Many coding challenges have multiple numbers in the same line, often with a first line telling how many numbers there are in the multi-number line:
4
31 415 9 26
Usually I just read the entire line, then .split() it and map the strings to numbers.
But is there a good way to not read the entire line at once, instead read one number at a time? In order to save memory, either because I can't or don't want to read the whole line into memory. I'd like to only use O(1) space (let's assume numbers are small/bounded so they're O(1) size). Doesn't have to be absolutely minimal, for example if a solution internally reads a full 4 KB memory page at a time, that's ok, still O(1) and relatively small. For use case, think millions of numbers on one line, and a memory limit let's say below 1 MB.
In C++ I'd do it like this:
int n;
std::cin >> n;
while (n--) {
int value;
std::cin >> value;
// now do something with the value
}
I wrote this generator that takes a file object and gives me an iterator of strings. For the above example, it yields the strings '4', '31', '415', '9' and '26'. It reads one character at a time, and splits at space characters as determined by .isspace():
def split(file):
value = []
while char := file.read(1):
if char.isspace():
if value:
yield ''.join(value)
value.clear()
else:
value.append(char)
if value:
yield ''.join(value)
But that's of course ridiculously complicated and slow, and I don't even know whether this str.isspace usage is equivalent to what str.split considers whitespace. It just illustrates one way to achieve what I want.
Edit: Here's a simpler way, but still more complicated and slow than I'd like. I'm looking for some built-in way that does the low-level work for me, at C speed.
from itertools import groupby
def split(file):
groups = groupby(iter(lambda: file.read(1), ''), str.isspace)
for isspace, chars in groups:
if not isspace:
yield ''.join(chars)
