In continuation of this issue comparison Mann-Whitney test between groups, I decided to create a new topic.
Solution of Rui Barradas helped me calculate Mann-Whitney for group 1-2 and 1-3.
lst <- split(mydat, mydat$group)
lapply(lst[-1], function(DF) wilcox.test(DF$var, lst[[1]]$var, exact = FALSE))
So now i want get the descriptive statistics. I use library:psych
describeBy(mydat$var,mydat$group)
So i get the following output
group: 1
vars n mean sd median trimmed mad min max range skew kurtosis se
X1 1 4 23.5 0.58 23.5 23.5 0.74 23 24 1 0 -2.44 0.29
--------------------------------------------------------------------------------------
group: 2
vars n mean sd median trimmed mad min max range skew kurtosis se
X1 1 4 23.5 0.58 23.5 23.5 0.74 23 24 1 0 -2.44 0.29
--------------------------------------------------------------------------------------
group: 3
vars n mean sd median trimmed mad min max range skew kurtosis se
X1 1 4 23.5 0.58 23.5 23.5 0.74 23 24 1 0 -2.44 0.29
It is inconvenient. I need only for each group mean,sd,median and p-value of wilcox.test.
I.E. i want these output
mean sd median p-value
group1 23,5 0,58 23,5 -
group2 23,5 0,58 23,5 1
group3 23,5 0,58 23,5 1
How can i performe it?
structure(list(`1` = structure(list(vars = 1, n = 4, mean = 23.5,
sd = 0.577350269189626, median = 23.5, trimmed = 23.5, mad = 0.7413,
min = 23, max = 24, range = 1, skew = 0, kurtosis = -2.4375,
se = 0.288675134594813), .Names = c("vars", "n", "mean",
"sd", "median", "trimmed", "mad", "min", "max", "range", "skew",
"kurtosis", "se"), row.names = "X1", class = c("psych", "describe",
"data.frame")), `2` = structure(list(vars = 1, n = 4, mean = 23.5,
sd = 0.577350269189626, median = 23.5, trimmed = 23.5, mad = 0.7413,
min = 23, max = 24, range = 1, skew = 0, kurtosis = -2.4375,
se = 0.288675134594813), .Names = c("vars", "n", "mean",
"sd", "median", "trimmed", "mad", "min", "max", "range", "skew",
"kurtosis", "se"), row.names = "X1", class = c("psych", "describe",
"data.frame")), `3` = structure(list(vars = 1, n = 4, mean = 23.5,
sd = 0.577350269189626, median = 23.5, trimmed = 23.5, mad = 0.7413,
min = 23, max = 24, range = 1, skew = 0, kurtosis = -2.4375,
se = 0.288675134594813), .Names = c("vars", "n", "mean",
"sd", "median", "trimmed", "mad", "min", "max", "range", "skew",
"kurtosis", "se"), row.names = "X1", class = c("psych", "describe",
"data.frame"))), .Dim = 3L, .Dimnames = structure(list(group = c("1",
"2", "3")), .Names = "group"), call = by.default(data = x, INDICES = group,
FUN = describe, type = type), class = c("psych", "describeBy"
))
答案 0 :(得分：2)
使用链接到问题的数据和上面的split指令，以下内容将产生所需的输出。
我重复测试以将其结果分配给wt_list。
wt_list <- lapply(lst[-1], function(DF) wilcox.test(DF$var, lst[[1]]$var, exact = FALSE))
mu <- tapply(mydat$var, mydat$group, mean)
s <- tapply(mydat$var, mydat$group, sd)
md <- tapply(mydat$var, mydat$group, median)
pval <- c(NA, sapply(wt_list, '[[', "p.value"))
