如何在Knitr中对齐列标题和内容

Question

只需尝试将列标题与其在knitr中的内容对齐。有什么作用？当我使用position = 'c'或DF_3 <-structure(list(. = structure(c(11L, 6L, 7L, 8L, 13L, 5L, 4L, 2L, 12L, 3L, 1L, 10L, 9L), .Label = c("B", "D", "E", "F", "G", "H", "H", "H", "I", "J", "M", "N", "V"), class = "f"), HEV.pos = structure(c(3L, 8L, 1L, 6L, 2L, 9L, 8L, 10L, 7L, 3L, 8L, 4L, 5L), .Label = c("1 (2)", "15", "16", "19", "20", "24", "26", "4 (7)", "6 (11)", "7 (13)"), class = "factor"), HEV.neg = structure(c(9L, 2L, 7L, 1L, 6L, 1L, 7L, 10L, 5L, 4L, 8L, 3L, 5L), .Label = c("10 (28)", "12 ", "15 ", "17", "18 ", "19 ", "2 ", "4", "7", "9"), class = "factor")), class = "data.frame", row.names = c(NA, -13L)) library(knitr) library(kableExtra) library(dplyr) kable(DF_3, format = "html") %>% kable_styling(bootstrap_options = "striped") %>% row_spec(1, bold = T) %>% row_spec(6, bold = T) %>% column_spec(1, bold = T) 时，它给我一个错误：

  

未使用的参数（align =“ c”）。

这是我的代码：

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Answer 1

不清楚您在哪里尝试使用align参数。但是，必须将其提供给kable函数。功能文档统计：

  

*列对齐：由“ l”（左），“ c”（中心）和/或“ r”（右）组成的字符向量。默认情况下，或者如果align = NULL，则数字列为右对齐，而其他列为左对齐。如果length（align）== 1L，则字符串将扩展为单个字母的向量，例如除非输出格式为LaTeX，否则'clc'变为c（'c'，'l'，'c'）。

例如：

knitr::kable(mtcars[1:5, 1:5], caption = "Left Aligned")

knitr::kable(mtcars[1:5, 1:5], align = "c", caption = "Center Aligned")

knitr::kable(mtcars[1:5, 1:5], align = c("c", "l", "r", "c", "c"), caption = "Mixed Aligned")