R- knitr:kable - 如何显示没有列名的表格?

时间:2017-06-23 05:31:28

标签: r knitr flexdashboard

目前,我有这个数据框(PS):

Table with column header

我显示此表的代码是:

kable(PS) %>%
kable_styling(bootstrap_options = c("striped", "hover", "condensed", "responsive"))

我想显示没有列名的表,如下所示: enter image description here

问题是

1)列名称应为非空,并且尝试使用空名称将得到不支持的结果

2)如果我转换数据框并删除列名,然后像这样使用kable:

PS.mat <- as.matrix(PS)
colnames(PS.mat) <- NULL
kable(PS) %>%
kable_styling(bootstrap_options = c("striped", "hover", "condensed", "responsive"))

我收到以下错误

Error in kable_info$colnames[[length(kable_info$colnames)]] : attempt to select less than one element in integerOneIndex

我也尝试了以下参数但没有结果

kable(PS, col.names = NA) 

编辑1:

可重现的例子:

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"))

1 个答案:

答案 0 :(得分:9)

根据您所需的输出格式,您可以使用此类功能。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)