让我们从一个数据开始:
structure(list(Group = c("Mark", "Matt", "Tim", "Tom"), `1` = c(0.749552072382562,
1.06820497349356, 1.00116263663573, 0.864987635002866), `2` = c(1.00839505250436,
0.796306651704629, 1.02603677593328, 1.00321936833133), `3` = c(0.736638669191169,
0.973483626272054, 1.14805519301778, 0.899272693725192), `4` = c(0.728882841159455,
0.871211836418332, 1.0442119745299, 0.859935708928745), `5` = c(0.749552072382562,
1.06820497349356, 1.00116263663573, 0.864987635002866), `6` = c(1.00839505250436,
0.796306651704629, 1.02603677593328, 1.00321936833133), `7` = c(0.736638669191169,
0.973483626272054, 1.14805519301778, 0.899272693725192), `8` = c(0.728882841159455,
0.871211836418332, 1.0442119745299, 0.859935708928745), `9` = c(0.749552072382562,
1.06820497349356, 1.00116263663573, 0.864987635002866), `10` = c(1.00839505250436,
0.796306651704629, 1.02603677593328, 1.00321936833133), `11` = c(0.736638669191169,
0.973483626272054, 1.14805519301778, 0.899272693725192), `12` = c(0.728882841159455,
0.871211836418332, 1.0442119745299, 0.859935708928745), `13` = c(0.749552072382562,
1.06820497349356, 1.00116263663573, 0.864987635002866), `14` = c(1.00839505250436,
0.796306651704629, 1.02603677593328, 1.00321936833133), `15` = c(0.736638669191169,
0.973483626272054, 1.14805519301778, 0.899272693725192), `16` = c(0.728882841159455,
0.871211836418332, 1.0442119745299, 0.859935708928745), `17` = c(0.766036811789943,
0.871085862829362, 1.02210371210681, 0.937452345474458), `18` = c(1.0357237385154,
1.02805558505417, 0.946794300033338, 1.04688545274238), `19` = c(0.763210436944137,
0.801397021884422, 0.952553568039278, 0.990226493248718), `20` = c(0.789338028300063,
0.822815644347233, 0.958462750269733, 1.04183361434861), `21` = c(0.766036811789943,
0.871085862829362, 1.02210371210681, 0.937452345474458), `22` = c(1.0357237385154,
1.02805558505417, 0.946794300033338, 1.04688545274238), `23` = c(0.763210436944137,
0.801397021884422, 0.952553568039278, 0.990226493248718), `24` = c(0.789338028300063,
0.822815644347233, 0.958462750269733, 1.04183361434861), `25` = c(0.766036811789943,
0.871085862829362, 1.02210371210681, 0.937452345474458), `26` = c(1.0357237385154,
1.02805558505417, 0.946794300033338, 1.04688545274238), `27` = c(0.763210436944137,
0.801397021884422, 0.952553568039278, 0.990226493248718), `28` = c(0.789338028300063,
0.822815644347233, 0.958462750269733, 1.04183361434861), `29` = c(0.766036811789943,
0.871085862829362, 1.02210371210681, 0.937452345474458), `30` = c(1.0357237385154,
1.02805558505417, 0.946794300033338, 1.04688545274238), `31` = c(0.763210436944137,
0.801397021884422, 0.952553568039278, 0.990226493248718), `32` = c(0.789338028300063,
0.822815644347233, 0.958462750269733, 1.04183361434861), `33` = c(0.937894856206067,
NA, 1.00383773624603, 1.04181193834546), `34` = c(1.03944921519508,
NA, 0.983868286249464, 1.10409633668759), `35` = c(0.949802513948967,
NA, 1.06522152108054, 1.04376827636719), `36` = c(0.965871712940006,
NA, 1.18437146805406, 1.01355356488254), `37` = c(0.937894856206067,
NA, 1.00383773624603, 1.04181193834546), `38` = c(1.03944921519508,
NA, 0.983868286249464, 1.10409633668759), `39` = c(0.949802513948967,
NA, 1.06522152108054, 1.04376827636719), `40` = c(0.965871712940006,
