CODEWITHSUNDEEP
rsubset
user3614783
user3614783

Reputation: 841

Subset data based on partial match of column names

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.

Upvotes: 22

Views: 52273

Answers (4)

GGAnderson
GGAnderson

Reputation: 2210

The question asked is how to retrieve specific column names, given only partial strings. Let me offer a simple grepl solution.

#example data
 df <- data.frame(col1_sse = paste0(1:5, LETTERS[1:5]),
            col2_swl = runif(5, max = 10), 
            col3_sdz = runif(5, max = 1000),
            col4_swl = paste0(letters[1:5]))

#assume partial names are complex
 partial_names <- c("2_sw", "sdz")

#create a "keepers" list of column names
 keepers <- names(df)[grepl(paste0(partial_names, collapse = "|"), names(df))]

#use "keepers" to extract cols from original data 
 new_df <- df[,keepers]

Upvotes: 0

Rich Scriven
Rich Scriven

Reputation: 99331

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

Upvotes: 19

Silence Dogood
Silence Dogood

Reputation: 3597

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))

Upvotes: 10

Matthew Lundberg
Matthew Lundberg

Reputation: 42639

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.

Upvotes: 4

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