kable-如何显示没有列名的表？
kable-如何显示没有列名的表？
提问于 2017-06-23 13:31:28
目前，我有这个数据帧(PS)：
我用来显示这个表的代码是：
kable(PS) %>%
kable_styling(bootstrap_options = c("striped", "hover", "condensed", "responsive"))我想显示不带列名的表，如下所示：
问题是
1)列名不能为空，尝试使用空名将产生不受支持的结果
2)如果我转换数据框并删除列名，然后使用kable，如下所示：
PS.mat <- as.matrix(PS)
colnames(PS.mat) <- NULL
kable(PS) %>%
kable_styling(bootstrap_options = c("striped", "hover", "condensed", "responsive"))我得到以下错误
Error in kable_info$colnames[[length(kable_info$colnames)]] : attempt to select less than one element in integerOneIndex我还尝试了以下参数，但没有结果
kable(PS, col.names = NA) 编辑1：
一个可重复的例子：
if (!require(pacman)) install.packages("pacman")
p_load("lubridate","knitr","kableExtra","scales")
Statistics <- c("AUM",
"Minimum Managed Account Size",
"Liquidity",
"Average Margin / Equity",
"Roundturns / $ Million / Year",
"Incentive Fees",
"Instruments Traded")
Value <- c("$30K","$30K","Daily","50%","6,933","25%","ES")
AI <- data.frame(Statistics,Value);
kable(AI) %>%
kable_styling(bootstrap_options = c("striped", "hover", "condensed", "responsive"))回答 1
Stack Overflow用户
回答已采纳
发布于 2017-06-23 16:28:57
根据您想要的输出格式，您可以使用这些函数。对于pandoc：
x = kable(AI, format="pandoc") %>%
kable_styling(bootstrap_options = c("striped", "hover", "condensed", "responsive"))
cat(x[3:9], sep="\n")对于html：
x = kable(AI, format="html") %>%
kable_styling(bootstrap_options = c("striped", "hover", "condensed", "responsive"))
gsub("<thead>.*</thead>", "", x)页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/44713731
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