Convert from K to Thousand (1000) in R

Convert from K to thousand (1000) in R

Another way similar to what @RichardScriven has suggested:

x <- c("1.4k", "14k")

as.numeric(sub("k", "e3", x, fixed = TRUE))

## [1]  1400 14000

How do I replace k and m with thousands and millions?

Using stringr and dplyr from tidyverse

library(tidyverse)
df %>%
  mutate(students = case_when(
    str_detect(students, "m") ~ as.numeric(str_extract(students, "[\\d\\.]+")) * 1000000,
    str_detect(students, "k") ~ as.numeric(str_extract(students, "[\\d\\.]+")) * 1000,
  ))
# A tibble: 3 x 2
  uni     students
  <chr>      <dbl>
1 Yale    16000000
2 Toronto   240000
3 NYU         7500

Convert a factor column with numbers in k format into numeric without losing any data

First detect which records with a "k".

df$is_k <- grepl("k", df$Likes)

Strip the "k", and then convert to numeric. If the record had a "k" then multiple my 1000, else multiple by 1.

df$Likes_num <- as.numeric(gsub("k", "", df$Likes)) * ifelse(df$is_k, 1000, 1)

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Edit

For multiple units, I adapted something I had elsewhere for a more complex problem. This shows the steps and is simple enough, though I am not sure how robust it is.

Function

convert_units <- function(x) {
  
  if (class(x) == "numeric") return(x)
  
  # named vector of scalings (you can add to this)
  unit_scale <- c("k" = 1e3, "m" = 1e6)
  
  # clean up some potential nuisances with the input
  x_str <- gsub(",", "", trimws(tolower(as.character(x))))
  
  # extract out the letters
  unit_char <- gsub("[^a-z]", "", x_str)
  
  # extract out the numbers and convert to numeric
  x_num <- as.numeric(gsub("[a-z]", "", x_str), "", x_str)
  
  # develop a vector of multipliers
  multiplier <- unit_scale[match(unit_char, names(unit_scale))]
  multiplier[is.na(multiplier)] <- 1
  
  # multiply
  x_num * multiplier
}

Application

df$Likes2 <- convert_units(df$Likes)

Sample Result

  ID Likes Likes2
1  1   99k  99000
2  2   997    997
3  3 15.5k  15500
4  4 9.25k   9250
5  5   575    575
6  6   800    800
7  7  8.5k   8500
8  8 2,400   2400

How to replace K for thousands and M for Millions in the same column in R

Supposing you have data that looks like this:

 fifa2 <- data.frame(Value = c("€565K", "€5.65M", "€777777"))

you can do this:

library(dplyr)
fifa2 %>%
  mutate(Value1 = as.numeric(gsub("[€MK]", "", Value)),
         Value1 = ifelse(grepl("K$", Value), Value1 * 1000, 
                                   ifelse(grepl("M$", Value), Value1 * 1000000, 
                                          Value1)))
    Value  Value1
1   €565K  565000
2  €5.65M 5650000
3 €777777  777777

Format thousand to Ks in R

You can also have a look at function scales::label_number_si which rounds the number.

a <- c(465456.6789, 3567.5, 1465458.12)
scales::label_number_si(accuracy = 0.1)(a)

#[1] "465.5K" "3.6K"   "1.5M"  

How to replace K with 1000 in a string using regular expression

This is a quick and dirty way which matches one or more digits followed by K and appends 000 to it:

data %>% 
  mutate(comment3 = str_replace(comment3 , "(\\d+)K", "\\1000"))

Where your data are placed in `data'.

Using \1 (or \\1 when escaped) to include the contents of the matched group (here, \\d+) seems to be the piece that you were missing in your attempt.

Results from your sample data:

# A tibble: 7 x 1
  comment3                              
  <chr>                                 
1 3.22%-1ST $100000/1.15% BAL           
2 3.25% ON 1ST $100000/1.16% ON BAL     
3 3.22% 1ST 100000/1.16 ON BAL          
4 3.22% 1ST 100000/1.15% ON BAL         
5 3.26% 1ST 100000/1.16% ON BAL         
6 3.20% 1ST 100000/1.15% ON BAL         
7 3.22% ON 1ST 100000 & 1.15% ON BALANCE

Format a number 1000 as 1k, 1000000 as 1m etc. in R

Using dplyr::case_when:

so_formatter <- function(x) {
  dplyr::case_when(
      x < 1e3 ~ as.character(x),
      x < 1e6 ~ paste0(as.character(x/1e3), "K"),
      x < 1e9 ~ paste0(as.character(x/1e6), "M"),
      TRUE ~ "To be implemented..."
  )
}

test <- c(1, 999, 1000, 999000, 1000000, 1500000, 1000000000, 100000000000)
so_formatter(test)

# [1] "1"                   
# [2] "999"                 
# [3] "1K"                  
# [4] "999K"                
# [5] "1M"                  
# [6] "1.5M"                
# [7] "To be implemented..."
# [8] "To be implemented..."

converting k and M to thousands and millions using mutate across and an ifelse statement

Try this: if you account for the string size you can decide how many 0's to append on to your data. in the example below all I am doing is adding an additional condition to the ifelse statement where if it is a length of 4 and contains 'M' append 5 0's else append 4 0's. You can add more ifelse statements to decided how many 0's to add on.

library(tidyverse)

dat <- data.frame(
  col1 = c('1.3M', '1.47k', '900k'),
  col2 = c('1.31M', '20k', '999k'),
  col3 = c('2.20M', '2.2M', '37M')
)

dat %>% 
  mutate(across(contains("col"), ~ifelse(grepl('k$',.), gsub('k','000',.),
                                         ifelse(grepl('M$',.) & nchar(.) ==4,gsub('M','00000',.),gsub('M','0000',.))))) %>% 
  mutate(across(contains("col"), ~str_remove_all(., '\\.')))

Edit:
This might be a cleaner way of doing the same thing (and more dynamic). I would just remove the K and M and then multiply by a 1000 or a 1000000 to get the full number.

dat %>% 
  mutate(across(contains("col"), ~ifelse(grepl('k$',.), as.numeric(gsub('k','',.))*1000,
                                  ifelse(grepl('M$',.) ,as.numeric(gsub('M','',.))*1000000,.)))) 

How do I format a number in thousands in R

How's sprintf("%.0f", 123542.52/1000) ?

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