如何选择在数据框中共享特定名称的列

Question

我很抱歉提出一个非常简单的问题.我有一个数据框,我尝试将dput放在下面.

mydf <- structure(list(br.Id = c(1992.0001, 1992.0002, 1992.0003, 1992.0004, 
1992.0005, 1992.0006, 1992.0007, 1992.0008, 1992.0009, 1992.001
), si.month = c(4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L), br.day = c(23L, 
23L, 23L, 23L, 23L, 23L, 23L, 23L, 23L, 23L), br.year = c(1992L, 
1992L, 1992L, 1992L, 1992L, 1992L, 1992L, 1992L, 1992L, 1992L
), branch = 1:10, br.location = c(160170L, 160170L, 160170L, 160170L, 
160170L, 160170L, 160170L, 160170L, 160170L, 160170L), si.length = c(90L, 
128L, 112L, 68L, 56L, 58L, 111L, 111L, 115L, 65L), si.weight = c(9.3, 
32.5, 19, 4.4, 2.1, 2.8, 16.1, 17.9, 22.7, 3.4), si.sex = structure(c(2L, 
1L, 2L, 2L, 3L, 2L, 2L, 2L, 2L, 1L), .Label = c("female", "male", 
"unknown"), class = "factor"), maturity = structure(c(7L, 7L, 
7L, 7L, 10L, 7L, 7L, 7L, 7L, 2L), .Label = c("developing", "immature", 
"mature", "nearly.ripe", "nearly.spent", "recovering", "ripe", 
"running", "spent", "unknown", "yoy"), class = "factor"), age = c(NA, 
NA, NA, NA, NA, NA, NA, NA, NA, 1L)), .Names = c("br.Id", "si.month", 
"br.day", "br.year", "branch", "br.location", "si.length", "si.weight", 
"si.sex", "maturity", "age"), row.names = c(NA, 10L), class = "data.frame")

我想要做的是选择与br具有相同名称的特定列.所以输出应该是这样的

      br.Id br.day br.year branch br.location
1  1992.000     23    1992      1      160170
2  1992.000     23    1992      2      160170
3  1992.000     23    1992      3      160170
4  1992.000     23    1992      4      160170
5  1992.001     23    1992      5      160170
6  1992.001     23    1992      6      160170
7  1992.001     23    1992      7      160170
8  1992.001     23    1992      8      160170
9  1992.001     23    1992      9      160170
10 1992.001     23    1992     10      160170

我想也许grep可以用来获取这些列,但我无法弄清楚如何使用它.我感谢你的帮助

Answer 1

你可以用这个吗？
一种方法是使用dplyr包.你需要加载它library(dplyr).然后select函数是一个已知的dplyr函数,用于选择变量,然后你可以使用contains来获取那些带有特定字母的列.

mydf %>% select(contains("br"))

#br.Id br.day br.year branch br.location
#1  1992.000     23    1992      1      160170
#2  1992.000     23    1992      2      160170
#3  1992.000     23    1992      3      160170
#4  1992.000     23    1992      4      160170
#5  1992.001     23    1992      5      160170
#6  1992.001     23    1992      6      160170
#7  1992.001     23    1992      7      160170
#8  1992.001     23    1992      8      160170
#9  1992.001     23    1992      9      160170
#10 1992.001     23    1992     10      160170