How do I read the next integer ignoring whitespace in Python?

Question

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())

Do you want to keep reading from command line for a potentially infinite amount of time? — inspectorG4dget, Oct 21 '13 at 21:10
You don't want to use input(), that will `eval` your code. Use raw_input instead. http://docs.python.org/2/library/functions.html#input — Murph, Oct 21 '13 at 21:11
Related: http://stackoverflow.com/questions/14905866/python-stream-extraction — Jon Clements, Oct 21 '13 at 21:11
@JonClements: OP is using Python3, in which `input` [works differently](http://docs.python.org/3.3/library/functions.html#input) — inspectorG4dget, Oct 21 '13 at 21:13
Neil, tell us why you think your example is insufficient. What is making you search for another solution? — Robᵩ, Oct 21 '13 at 21:37
@Robᵩ: I would like to actually have something like java's scanner where I can "get next int" or "get next float" — Neil G, Oct 21 '13 at 21:47

score 3 · Answer 1 · answered Oct 21 '13 at 21:17

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))

I deleted my attempt because I prefer this method. +1 – Burhan Khalid Oct 21 '13 at 21:21 — Burhan Khalid, Oct 21 '13 at 21:21

dawg · Answer 2 · 2013-10-22T00:09:04.660

0

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....)

edited Oct 22 '13 at 00:09

answered Oct 21 '13 at 21:30

dawg

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I'd probably go for `iter(input, '')` here instead of the `while True` and `sys.stdin.readline()` – Jon Clements Oct 21 '13 at 21:53
@JonClements: I think just using `for line in sys.stdin:` instead of `While True` is even better. Thanks! – dawg Oct 21 '13 at 22:30
I would replace []-s to ()-s in the generator expressions. – SzieberthAdam Oct 21 '13 at 22:48

How do I read the next integer ignoring whitespace in Python?

2 Answers2