目前,我有以下数据帧(PS):

R Knitr :kable - How to display table without column names?-LMLPHP

我显示此表的代码是:

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

我想显示没有这样的列名的表:
R Knitr :kable - How to display table without column names?-LMLPHP

问题是

1)列名称应为非空,并且尝试使用空名称将导致不支持的结果

2)如果我转换数据框并删除列名,然后使用像这样的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:

一个可重现的示例:
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"))

最佳答案

根据所需的输出格式,您可以使用这些功能。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)

关于R Knitr :kable - How to display table without column names?,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/44713731/

10-12 20:00