Count all +1's in the file python

Question

I have the following data:

  1 3 4 2 6 7 8 8 93 23 45 2 0 0 0 1
  0 3 4 2 6 7 8 8 90 23 45 2 0 0 0 1
  0 3 4 2 6 7 8 6 93 23 45 2 0 0 0 1
  -1 3 4 2 6 7 8 8 21 23 45 2 0 0 0 1
  -1 3 4 2 6 7 8 8 0 23 45 2 0 0 0 1

The above data is in a file. I want to count the number of 1's,0's,-1's but only in 1st column. I am taking the file in standard input but the only way I could think of is to do like this:

  cnt = 0
  cnt1 = 0
  cnt2 = 0
  for line in sys.stdin:
      (t1, <having 15 different variables as that many columns are in files>) = re.split("\s+", line.strip())
      if re.match("+1", t1):
         cnt = cnt + 1
      if re.match("-1", t1):
         cnt1 = cnt1 + 1
      if re.match("0", t1):
         cnt2 = cnt2 + 1

How can I make it better especially the 15 different variables part as thats the only place where I will be using those variables.

Answer 1

Use collections.Counter:

from collections import Counter
with open('abc.txt') as f:
    c = Counter(int(line.split(None, 1)[0]) for line in f)
    print c

Output:

Counter({0: 2, -1: 2, 1: 1})

Here str.split(None, 1) splits the line just once:

>>> s = "1 3 4 2 6 7 8 8 93 23 45 2 0 0 0 1"                                                
>>> s.split(None, 1)
['1', '3 4 2 6 7 8 8 93 23 45 2 0 0 0 1']

---

Numpy makes it even easy:

>>> import numpy as np
>>> from collections import Counter                                                         
>>> Counter(np.loadtxt('abc.txt', usecols=(0,), dtype=np.int))                                     
Counter({0: 2, -1: 2, 1: 1})