用dplyr调用R中的prop.test函数 [英] Calling prop.test function in R with dplyr

问题描述

我正在尝试计算几个二项式比例置信区间。我的数据在数据框中，尽管我可以从 prop.test 返回的对象中成功提取估计 ， conf.int 变量在数据帧上运行时似乎为空。

I am trying to calculate several binomial proportion confidence intervals. My data are in a data frame, and though I can successfully extract the estimate from the object returned by prop.test, the conf.int variable seems to be null when run on the data frame.

library(dplyr)

cases <- c(50000, 1000, 10, 2343242)
population <- c(100000000, 500000000, 100000, 200000000)

df <- as.data.frame(cbind(cases, population))
df %>% mutate(rate = prop.test(cases, population, conf.level=0.95)$estimate)

这将适当地返回

    cases population       rate
1   50000      1e+08 0.00050000
2    1000      5e+08 0.00000200
3      10      1e+05 0.00010000
4 2343242      2e+08 0.01171621

但是，当我跑步时

df %>% mutate(confint.lower= prop.test(cases, pop, conf.level=0.95)$conf.int[1])

我很难过ly get

I sadly get

Error in mutate_impl(.data, dots) : 
  Column `confint.lower` is of unsupported type NULL

有什么想法吗？我知道计算二项式比例置信区间的其他方法，但我真的很想学习如何很好地使用 dplyr 。

Any thoughts? I know alternative ways to calculate the binomial proportion confidence interval, but I would really like to learn how to use dplyr well.

谢谢！

推荐答案

您可以使用 dplyr :: rowwise（）按行分组：

df %>%
    rowwise() %>%
    mutate(lower_ci = prop.test(cases, pop, conf.level=0.95)$conf.int[1])

默认情况下， dplyr 接受列名并将其视为向量。因此，像上面提到的@Jake Fisher一样，矢量化函数无需添加 rowwise（）即可工作。

By default dplyr takes the column names and treats them like vectors. So vectorized functions, like @Jake Fisher mentioned above, just work without rowwise() added.

这就是我会立即捕获所有置信区间成分：

This is what I would do to catch all of the confidence interval components at once:

df %>%
    rowwise %>%
    mutate(tst = list(broom::tidy(prop.test(cases, pop, conf.level=0.95)))) %>%
    tidyr::unnest(tst)

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