Find all occurrences of integer within text in Python

Question

my purpose of this code is to extract all the integers from the text and sum them up together.

I have been looking for solutions to pluck out all the integers in a line of text. I saw some solutions suggesting to use \\D and \\b , I just got started with regular expression and still unfamiliar with how it can fit into my code. Please help :(

import re
import urllib2

data = urllib2.urlopen("http://python-data.dr-chuck.net/regex_sum_179860.txt")
aList = []

for word in data:
    data = (str(w) for w in data)
    s = re.findall(r'[\d]+', word)
    if len(s) != 1: continue
    num = int(s[0])
    aList.append(num)

print aList

Answer 1

1. You need call read of the return value of the urllib2.urlopen ; The return value of urllib2.urlopen is not a string, but a connection object (file-like object)
2. Just apply re.findall to the data .
3. Square brackets around \\d are not necessary.

---

import re
import urllib2

data = urllib2.urlopen("http://python-data.dr-chuck.net/regex_sum_179860.txt").read()
int_list = map(int, re.findall(r'\d+', data))

---

>>> int_list
[3524, 9968, 6177, 3133, 6508, 7940, 3738, 1112, 6179, 4570, 6127, 9150,
 9883, 418, 3538, 2992, 8527, 1150, 2049, 2834, 2630, 3840, 2638, 3800,
 9144, 5866, 6742, 588, 6918, 7802, 8229, 7947, 8992, 1339, 2119, 846,
 3820, 4070, 9356, 9708, 3238, 9380, 5572, 9491, 3038, 7434, 7771, 288,
 8632, 3962, 9136, 8106, 7295, 3699, 4136, 3459, 8120, 6018, 8963, 5779,
 3635, 3984, 4850, 9633, 2588, 7631, 9591, 1067, 7182, 1301, 8041, 1361,
 5425, 8326, 7094, 8155, 2581, 7199, 6125, 42]

Answer 2

You can do it line by line, call findall using the pattern "\\d+" for one or more digits and extending your output list:

import re
import urllib2

data = urllib2.urlopen("http://python-data.dr-chuck.net/regex_sum_179860.txt")
r = re.compile("\d+")
l = []
for line in data:
    l.extend(map(int,r.findall(line)))

Output:

[3524, 9968, 6177, 3133, 6508, 7940, 3738, 1112, 6179, 4570, 6127, 9150, 9883, 418, 3538, 2992, 8527, 1150, 2049, 2834, 2630, 3840, 2638,  3800, 9144, 5866, 6742, 588, 6918, 7802, 8229, 7947, 8992, 1339, 
2119,  846, 3820, 4070, 9356, 9708, 3238, 9380, 5572, 9491, 3038, 
7434, 7771, 288, 8632, 3962, 9136, 8106, 7295, 3699, 4136, 3459, 8120,
6018, 8963, 5779, 3635, 3984, 4850, 9633, 2588, 7631, 9591, 1067, 
7182, 1301, 8041, 1361, 5425, 8326, 7094, 8155, 2581, 7199, 6125, 42]

You could also use str.isdigit :

l = []
for line in data:
     l.extend(map(int,(w for w in line.split() if w.isdigit())))

If you just want to sum the numbers, you don't need to store all the numbers at all:

print(sum(sum(map(int,(w for w in line.split() if w.isdigit()))) for line in data))

Output:

435239

Or using a regex:

 print(sum(sum(map(int,r.findall(line))) for line in data))

Probably irrelevant in your case but if you wanted to avoid any intermediary lists using python2 you could use itertools.imap :

from itertools import imap
print(sum(sum(imap(int,r.findall(line))) for line in data))

Answer 3

Since you mentioned you wanted to sum all integers, this will work in Python 3 (as urllib2 has been split across several modules in Python 3 named urllib.request and urllib.error ):

from urllib import request
import re


data = request.urlopen("http://python-data.dr-chuck.net/regex_sum_179860.txt")

result = 0

for word in data:
    result += sum([int(x) for x in re.findall(r'\d+', str(word))])

print(result)