#> 9 2022.11 Mean A B 0.783 1 0.783 ns T-test#> 10 2022.12 Mean A B 0.399 1 0.399 ns T-test#> 11 2023.01 Mean A B 0.255 1 0.255 ns T-test
<sup>Created on 2023-06-01 with reprex v2.0.2</sup>
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