在dplyr summarise（）中按组计算上下置信区间 [英] Calculating upper and lower confidence intervals by group in dplyr summarise()

问题描述

我正在尝试创建一个表格，该表格显示 N （观察次数），频率百分比（答案> 0）以及百分比频率的上下置信区间，我想按类型分组。

I am trying to make a table that shows N (number of observations), percent frequency (of answers > 0), and the lower and upper confidence intervals for percent frequency, and I want to group this by type.

数据示例

dat <- data.frame(
  "type" = c("B","B","A","B","A","A","B","A","A","B","A","A","A","B","B","B"),
  "num" = c(3,0,0,9,6,0,4,1,1,5,6,1,3,0,0,0)
)

预期输出（使用值填充）：

Type   N   Percent   Lower 95% CI   Upper 95% CI
A
B

尝试

library(dplyr)
library(qwraps2)

table<-dat %>%
  group_by(type) %>%
  summarise(N=n(),
            mean.ci = mean_ci(dat$num),
            "Percent"=n_perc(num > 0))

这可以得到N和百分比频率，但返回错误：当我在mean_ci中添加时，列的长度必须为1（一个汇总值），而不是3

This worked to get N and percent frequency, but returned an error: "Column must be length 1 (a summary value), not 3" when I added in mean_ci

我尝试的第二个代码找到了此处：

The second code I tried, found here:

table2<-dat %>%
  group_by(type) %>%
  summarise(N.num=n(),
            mean.num = mean(dat$num),
            sd.num = sd(dat$num),
            "Percent"=n_perc(num > 0)) %>%
  mutate(se.num = sd.num / sqrt(N.num),
         lower.ci = 100*(mean.num - qt(1 - (0.05 / 2), N.num - 1) * se.num),
         upper.ci = 100*(mean.num + qt(1 - (0.05 / 2), N.num - 1) * se.num))

# A tibble: 2 x 8
#  type  N.num mean.num sd.num Percent        se.num lower.ci upper.ci
# <fct> <int>    <dbl>  <dbl> <chr>           <dbl>    <dbl>    <dbl>
#1 A         8     2.44   2.83 "6 (75.00\\%)"   1.00     7.35     480.
#2 B         8     2.44   2.83 "4 (50.00\\%)"   1.00     7.35     480.

这给了我一个输出，但是置信区间是不合逻辑的。

This gave me an output, but the confidence intervals are not logical.

推荐答案

mean_ci 的输出是长度为3的向量。这可能是意料之外的，因为该软件包添加了打印方法，因此当您在控制台中看到此方法时，它看起来像一个字符值，而不是数字长度> 1个向量。但是，您可以通过查看 str 来查看基础数据结构。

The output of mean_ci is a vector of length 3. This is maybe unexpected because the package has added a print method so that when you see this in the console it looks like a single character value and not a numeric length > 1 vector. But, you can see the underlying data structure by looking at str.

mean_ci(dat$num) %>% str
 # 'qwraps2_mean_ci' Named num [1:3] 2.44 1.05 3.82
 # - attr(*, "names")= chr [1:3] "mean" "lcl" "ucl"
 # - attr(*, "alpha")= num 0.05

总而言之，输出的每一列的每个元素都必须为长度1，因此为摘要提供一个长度为3的对象以放入单个单元格（列元素）会导致错误。一种解决方法是将长度为3的向量放入列表中，以便现在为长度为1的列表。然后，您可以使用 unnest_wider 将其分为3列（并因此使表更宽）

In summarize, each element of each column of the output needs to be length 1, so providing a length 3 object for summarize to put in a single "cell" (column element) results in an error. A workaround is to put the length 3 vector in a list, so that it is now a length 1 list. Then you can use unnest_wider to separate it into 3 columns (and therefore making the table "wider")

library(tidyverse)

dat %>%
  group_by(type) %>%
  summarise( N=n(),
            mean.ci = list(mean_ci(num)),
            "Percent"= n_perc(num > 0)) %>% 
  unnest_wider(mean.ci)
# # A tibble: 2 x 6
#   type      N  mean   lcl   ucl Percent       
#   <fct> <int> <dbl> <dbl> <dbl> <chr>         
# 1 A         8  2.25 0.523  3.98 "6 (75.00\\%)"
# 2 B         8  2.62 0.344  4.91 "4 (50.00\\%)"

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