CODEWITHSUNDEEP
rstringgsubgrepl
eraysahin
eraysahin

Reputation: 119

Adding leading zeros in the character and numeric mixed column names in R

I have a dataframe with some column names like; "Sample_ID", "Time00", "X7236Nr1", "Y844Nr1856", "X9834Nr21", "S844Nr567"

I want to add leading zeros to the digits after "Nr", so that I can convert it all to 4 digit numbers; "Sample_ID", "Time00", "X7236Nr0001", "Y844Nr1856", "X9834Nr0021", "S844Nr0567"

I tried to use rename_at to select the columns and apply the appropriate function such as sprintf,

df %>% rename_at(vars(starts_with("[A-B][0-9]")), ~ FUNCTION)

but could not build correct function. Can you please advise any way to deal with that kind of mixed strings?

Thanks in advance

Upvotes: 3

Views: 136

Answers (4)

ThomasIsCoding
ThomasIsCoding

Reputation: 101335

You can try

  • Option 1
idx <- grepl("Nr\\d+$", s)
x <- s[idx]
s[idx] <- paste0(sub("\\d+$", "", x), sprintf("%.4i", as.integer(sub(".*Nr", "", x))))
  • Option 2
sapply(
  strsplit(s, "(?<=Nr)(?=\\d)", perl = TRUE),
  \(x) {
    if (length(x) == 2) {
      paste0(x[1], strrep("0", max(0, 4 - nchar(x[2]))), x[2])
    } else {
       x
    }
  }
)
  • Option 3
library(gsubfn)
gsubfn("(.*Nr)(\\d+)$", ~ paste0(x, sprintf("%.4i", as.integer(y))), s)

or

gsubfn("Nr(\\d+)$" ~ sprintf("Nr%.4i", as.integer(x)), s)

which gives

> s
[1] "Sample_ID"   "Time00"      "X7236Nr0001" "Y844Nr1856"  "X9834Nr0021"
[6] "S844Nr0567"

Data

s <- c("Sample_ID", "Time00", "X7236Nr1", "Y844Nr1856", "X9834Nr21", "S844Nr567")

Upvotes: 4

dog
dog

Reputation: 2911

In one row

# your data.frame
df <- data.frame(Sample_ID = 1, Time00 = 1, X7236Nr1 = 1, Y844Nr1856 = 1, X9834Nr21 = 1, S844Nr567 = 1)
# one row only base R for the enthusiasts w/o any explanation
df <- do.call(data.frame, lapply(names(df), function(x) setNames(list(df[[x]]), if(grepl("Nr(\\d+)", x)) paste0(sub("Nr(\\d+)", "", x), "Nr", sprintf("%04d", as.numeric(sub('.+Nr(.+)', '\\1', x)))) else x)))

Accepted Answer

I have a dataframe with some column names like; "Sample_ID", "Time00", "X7236Nr1", "Y844Nr1856", "X9834Nr21", "S844Nr567"

you can do it by using str_replace_all with a str_match that finds the "Nr" + number and str_pad()s the number to 4 digits with zeroes.

library(dplyr)
library(stringr)

# your data.frame
df <- data.frame(Sample_ID = 1, Time00 = 1, X7236Nr1 = 1, Y844Nr1856 = 1, X9834Nr21 = 1, S844Nr567 = 1)

df <- df %>%
  rename_with(~ str_replace_all(., "Nr(\\d+)", function(x) {
    match <- str_match(x, "Nr(\\d+)")
    if (!is.na(match[2])) {
      paste0("Nr", str_pad(match[2], 4, pad = "0")) # only do if "Nr" is found
    } else {
      x
    }
  }))

### Result
> colnames(df)
"Sample_ID"   "Time00"      "X7236Nr0001" "Y844Nr1856"  "X9834Nr0021" "S844Nr0567" 

# Explanations
> str_match("Y844Nr0856", "Nr(\\d+)")
     [,1]     [,2]  
[1,] "Nr0856" "0856"
 
> str_match("Time00", "Nr(\\d+)") # has NA as match[,2], therefore we will not replace anything
     [,1] [,2]
[1,] NA   NA  
 
> str_pad("856", 4, pad = "0") # could also use sprintf()
[1] "0856"

Upvotes: 1

jay.sf
jay.sf

Reputation: 72803

Here a way using sprintf

> f <- \(x) {
+   sts <- strsplit(x, '(?<=Nr)', perl=TRUE)
+   nbs <- sapply(sts[u <- lengths(sts) > 1], `[[`, 2)
+   sts[u] <- Map(c, lapply(sts[u], `[[`, 1), 
+       sprintf(paste0('%0', max(nchar(nbs)), 'd'), as.integer(nbs))
+   ) 
+   sts |> sapply(paste, collapse='')
+ }
> f(x)
[1] "Sample_ID"   "Time00"      "X7236Nr0001" "Y844Nr1856"  "X9834Nr0021" "S844Nr0567"

Upvotes: 0

Onyambu
Onyambu

Reputation: 79218

Use str_Replace_all as shown below:

str_replace_all(names(df), "(?<=Nr)\\d+", ~sprintf("%04d", as.numeric(.x)))

[1] "Sample_ID"   "Time00"      "X7236Nr0001" "Y844Nr1856"  "X9834Nr0021"
[6] "S844Nr0567"

The regex is even simpler if you can use rename_with:

df %>% 
rename_with(~str_replace_all(., "\\d+", ~sprintf("%04d", as.numeric(.))), matches("Nr\\d+"))

Sample_ID Time00 X7236Nr0001 Y0844Nr1856 X9834Nr0021 S0844Nr0567
1         1      A           1           4           7          10
2         2      B           2           5           8          11
3         3      C           3           6           9          12

in base R, this will directly change the names of the df:

m <- regexpr("(?<=Nr)\\d+", names(df), perl = TRUE)
regmatches(names(df), m) <- sprintf("%04d", as.numeric(regmatches(names(df), m)))

df
  Sample_ID Time00 X7236Nr0001 Y844Nr1856 X9834Nr0021 S844Nr0567
1         1      A           1          4           7         10
2         2      B           2          5           8         11
3         3      C           3          6           9         12

Upvotes: 3

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