在一列中保留 emmeans::contrast 的 p 值格式，在其他列中保留合理数量的小数点
Keep p-value format from emmeans::contrast in one column and a reasonable number of decimal points at other columns
我的 .Rprofile 文件中有 option(scipen=999)，我的 .Rmd 文件开头有 options(digits = 7)，以适应我的大比率和范围广泛的比率对比表。
我想在ratio、SE和t.ratio列中保留2位小数，除非值太小，并保持<.0001格式在p.value column。当值太小时，在我的例子中是 0.0004，我不想将它四舍五入为 0。我正在寻找比我想出的更圆滑的解决方案。
这是我要编辑的dataframe
tab1 <- contrast(model, list("..."), infer = c(F,T)) %>% print(export = T) %>% as.data.frame()
虽然 print 命令使输出 而不是 dataframe，但保持 p.value 格式，as.data.frame 命令允许我稍后修改其他列。 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", "51", "51", "51", "51", "51", "51"), null = c(" 1",
" 1", " 1", " 1", " 1", " 1", " 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", " <.0001", " <.0001", " 0.0071", " <.0001", " 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)] <- sapply(tab1[c(2,3,6)],as.numeric)
第一个 mutate 命令将数字四舍五入到小数点后 4 位以以非科学格式显示，第二个 mutate 删除了 ratio、[=19] 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，这是我对 mutate(across(where(is.numeric), round, 4)) 的期望。
使用 mutate(across(where(is.numeric), round, 5)) 返回 0.00039。
我知道 digits = 命令使用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而不是试图搞乱 digits
感谢您的关注！！
一个选项是使用 formattable::comma 并指定 digits，这也会将列保留为 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
或转换为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"
我的 .Rprofile 文件中有 option(scipen=999)，我的 .Rmd 文件开头有 options(digits = 7)，以适应我的大比率和范围广泛的比率对比表。
我想在ratio、SE和t.ratio列中保留2位小数，除非值太小，并保持<.0001格式在p.value column。当值太小时，在我的例子中是 0.0004，我不想将它四舍五入为 0。我正在寻找比我想出的更圆滑的解决方案。
这是我要编辑的dataframe
tab1 <- contrast(model, list("..."), infer = c(F,T)) %>% print(export = T) %>% as.data.frame()
虽然 print 命令使输出 而不是 dataframe，但保持 p.value 格式，as.data.frame 命令允许我稍后修改其他列。 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", "51", "51", "51", "51", "51", "51"), null = c(" 1",
" 1", " 1", " 1", " 1", " 1", " 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", " <.0001", " <.0001", " 0.0071", " <.0001", " 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)] <- sapply(tab1[c(2,3,6)],as.numeric)
第一个 mutate 命令将数字四舍五入到小数点后 4 位以以非科学格式显示，第二个 mutate 删除了 ratio、[=19] 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，这是我对 mutate(across(where(is.numeric), round, 4)) 的期望。
使用 mutate(across(where(is.numeric), round, 5)) 返回 0.00039。
我知道 digits = 命令使用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而不是试图搞乱 digits
感谢您的关注！！
一个选项是使用 formattable::comma 并指定 digits，这也会将列保留为 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
或转换为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"