How do I read the next integer ignoring whitespace in Python 3? Is there a way to make a generator that returns integers from a stream of whitespace-separated integers?
The best I have is this:
def ints():
while True:
yield from (int(x) for x in input().split())
Complete ignoring white space and reading from a file (treating an integer purely as consecutive digits):
import mmap
import re
with open('somefile', 'rb') as fin:
mf = mmap.mmap(fin.fileno(), 0, access=mmap.ACCESS_READ)
for digits in re.finditer('(\d+)', mf):
print(digits.group(1))
Or, if you've already got everything in a string, then adapt the finditer appropriately, maybe something like:
yield from (match.group(1) for match in re.finditer('\d+', some_string))
If you are reading from stdin, you can do this:
import sys
def ints():
for line in sys.stdin:
rtr=[int(d) for d in line.split() if d.isdigit()]
if rtr:
for n in rtr:
yield n
print(list(ints()))
Or, if your Py 3.3+ supports generator delegation:
import sys
def ints():
for line in sys.stdin:
yield from (int(d) for d in line.split() if d.isdigit())
You can also use the very slick fileinput which has the advantage of supporting both a file name (and the file by that name will then be opened and read) or input from the redirected stdin:
import fileinput
def ints():
for line in fileinput.input():
yield from (int(d) for d in line.split() if d.isdigit())
(If you test fileinput directly or manually, know that you need to end the input with two CtlD's -- not just one....)
iter(input, '') here instead of the while True and sys.stdin.readline()
for line in sys.stdin: instead of While True is even better. Thanks!
evalyour code. Use raw_input instead. docs.python.org/2/library/functions.html#inputinputworks differently