如何选择可能存在或不存在的列？

I have a data frame that may or may not have some particular columns present. I want to select columns using dplyr if they do exist and, if not, just ignore that I tried to select them. Here's an example:

我有一个数据框,可能有也可能没有某些特定的列。我想使用dplyr选择列,如果它们确实存在,如果不存在,只是忽略我尝试选择它们。这是一个例子:

# Load libraries
library(dplyr)

# Create data frame
df <- data.frame(year = 2000:2010, foo = 0:10, bar = 10:20)

# Pull out some columns
df %>% select(year, contains("bar"))

# Result
#    year bar
# 1  2000  10
# 2  2001  11
# 3  2002  12
# 4  2003  13
# 5  2004  14
# 6  2005  15
# 7  2006  16
# 8  2007  17
# 9  2008  18
# 10 2009  19
# 11 2010  20

# Try again for non-existent column
df %>% select(year, contains("boo"))

# Result
#data frame with 0 columns and 11 rows

In the latter case, I just want to return a data frame with the column year since the column boo doesn't exist. My question is why do I get an empty data frame in the latter case and what is a good way of avoiding this and achieving the desired result?

在后一种情况下,我只想返回一个包含列年份的数据框,因为列boo不存在。我的问题是为什么在后一种情况下我得到一个空的数据框?什么是避免这种情况并实现预期结果的好方法?

EDIT: Session info

编辑:会话信息

R version 3.3.3 (2017-03-06)
Platform: x86_64-w64-mingw32/x64 (64-bit)
Running under: Windows 7 x64 (build 7601) Service Pack 1

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

other attached packages:
[1] dplyr_0.5.0

loaded via a namespace (and not attached):
[1] lazyeval_0.2.0   magrittr_1.5     R6_2.2.0         assertthat_0.2.0 DBI_0.6-1        tools_3.3.3     
[7] tibble_1.3.0     Rcpp_0.12.10

1 个解决方案

#1

In the devel version of dplyr

在dplyr的devel版本中

df %>%
   select(year, contains("boo"))
#     year
#1  2000
#2  2001
#3  2002
#4  2003
#5  2004
#6  2005
#7  2006
#8  2007
#9  2008
#10 2009
#11 2010

gives the expected output

给出预期的输出

Otherwise one option would be to use one_of

否则一个选项是使用one_of

df %>%
   select(one_of("year", "boo"))

It returns a warning message if the column is not available

如果列不可用,它将返回警告消息

Other option is matches

其他选项是匹配

df %>%
  select(matches("year|boo"))

#1