Given a random string consisting of numbers, letters, symbols, you need to sum up the numbers in the string. 

Note:

- Consecutive integers should be treated as a single number. eg, `2015` should be treated as a single number `2015`, NOT four numbers
- All the numbers should be treaded as positive integer. eg, `11-14` should be treated as two numbers `11` and `14`. Same as `3.14`, should be treated as two numbers `3` and `14`
- If no number was given in the string, it should return `0`


Example:

```
str = "In 2015, I want to know how much does iPhone 6+ cost?"
```

 The numbers are `2015`, `6`

Sum is `2021`.

Difficulty: introductory

Starter code:

def sum_from_string(string):
	

Solutions:
import re
def sum_from_string(string):
    d = re.findall("\d+",string)
    return sum(int(i) for i in d)

import re
def sum_from_string(s):
    return sum(map(int, re.findall(r'\d+', s)))

from re import findall


def sum_from_string(string):
    return sum(int(a) for a in findall(r'\d+', string))


import re

def sum_from_string(str_):
    return sum(map(int, re.findall(r'\d+', str_)))

def sum_from_string(s):
    return sum(map(int, "".join(c if c.isdigit() else " " for c in s).split()))

sum_from_string=lambda s:sum([int(k) for k in __import__('re').findall(r'\d+',s)])

def sum_from_string(string):
    str_nr, all_nr, have_nr = '', [], False
    for item in string + '_':
        if '0' <= item <= '9':
            str_nr += item
            have_nr = True
        else:
            if have_nr:
                all_nr.append(int(str_nr))
                have_nr , str_nr = False, ''
    return sum(all_nr)

search = __import__("re").compile(r"\d+").findall

def sum_from_string(string):
    return sum(map(int, search(string)))

from itertools import groupby

def sum_from_string(string):
  return sum(int(''.join(gp)) for b, gp in groupby(string, key = str.isdigit) if b)

import re

def sum_from_string(string):
    result = 0
    res = re.findall(r'\d+', string)
    for x in res:
        result += int(x)
    return result