df_smry <- data.frame(mean = mu, sd = s, median = md, p.value = pval)
df_smry
# mean sd median p.value
#1 23.5 0.5773503 23.5 NA
#2 23.5 0.5773503 23.5 1
#3 23.5 0.5773503 23.5 1
答案 1 :(得分：2)
您可以将tidyverse与broom一起使用。
tidy()将测试结果作为data.frame给出。我们使用group添加缺失的complete值。然后，我们使用dplyr的{{1}}和group_by计算描述性统计量，并将结果合并到p.values中。如有必要，您可以最后过滤。
summarise_all然后您可以过滤期望的输出
library(tidyverse)
mydat %>%
with(.,pairwise.wilcox.test(var, group, exact =F)) %>%
broom::tidy() %>%
complete(group1 = factor(mydat$group)) %>%
left_join(mydat %>%
group_by(group=as.character(group)) %>%
summarise_all(c("mean", "sd", "median")),
by=c("group1"="group"))
# A tibble: 4 x 6
group1 group2 p.value mean sd median
<chr> <chr> <dbl> <dbl> <dbl> <dbl>
1 1 NA NA 23.5 0.577 23.5
2 2 1 1 23.5 0.577 23.5
3 3 1 1 23.5 0.577 23.5
4 3 2 1 23.5 0.577 23.5
答案 2 :(得分：1)
这对您有用吗？如上所述，您的dput存在问题。
我必须为每个组使用unlist才能使用rbind，然后从dplyr中进行简单选择。
dat <- structure(list(`1` = structure(list(vars = 1, n = 4, mean = 23.5,
sd = 0.577350269189626, median = 23.5, trimmed = 23.5, mad = 0.7413,
min = 23, max = 24, range = 1, skew = 0, kurtosis = -2.4375,
se = 0.288675134594813), .Names = c("vars", "n", "mean",
"sd", "median", "trimmed", "mad", "min", "max", "range", "skew",
"kurtosis", "se"), row.names = "X1", class = c("psych", "describe",
"data.frame")), `2` = structure(list(vars = 1, n = 4, mean = 23.5,
sd = 0.577350269189626, median = 23.5, trimmed = 23.5, mad = 0.7413,
min = 23, max = 24, range = 1, skew = 0, kurtosis = -2.4375,
se = 0.288675134594813), .Names = c("vars", "n", "mean",
"sd", "median", "trimmed", "mad", "min", "max", "range", "skew",
"kurtosis", "se"), row.names = "X1", class = c("psych", "describe",
"data.frame")), `3` = structure(list(vars = 1, n = 4, mean = 23.5,
sd = 0.577350269189626, median = 23.5, trimmed = 23.5, mad = 0.7413,
min = 23, max = 24, range = 1, skew = 0, kurtosis = -2.4375,
se = 0.288675134594813), .Names = c("vars", "n", "mean",
"sd", "median", "trimmed", "mad", "min", "max", "range", "skew",
"kurtosis", "se"), row.names = "X1", class = c("psych", "describe",
"data.frame"))), .Dim = 3L, .Dimnames = structure(list(group = c("1",
"2", "3")), .Names = "group"), class = c("psych", "describeBy"
))
require(tidyverse)
rbind(unlist(dat[[1]]),unlist(dat[[2]]),unlist(dat[[3]])) %>%
as.data.frame() %>%
select(mean, sd, median)
答案 3 :(得分：1)
另一种实现此目的的方法是在mat上添加参数describeBy
describeBy(mydat$var, mydat$group, mat = TRUE)
# So first I've used the data and the code form the link:
lst <- split(mydat, mydat$group)
.ls <- lapply(lst[-1], function(DF) wilcox.test(DF$var, lst[[1]]$var, exact = FALSE))
# Then I extracted values of p.values
.ls <- c("-", sapply(.ls, '[[', "p.value"))
# And finally I combined desired columns with extracted p.values
cbind(describeBy(mydat$var, mydat$group, mat = TRUE)[c(5, 6, 7)], "p.value" =.ls)
# And the output:
mean sd median p.value
11 23.5 0.5773503 23.5 -
12 23.5 0.5773503 23.5 1
13 23.5 0.5773503 23.5 1