Splitting a String by a Character

Split string into array of character strings


This will produce

array ["c", "a", "t"]

Python: split a string by the position of a character

You can do this with strings (and lists) using slicing:

string = "hello world!"
splitat = 4
left, right = string[:splitat], string[splitat:]

will result in:

>>> left
>>> right
o world!

How do I split a string in Java?

Use the appropriately named method String#split().

String string = "004-034556";
String[] parts = string.split("-");
String part1 = parts[0]; // 004
String part2 = parts[1]; // 034556

Note that split's argument is assumed to be a regular expression, so remember to escape special characters if necessary.

there are 12 characters with special meanings: the backslash \, the caret ^, the dollar sign $, the period or dot ., the vertical bar or pipe symbol |, the question mark ?, the asterisk or star *, the plus sign +, the opening parenthesis (, the closing parenthesis ), and the opening square bracket [, the opening curly brace {, These special characters are often called "metacharacters".

For instance, to split on a period/dot . (which means "any character" in regex), use either backslash \ to escape the individual special character like so split("\\."), or use character class [] to represent literal character(s) like so split("[.]"), or use Pattern#quote() to escape the entire string like so split(Pattern.quote(".")).

String[] parts = string.split(Pattern.quote(".")); // Split on the exact string.

To test beforehand if the string contains certain character(s), just use String#contains().

if (string.contains("-")) {
// Split it.
} else {
throw new IllegalArgumentException("String " + string + " does not contain -");

Note, this does not take a regular expression. For that, use String#matches() instead.

If you'd like to retain the split character in the resulting parts, then make use of positive lookaround. In case you want to have the split character to end up in left hand side, use positive lookbehind by prefixing ?<= group on the pattern.

String string = "004-034556";
String[] parts = string.split("(?<=-)");
String part1 = parts[0]; // 004-
String part2 = parts[1]; // 034556

In case you want to have the split character to end up in right hand side, use positive lookahead by prefixing ?= group on the pattern.

String string = "004-034556";
String[] parts = string.split("(?=-)");
String part1 = parts[0]; // 004
String part2 = parts[1]; // -034556

If you'd like to limit the number of resulting parts, then you can supply the desired number as 2nd argument of split() method.

String string = "004-034556-42";
String[] parts = string.split("-", 2);
String part1 = parts[0]; // 004
String part2 = parts[1]; // 034556-42

Splitting a string by a character

Using vectors, strings and stringstream. A tad cumbersome but it does the trick.

#include <string>
#include <vector>
#include <sstream>

std::stringstream test("this_is_a_test_string");
std::string segment;
std::vector<std::string> seglist;

while(std::getline(test, segment, '_'))

Which results in a vector with the same contents as

std::vector<std::string> seglist{ "this", "is", "a", "test", "string" };

Java: How to split a string by a number of characters?

I think that what he wants is to have a string split into substrings of size 4. Then I would do this in a loop:

List<String> strings = new ArrayList<String>();
int index = 0;
while (index < text.length()) {
strings.add(text.substring(index, Math.min(index + 4,text.length())));
index += 4;

python - splitting string by characters

This can easily be solved with a simple regex : \d+

import re

my_str = '15(1)(635)(634)(582)(583)'
print([int(i.group()) for i in re.finditer(r'\d+', my_str)])


[15, 1, 635, 634, 582, 583]

Split a string at every 4-th character?

This ought to do it:

String[] split = myString.split("(?=(....)+$)");
// or
String[] split = myString.split("(?=(.{4})+$)");

What it does is this: split on the empty string only if that empty string has a multiple of 4 chars ahead of it until the end-of-input is reached.

Of course, this has a bad runtime (O(n^2)). You can get a linear running time algorithm by simply splitting it yourself.

As mentioned by @anubhava:

(?!^)(?=(?:.{4})+$) to avoid empty results if string length is in multiples of 4

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