1

我正在尝试制作一个表格,显示N(观察次数)、百分比频率(答案 > 0)以及百分比频率的上下置信区间,我想按类型对其进行分组。

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(填写值):

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

试图

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,但返回一个错误:“列必须是长度 1(汇总值),而不是 3”,当我在 mean_ci 中添加时

我尝试的第二个代码,在这里找到:

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,但置信区间不合逻辑。

4

2 回答 2

5

的输出mean_ci是一个长度为 3 的向量。这可能是出乎意料的,因为该包添加了一个打印方法,因此当您在控制台中看到它时,它看起来像一个单个字符值,而不是一个长度 > 1 的数字向量。但是,您可以通过查看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

总之,输出的每一列的每个元素都需要长度为 1,因此为汇总提供一个长度为 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 的向量放在一个列表中,这样它现在是一个长度为 1 的列表。然后您可以使用unnest_wider将其分成 3 列(从而使表格“更宽”)

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\\%)"
于 2020-01-22T14:56:26.950 回答
1

IceCreamToucan 的回答非常好。我发布这个答案是为了提供一种不同的方式来呈现信息。

library(dplyr)
#> 
#> Attaching package: 'dplyr'
#> The following objects are masked from 'package:stats':
#> 
#>     filter, lag
#> The following objects are masked from 'package:base':
#> 
#>     intersect, setdiff, setequal, union
library(qwraps2)

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))

在构建dplyr::summarize调用时,您可以使用qwraps2::frmtci 调用将输出格式化qwraps2::mean_ci为长度为 1 的字符串。

我还建议使用数据代词.data,这样您就可以明确了解要总结的变量。

dat %>%
  dplyr::group_by(type) %>%
  dplyr::summarize(N = n(),
                   mean.ci = qwraps2::frmtci(qwraps2::mean_ci(.data$num)),
                   Percent = qwraps2::n_perc(.data$num > 0))
#> `summarise()` ungrouping output (override with `.groups` argument)
#> # A tibble: 2 x 4
#>   type      N mean.ci           Percent       
#>   <chr> <int> <chr>             <chr>         
#> 1 A         8 2.25 (0.52, 3.98) "6 (75.00\\%)"
#> 2 B         8 2.62 (0.34, 4.91) "4 (50.00\\%)"

reprex 包于 2020-09-15 创建(v0.3.0)

devtools::session_info()
#> ─ Session info ───────────────────────────────────────────────────────────────
#>  setting  value                       
#>  version  R version 4.0.2 (2020-06-22)
#>  os       macOS Catalina 10.15.6      
#>  system   x86_64, darwin17.0          
#>  ui       X11                         
#>  language (EN)                        
#>  collate  en_US.UTF-8                 
#>  ctype    en_US.UTF-8                 
#>  tz       America/Denver              
#>  date     2020-09-15                  
#> 
#> ─ Packages ───────────────────────────────────────────────────────────────────
#>  package     * version date       lib source        
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于 2020-03-02T17:00:33.920 回答