NA, 1.18437146805406, 1.01355356488254), `41` = c(0.937894856206067,
NA, 1.00383773624603, 1.04181193834546), `42` = c(1.03944921519508,
NA, 0.983868286249464, 1.10409633668759), `43` = c(0.949802513948967,
NA, 1.06522152108054, 1.04376827636719), `44` = c(0.965871712940006,
NA, 1.18437146805406, 1.01355356488254), `45` = c(0.937894856206067,
NA, 1.00383773624603, 1.04181193834546), `46` = c(1.03944921519508,
NA, 0.983868286249464, 1.10409633668759), `47` = c(0.949802513948967,
NA, 1.06522152108054, 1.04376827636719)), .Names = c("Group",
"1", "2", "3", "4", "5", "6", "7", "8", "9", "10", "11", "12",
"13", "14", "15", "16", "17", "18", "19", "20", "21", "22", "23",
"24", "25", "26", "27", "28", "29", "30", "31", "32", "33", "34",
"35", "36", "37", "38", "39", "40", "41", "42", "43", "44", "45",
"46", "47"), row.names = c(NA, 4L), class = "data.frame")
每行是我从两组比较中得到的比率的集合。我想知道这些比率是否与 1 显着不同。因此，我想通过使用标题中提到的两个测试来测试每行(向量)是否与 1 不同。如何将这些测试应用于我的数据？请考虑每一行的长度可能不同。
NAs 应该被忽略。作为和输出，我想要一个包含 3 列的表: Group name 、 p-value t-test 、 p.value Wilcoxon 。有人可以帮助我吗？
最佳答案
可能有一种方法可以使用您拥有的原始数据框的行，但我强烈建议使用列(整洁的数据框)。
library(dplyr)
library(tidyr)
# assuming this is the name of your original dataset
dt
# reshape to create a column for each name
dt2 = data.frame(t(dt), stringsAsFactors = F)
names(dt2) = dt2[1,]
dt2 = dt2[-1,]
dt2[,names(dt2)] = sapply(dt2[,names(dt2)], as.numeric)
# reshape to create a column of names and values
dt3 = dt2 %>%
gather(name,value,Mark:Tom) %>%
filter(!is.na(value)) # remove NAs
dt3 %>%
group_by(name) %>% # for each name
summarise(pval_ttest = t.test(value, mu=1, data=.)$p.value, # calculate t test p value
pval_wilc = wilcox.test(value, mu=1, data=.)$p.value) # calculate Wilcoxon p value
# # A tibble: 4 × 3
# name pval_ttest pval_wilc
# <chr> <dbl> <dbl>
# 1 Mark 4.408038e-09 1.020895e-06
# 2 Matt 6.679416e-06 2.502045e-04
# 3 Tim 1.777060e-02 6.932590e-02
# 4 Tom 2.433548e-01 5.148382e-01
关于配对 t 检验如何“理解”您给出的测量值以及为什么差异和比率可能会给出不同结果的一些附加信息。
考虑以下示例:
# paired t test of 2 vectors of same size (before and after treatment)
# it compares the means of those vectors
t.test(1:10, 13:4, paired = T)
# Paired t-test
#
# data: 1:10 and 13:4
# t = -1.5667, df = 9, p-value = 0.1516
# alternative hypothesis: true difference in means is not equal to 0
# 95 percent confidence interval:
# -7.331701 1.331701
# sample estimates:
# mean of the differences
# -3
# t test that compares one vector's mean to 0
# that vector is the differences of the two initial vectors
t.test(1:10 - 13:4, mu=0)
# One Sample t-test
#
# data: 1:10 - 13:4
# t = -1.5667, df = 9, p-value = 0.1516
# alternative hypothesis: true mean is not equal to 0
# 95 percent confidence interval:
# -7.331701 1.331701
# sample estimates:
# mean of x
# -3
# t test that compares one vector's mean to 1
# that vector is the ratios of the two initial vectors
t.test(1:10 / 13:4, mu=1)
# One Sample t-test
#
# data: 1:10/13:4
# t = -0.46036, df = 9, p-value = 0.6562
# alternative hypothesis: true mean is not equal to 1
# 95 percent confidence interval:
# 0.3229789 1.4480623
# sample estimates:
# mean of x
# 0.8855206
您可以看到配对 t 检验是差异向量的简单 t 检验，这是可能的，因为您有 2 个相同长度的向量(处理前)。这与比率向量的简单 t 检验不同。
所以，有不同的结果是合理的，但在某些应用中，比率测试更好。检查你的引用书目。
关于r - 如何将配对 t.test 或 Wilcoxon 检验应用于我的数据，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/42022067/