Extract First N Digits from a String

Extract first N digits from a string

Solution using regex \\D to match non-digit characters and \\d{2} to match first two digits.

as.numeric(sub("\\D*(\\d{2}).*", "\\1", INPUT))
# [1] 55 19 24 24

data:

INPUT <- c("ABC Conference Room Monitor - Z5580J",
           "ABC 19 Monitor",
           "ABC 24 Monitor for Video-Conferencing",
           "ABC UltraSharp 24 Monitor -QU2482Z")

How do I get the first n characters of a string without checking the size or going out of bounds?

Here's a neat solution:

String upToNCharacters = s.substring(0, Math.min(s.length(), n));

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^{Opinion: while this solution is "neat", I think it is actually less readable than a solution that uses if / else in the obvious way. If the reader hasn't seen this trick, he/she has to think harder to understand the code. IMO, the code's meaning is more obvious in the if / else version. For a cleaner / more readable solution, see @paxdiablo's answer.}

Get the first numbers from a String on Java

This method can be used to return a number before another character is found

static int getNumbers(String s) {

    String[] n = s.split(""); //array of strings
    StringBuffer f = new StringBuffer(); // buffer to store numbers

    for (int i = 0; i < n.length; i++) {
        if((n[i].matches("[0-9]+"))) {// validating numbers
            f.append(n[i]); //appending
        }else {
            //parsing to int and returning value
            return Integer.parseInt(f.toString()); 
        }   
    }
    return 0;
 }

Usage:

getNumbers(s);
getNumbers(str);

Output:

5634
78695

How to extract first 8 characters from a string in pandas

You are close, need indexing with str which is apply for each value of Series:

data['Order_Date'] = data['Shipment ID'].str[:8]

For better performance if no NaNs values:

data['Order_Date'] = [x[:8] for x in data['Shipment ID']]

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print (data)
        Shipment ID Order_Date
0  20180504-S-20000   20180504
1  20180514-S-20537   20180514
2  20180514-S-20541   20180514
3  20180514-S-20644   20180514
4  20180514-S-20644   20180514
5  20180516-S-20009   20180516
6  20180516-S-20009   20180516
7  20180516-S-20009   20180516
8  20180516-S-20009   20180516

If omit str code filter column by position, first N values like:

print (data['Shipment ID'][:2])
0    20180504-S-20000
1    20180514-S-20537
Name: Shipment ID, dtype: object

Extract the first 2 Characters in a string

You can just use the substr function directly to take the first two characters of each string:

x <- c("75 to 79", "80 to 84", "85 to 89")
substr(x, start = 1, stop = 2)
# [1] "75" "80" "85"

You could also write a simple function to do a "reverse" substring, giving the 'start' and 'stop' values assuming the index begins at the end of the string:

revSubstr <- function(x, start, stop) {
  x <- strsplit(x, "")
  sapply(x, 
         function(x) paste(rev(rev(x)[start:stop]), collapse = ""), 
         USE.NAMES = FALSE)
}
revSubstr(x, start = 1, stop = 2)
# [1] "79" "84" "89" 

How to get first n characters from a string in R

library(stringr)
library(dplyr)

df$name %>% 
  str_extract_all("(?<=(^|[:space:]))[:alpha:]{3}") %>% 
  map_chr(~ str_c(.x, collapse = "_"))

The stringr cheatsheet is very useful for working through these types of problems.
https://www.rstudio.com/resources/cheatsheets/

^{Created on 2022-03-26 by the reprex package (v2.0.1)}

extract first N digits from string with regex

How about:

$str = preg_replace('/^(\d+).*$/', "$1", $str);

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