I use this to split string by a delimiter. The first puts the results in a pre-constructed vector, the second returns a new vector.
#include <string>
#include <sstream>
#include <vector>
#include <iterator>
template <typename Out>
void split(const std::string &s, char delim, Out result) {
std::istringstream iss(s);
std::string item;
while (std::getline(iss, item, delim)) {
*result++ = item;
}
}
std::vector<std::string> split(const std::string &s, char delim) {
std::vector<std::string> elems;
split(s, delim, std::back_inserter(elems));
return elems;
}
Note that this solution does not skip empty tokens, so the following will find 4 items, one of which is empty:
std::vector<std::string> x = split("one:two::three", ':');
Here's a generator that yields the chunks you want:
def chunks(lst, n):
"""Yield successive n-sized chunks from lst."""
for i in range(0, len(lst), n):
yield lst[i:i + n]
import pprint
pprint.pprint(list(chunks(range(10, 75), 10)))
[[10, 11, 12, 13, 14, 15, 16, 17, 18, 19],
[20, 21, 22, 23, 24, 25, 26, 27, 28, 29],
[30, 31, 32, 33, 34, 35, 36, 37, 38, 39],
[40, 41, 42, 43, 44, 45, 46, 47, 48, 49],
[50, 51, 52, 53, 54, 55, 56, 57, 58, 59],
[60, 61, 62, 63, 64, 65, 66, 67, 68, 69],
[70, 71, 72, 73, 74]]
If you're using Python 2, you should use xrange()
instead of range()
:
def chunks(lst, n):
"""Yield successive n-sized chunks from lst."""
for i in xrange(0, len(lst), n):
yield lst[i:i + n]
Also you can simply use list comprehension instead of writing a function, though it's a good idea to encapsulate operations like this in named functions so that your code is easier to understand. Python 3:
[lst[i:i + n] for i in range(0, len(lst), n)]
Python 2 version:
[lst[i:i + n] for i in xrange(0, len(lst), n)]
Best Answer
You can use the
std::string::find()
function to find the position of your string delimiter, then usestd::string::substr()
to get a token.Example:
The
find(const string& str, size_t pos = 0)
function returns the position of the first occurrence ofstr
in the string, ornpos
if the string is not found.The
substr(size_t pos = 0, size_t n = npos)
function returns a substring of the object, starting at positionpos
and of lengthnpos
.If you have multiple delimiters, after you have extracted one token, you can remove it (delimiter included) to proceed with subsequent extractions (if you want to preserve the original string, just use
s = s.substr(pos + delimiter.length());
):This way you can easily loop to get each token.
Complete Example
Output: