在一列中保留 emmeans::contrast 中的 p 值格式，并在其他列中保留合理数量的小数点

问题描述

我的 option(scipen=999) 文件中有 .Rprofile，options(digits = 7) 的开头有 .Rmd，以适应对比表中的大比率和范围广泛的比率.

我想在 ratio、SE 和 t.ratio 列中保留 2 个小数位，除非值太小，并在 <.0001 列中保留 p.value column 格式{1}}。当值太小（在我的情况下为 0.0004）时，我不想将其四舍五入为 0。我正在寻找比我想出的更简洁的解决方案。

这是我想编辑的 dataframe

tab1 <- contrast(model,list("..."),infer = c(F,T)) %>% print(export = T) %>% as.data.frame()

虽然 print 命令使输出 not 成为 dataframe，保持 p.value 格式，但 as.data.frame 命令允许我修改其他列稍后。 Keep p-value format from the `emmeans::joint_tests()` output in a knitted pdf

> dput(tab1)
structure(list(`contrast           ` = c("(2-year) / (3-year)","(2-year) / (4-year)","(3-year) / (4-year)","soybean / oat    ","soybean / alfalfa","soybean / corn   ","oat / alfalfa    ","oat / corn       ","alfalfa / corn   "),ratio = c(0.429,0.286,0.668,0.000388,0.00323,0.0331,8.34,85.3,10.2),SE = c(0.2,0.127,0.26,0.000181,0.00191,0.0138,5.21,39.8,6.03),df = c("51","51","51"),null = c("   1","   1","   1"
),t.ratio = c(-1.818,-2.829,-1.034,-16.857,-9.728,-8.18,3.392,9.541,3.944),p.value = c(" 0.1738"," 0.0180"," 0.5588"," <.0001"," 0.0071"," 0.0014"
)),row.names = c("X","X.1","X.2","X.3","X.4","X.5","X.6","X.7","X.8"),class = "data.frame")

将 ratio、SE 和 t.ratio 转换为数字： tab1[,c(2,3,6)]

第一个 mutate 命令将数字四舍五入到 4 个小数位以以非科学格式显示，第二个 mutate 删除了 ratio、SE 中的所有尾随零,和 t.ratio。

tab1_rounded <- tab1 %>% mutate(across(where(is.numeric),round,4)) %>%mutate(across(where(is.numeric),as.character))`

我对我的 tab1_rounded 并不完全满意，因为 soybean/oat 是科学形式，4e-04。我想要 0.0004，这正是我对 mutate(across(where(is.numeric),4)) 的期望。

使用 mutate(across(where(is.numeric),5)) 返回 0.00039。

我知道 digits = 命令使用该特定列中的最大值指示列中的总位数，并且可能呈现不一致的小数位数，因此建议使用 round 而不是试图惹digits R options(digits=2) function changes the total number of digits’ format. Looking for a way to change the digits after decimal point

感谢您的关注！！

解决方法

一个选项是使用带有 formattable::comma 的 digits，这也会使列保持为 numeric

library(dplyr)
tab1 %>%
    mutate(across(where(is.numeric),~ formattable::comma(.,digits = 4)))
    contrast              ratio      SE df null  t.ratio p.value
X   (2-year) / (3-year)  0.4290  0.2000 51    1  -1.8180  0.1738
X.1 (2-year) / (4-year)  0.2860  0.1270 51    1  -2.8290  0.0180
X.2 (3-year) / (4-year)  0.6680  0.2600 51    1  -1.0340  0.5588
X.3   soybean / oat      0.0004  0.0002 51    1 -16.8570  <.0001
X.4   soybean / alfalfa  0.0032  0.0019 51    1  -9.7280  <.0001
X.5   soybean / corn     0.0331  0.0138 51    1  -8.1800  <.0001
X.6   oat / alfalfa      8.3400  5.2100 51    1   3.3920  0.0071
X.7   oat / corn        85.3000 39.8000 51    1   9.5410  <.0001
X.8   alfalfa / corn    10.2000  6.0300 51    1   3.9440  0.0014

---

或转为tibble进行格式打印

tab1 %>%
     mutate(across(where(is.numeric),~ as.numeric(sprintf('%.4f',.)))) %>% 
     as_tibble
# A tibble: 9 x 7
  `contrast           `   ratio      SE df    null   t.ratio p.value  
  <chr>                   <dbl>   <dbl> <chr> <chr>    <dbl> <chr>    
1 "(2-year) / (3-year)"  0.429   0.2    51    "   1"   -1.82 " 0.1738"
2 "(2-year) / (4-year)"  0.286   0.127  51    "   1"   -2.83 " 0.0180"
3 "(3-year) / (4-year)"  0.668   0.26   51    "   1"   -1.03 " 0.5588"
4 "soybean / oat    "    0.0004  0.0002 51    "   1"  -16.9  " <.0001"
5 "soybean / alfalfa"    0.0032  0.0019 51    "   1"   -9.73 " <.0001"
6 "soybean / corn   "    0.0331  0.0138 51    "   1"   -8.18 " <.0001"
7 "oat / alfalfa    "    8.34    5.21   51    "   1"    3.39 " 0.0071"
8 "oat / corn       "   85.3    39.8    51    "   1"    9.54 " <.0001"
9 "alfalfa / corn   "   10.2     6.03   51    "   1"    3.94 " 0.0014"