Find strings that contain a sequence of characters regardless of the order in r

Question

I have a dataframe(df)

    V1    V2
1 "BCC"  Yes
2 "ABB"  Yes

I want to find all the strings that contain a certain sequence of characters, regardless of the order. For example if I have the string "CBC" or "CCB" I would like to get

    V1    V2
1 "BCC"  Yes

I've tried with grep, but It only finds the matching patterns

>df[grep("CBC", df$V1),]
1  V1   V2
<0 rows> (or 0-length row.names)

>df[grep("BCC", df$V1),]
   V1   V2
1 "BCC" Yes

Answer 1

We can create a logical index by splitting the column

i1 <- sapply(strsplit(df$V1, ""), function(x) all(c("B", "C") %in% x))
df[i1, , drop = FALSE]
#   V1  V2
#1 BCC Yes

---

if we have two datasets and one is a lookup table ('df2'), then split the column into characters,paste the sorted elements, and use %in% to create the logical vector for filtering the rows

v1n <- sapply(strsplit(df1$v1, ""), function(x) paste(sort(x), collapse=""))
v1l <- sapply(strsplit(df2$v1, ""), function(x) paste(sort(x), collapse=""))
df1[v1n %in% v1l, , drop = FALSE]

data

df1 <- data.frame(v1 = c("BCC", "CAB" , "ABB", "CBC", "CCB", "BAB", "CDB"),
     stringsAsFactors = FALSE)
df2 <- data.frame(v1 = c("CBC", "ABB"), stringsAsFactors = FALSE)

Answer 2

In the comments you mention a lookup table. If this is the case, an approach could be to join both sets together, then use the regex by Wiktor Stribiżew to indicate which are valid

As I'm joining data sets I'm going to use data.table

Method 1: Join everything

library(data.table)

## dummy data, and a lookup table
dt <- data.frame(V1 = c("BCC", "ABB"))
dt_lookup <- data.frame(V1 = c("CBC","BAB", "CCB"))

## convert to data.table
setDT(dt); setDT(dt_lookup)

## add some indexes to keep track of rows from each dt
dt[, idx := .I]
dt_lookup[, l_idx := .I]

## create a column to join on
dt[, key := 1L]
dt_lookup[, key := 1L]

## join EVERYTHING
dt <- dt[
    dt_lookup
    , on = "key"
    , allow.cartesian = T
]

#regex
dt[
    , valid := grepl(paste0("^[",i.V1,"]+$"), V1)
    , by = 1:nrow(dt)
]

#     V1 idx key i.V1 l_idx valid
# 1: BCC   1   1  CBC     1  TRUE
# 2: ABB   2   1  CBC     1 FALSE
# 3: BCC   1   1  BAB     2 FALSE
# 4: ABB   2   1  BAB     2  TRUE
# 5: BCC   1   1  CCB     3  TRUE
# 6: ABB   2   1  CCB     3 FALSE

---

Method 2: EACHI join

A slightly more memory-efficient approach might be to use this technique by Jaap as it avoids the 'join everything' step, and in stead joins it 'by each i' (row) at a time.

dt_lookup[
    dt, 
    {
        valid = grepl(paste0("^[",i.V1,"]+$"), V1)
        .(
            V1 = V1[valid]
            , idx = i.idx
            , match = i.V1
            , l_idx = l_idx[valid]
            )
    }
    , on = "key"
    , by = .EACHI
]

#    key  V1 idx match l_idx
# 1:   1 CBC   1   BCC     1
# 2:   1 CCB   1   BCC     3
# 3:   1 BAB   2   ABB     2

Answer 3

Here is one method using sapply, table, and identical.

# construct a named vector of integers with names in 
# alphabetical order: your match
myVal <- c("B"=1L, "C"=2L)
# run through character variable, perform check
sapply(strsplit(dat$V1, ""), function(x) identical(c(table(x)), myVal))
[1]  TRUE FALSE

Two key points related to the use of identical and the output of table:

1. the match vector, myVal must be integer.
2. You want to order the match vector alphabetically, yough you can do this ahead of time, you can also do it after the fact with order, names, and [.

Also, not that I wrapped the output of table in c to strip off undesired attributes, while maintaining the names.

Answer 4

You can use stringi::stri_count_regex to see if the number of occurrences in your string matches the table of strsplit(str_to_find, ''). The last reduce("|") means it's checking if there are any matches, so change | to & if you want to check if it matches all the strings in to.find.

set.seed(0)
df <- data.frame(a = replicate(20, paste0(sample(LETTERS[1:3], 3, T), collapse = ''))
                 , stringsAsFactors = F)

to.find <- c("CBB", "CCB")
to.find <- strsplit(to.find, '')

library(tidyverse)
library(stringi)
df$b <- 
sapply(df$a, function(x){
         lapply(to.find, function(y){
           imap(table(y), ~ .x == stri_count_regex(x, .y)) %>% 
             reduce(`&`)}) %>% 
          reduce(`|`)})

df

# a     b
# 1  CAB FALSE
# 2  BCA FALSE
# 3  CCB  TRUE
# 4  BAA FALSE
# 5  ACB FALSE
# 6  CBC  TRUE
# 7  CBC  TRUE
# 8  CAB FALSE
# 9  AAB FALSE
# 10 ABC FALSE
# 11 BBB FALSE
# 12 BAC FALSE
# 13 CCA FALSE
# 14 CBC  TRUE
# 15 BCB  TRUE
# 16 BCA FALSE
# 17 BCC  TRUE
# 18 BCB  TRUE
# 19 AAA FALSE
# 20 ABB FALSE
# 19 AAA FALSE
# 20 ABB FALSE

You can also do it all with map, but that's harder to read

df$b <- 
df$a %>% 
  map(~{x <- .x
        map(to.find, 
            ~imap(table(.x), ~ .x == stri_count_regex(x, .y)) %>% 
              reduce(`&`)) %>% 
          reduce(`|`)})