用na.rm = TRUE总结数据

Question

考虑下面的示例，它总结了使用dplyr的summarise管道来识别与某些CHAR关联的minimum DATE的数据框架。

library('tidyverse')
library('lubridate')

temp <- data.frame(
  CHAR = c(
    'A',
    'B',
    'C'
  ),
  DATE = c(
    '20090101',
    '20100101',
    NA
  ) %>% ymd(), # Turn character strings to dates
  stringsAsFactors = FALSE
) %>% group_by(
  CHAR
) %>% summarise(
  DATE = min(DATE, na.rm = TRUE) # Extract minimum date
) %>% ungroup()

识别minimum是否为NA将使用is.na进行测试。

temp %>% mutate(
  DATE_lgl = DATE %>% is.na() # Identify dates that are missing/NA
)

输出量

# A tibble: 3 x 3
  CHAR  DATE       DATE_lgl
  <chr> <date>     <lgl>   
1 A     2009-01-01 FALSE   
2 B     2010-01-01 FALSE   
3 C     NA         FALSE   

DATE_lgl错误地显示了FALSE，其中DATE是NA。为什么会这样呢？

删除na.rm = TRUE修复了这个问题，但不适用于以下配置，在这种配置中，需要na.rm = TRUE来消除缺少的条目：

temp <- data.frame(
  CHAR = c(
    'A',
    'B',
    'C',
    'C'
  ),
  DATE = c(
    '20090101',
    '20100101',
    NA,
    '20110101'
  ) %>% ymd(), # Turn character strings to dates
  stringsAsFactors = FALSE
) %>% group_by(
  CHAR
) %>% summarise(
  DATE = min(DATE, na.rm = TRUE) # Extract minimum date
) %>% ungroup()

> sessionInfo()
R version 3.5.0 (2018-04-23)
Platform: x86_64-w64-mingw32/x64 (64-bit)
Running under: Windows 7 x64 (build 7601) Service Pack 1

Matrix products: default

locale:
[1] LC_COLLATE=English_Canada.1252  LC_CTYPE=English_Canada.1252    LC_MONETARY=English_Canada.1252
[4] LC_NUMERIC=C                    LC_TIME=English_Canada.1252    

attached base packages:
[1] stats     graphics  grDevices utils     datasets  methods   base     

other attached packages:
 [1] bindrcpp_0.2.2  lubridate_1.7.4 forcats_0.3.0   stringr_1.3.1   dplyr_0.7.5     purrr_0.2.5    
 [7] readr_1.1.1     tidyr_0.8.1     tibble_1.4.2    ggplot2_2.2.1   tidyverse_1.2.1

loaded via a namespace (and not attached):
 [1] Rcpp_0.12.17     cellranger_1.1.0 pillar_1.2.3     compiler_3.5.0   plyr_1.8.4       bindr_0.1.1     
 [7] tools_3.5.0      jsonlite_1.5     nlme_3.1-137     gtable_0.2.0     lattice_0.20-35  pkgconfig_2.0.1 
[13] rlang_0.2.1      psych_1.8.4      cli_1.0.0        rstudioapi_0.7   yaml_2.1.19      parallel_3.5.0  
[19] haven_1.1.1      xml2_1.2.0       httr_1.3.1       hms_0.4.2        grid_3.5.0       tidyselect_0.2.4
[25] glue_1.2.0       R6_2.2.2         readxl_1.1.0     foreign_0.8-70   modelr_0.1.2     reshape2_1.4.3  
[31] magrittr_1.5     scales_0.5.0     rvest_0.3.2      assertthat_0.2.0 mnormt_1.5-5     colorspace_1.3-2
[37] utf8_1.1.4       stringi_1.1.7    lazyeval_0.2.1   munsell_0.4.3    broom_0.4.4      crayon_1.3.4    

Answer 1

问题是你在评估

min(NA, na.rm=TRUE)
# Inf

对于第3行，这将导致

dput(temp$DATE[3])
# structure(Inf, class = "Date")

将is.finite添加到mutate中

temp %>% 
   mutate(DATE_lgl = is.finite(DATE) | is.na(DATE)  # Identify dates that are missing/NA)

 # A tibble: 3 x 3
 #   CHAR  DATE       DATE_lgl
 #  <chr> <date>     <lgl>   
 # 1 A     2009-01-01 TRUE    
 # 2 B     2010-01-01 TRUE    
 # 3 C     NA         FALSE

打印NA可能是日期类的打印限制。

as.Date(Inf, origin="1970-01-01")
# NA
dput(as.Date(Inf, origin="1970-01-01"))
# structure(Inf, class = "Date")

Answer 2

解决方法是将Date列转换为字符，然后计算它是否为NA。

temp %>% mutate(
  DATE_lgl = is.na(as.character(DATE))
)

# # A tibble: 3 x 3
#   CHAR  DATE       DATE_lgl
#   <chr> <date>     <lgl>   
# 1 A     2009-01-01 FALSE   
# 2 B     2010-01-01 FALSE   
# 3 C     NA         TRUE