5

我正在使用summary()在文档中生成简单的日期变量摘要,knitr并希望将结果格式化。

summary(some_date)

产生我想要的结果(值显示为日期),但格式不正确。所以,我尝试了:

pander::pander(summary(some_date))

这会产生格式良好的结果,但值显示为整数而不是日期。(这适用于 Date、POSIXct 和 POSIXlt 日期)。

有什么方法可以很好地格式化文档summary()中日期的结果吗?knitr也许有一些包比pander哪个更适合这项任务。

以下示例:

summary(some_date)
##         Min.      1st Qu.       Median         Mean      3rd Qu. 
## "2014-05-01" "2015-02-15" "2015-06-17" "2015-05-05" "2015-09-08" 
##         Max. 
## "2015-11-21"

pander::pander(summary(some_date))
Min.    1st Qu. Median  Mean    3rd Qu. Max.
16191   16481   16603   16560   16686   16760

sessionInfo()
## R version 3.2.3 (2015-12-10)
## Platform: x86_64-pc-linux-gnu (64-bit)
## Running under: Ubuntu 14.04.3 LTS
## 
## locale:
##  [1] LC_CTYPE=en_AU.UTF-8       LC_NUMERIC=C              
##  [3] LC_TIME=en_AU.UTF-8        LC_COLLATE=en_AU.UTF-8    
##  [5] LC_MONETARY=en_AU.UTF-8    LC_MESSAGES=en_AU.UTF-8   
##  [7] LC_PAPER=en_AU.UTF-8       LC_NAME=C                 
##  [9] LC_ADDRESS=C               LC_TELEPHONE=C            
## [11] LC_MEASUREMENT=en_AU.UTF-8 LC_IDENTIFICATION=C       
## 
## attached base packages:
## [1] stats     graphics  grDevices utils     datasets  methods   base     
## 
## other attached packages:
## [1] pander_0.6.0  ggplot2_2.0.0 daff_0.1.4    magrittr_1.5  dplyr_0.4.3  
## [6] readr_0.2.2  
## 
## loaded via a namespace (and not attached):
##  [1] Rcpp_0.12.2      knitr_1.11       munsell_0.4.2    lattice_0.20-33 
##  [5] colorspace_1.2-6 R6_2.1.1         plyr_1.8.3       stringr_1.0.0   
##  [9] tools_3.2.3      parallel_3.2.3   grid_3.2.3       packrat_0.4.6-4 
## [13] nlme_3.1-122     gtable_0.1.2     mgcv_1.8-7       DBI_0.3.1       
## [17] htmltools_0.2.6  lazyeval_0.1.10  yaml_2.1.13      assertthat_0.1  
## [21] digest_0.6.8     Matrix_1.2-3     formatR_1.2.1    curl_0.9.4      
## [25] evaluate_0.8     rmarkdown_0.9    labeling_0.3     V8_0.9          
## [29] stringi_1.0-1    scales_0.3.0     jsonlite_0.9.19
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1 回答 1

5

感谢您的错误报告,它的开发版本pander现在应该可以正确呈现:

> pander::pander(summary(as.Date('2015-01-01') + 1:100))

-----------------------------------------------------------------
   Min.     1st Qu.     Median      Mean     3rd Qu.      Max.   
---------- ---------- ---------- ---------- ---------- ----------
2015-01-02 2015-01-26 2015-02-20 2015-02-20 2015-03-17 2015-04-11
-----------------------------------------------------------------
于 2015-12-29T08:52:32.653 回答