Subset data based on partial match of column names

问题:

I need to subset a df to include certain strings. Some of these are full column names, and the following works fine:

testData[,c("FullColName1","FullColName2","FullColName3")]

My problem is that I need to expand this to also include column names that contain specific strings that may partially match to some other column names. These strings include letters and symbols:

"PartString1()","PartString2()"

I tried putting wildcards around these. (I've indicated this below with the prefix "star" because the "*" symbol didn't render correctly.)

testData[ ,c("FullColName1","FullColName2","FullColName3",
             "starPartString1()star","starPartString2()star")]

But I'm getting an error message: undefined columns selected. I can't figure out if or how I need grep to make this work.

回答1:

You mentioned you may be looking for symbols, so for this particular example we can use [[:punct:]] as our regular expression. This will find all the strings with punctuation symbols in the column names.

d <- data.frame(1:3, 3:1, 11:13, 13:11, rep(1, 3))
names(d) <- c("FullColName1", "FullColName2", "FullColName3",
              "PartString1()","PartString2()")

d[grepl("[[:punct:]]", names(d))]
#   PartString1() PartString2()
# 1            13             1
# 2            12             1
# 3            11             1

This last part just illustrates another way to do this with other string processing functions from stringr

library(stringr)
d[str_detect(names(d), "[[:punct:]]")]
#   PartString1() PartString2()
# 1            13             1
# 2            12             1
# 3            11             1

ADD per OPs comment

d[grepl("ring[12()]", names(d))]

to get either of the substrings ring1() or ring2() from the names vector

回答2:

You can use grep to find indices of column names with partial match to a particular pattern

require(PerformanceAnalytics)
data(managers)

colnames(managers)
#[1] "HAM1"        "HAM2"        "HAM3"        "HAM4"        "HAM5"       
#[6] "HAM6"        "EDHEC LS EQ" "SP500 TR"    "US 10Y TR"   "US 3m TR"

suppose the pattern you want to match is "HAM", along with some fixed column names ("SP500 TR" "US 10Y TR" "US 3m TR")

head(managers[,c("SP500 TR","US 10Y TR","US 3m TR",colnames(managers)[grep("HAM",colnames(managers))])])
#           SP500 TR US 10Y TR US 3m TR    HAM1 HAM2    HAM3    HAM4 HAM5 HAM6
#1996-01-31   0.0340   0.00380  0.00456  0.0074   NA  0.0349  0.0222   NA   NA
#1996-02-29   0.0093  -0.03532  0.00398  0.0193   NA  0.0351  0.0195   NA   NA
#1996-03-31   0.0096  -0.01057  0.00371  0.0155   NA  0.0258 -0.0098   NA   NA
#1996-04-30   0.0147  -0.01739  0.00428 -0.0091   NA  0.0449  0.0236   NA   NA
#1996-05-31   0.0258  -0.00543  0.00443  0.0076   NA  0.0353  0.0028   NA   NA
#1996-06-30   0.0038   0.01507  0.00412 -0.0039   NA -0.0303 -0.0019   NA   NA

you can specify multiple patterns using, grep("pattern1 | pattern2 ", colnames(data))

回答3:

You can use grepl for a search by column name. It returns a logical vector indicating matches.

Here is an example:

d <- read.table(header=TRUE, check.names=FALSE,
                text="1PartString()2 1PartString()3 OtherCol
                1 2 3
                3 4 5")
d
##   1PartString()2 1PartString()3 OtherCol
## 1              1              2        3
## 2              3              4        5

d[,grepl("PartString\\(\\)", names(d))]
##   1PartString()2 1PartString()3
## 1              1              2
## 2              3              4

grepl check to see if the pattern is present anywhere in the name, so a wildcard is not required.