First items in inner list efficiently as possible [duplicate]

Answer 1

This should do it:

c = [x[0] for x in A]

It's a list comprehension that takes the first (sub-)element of every element of A.

Answer 2

For efficieny and extended slices, you can use numpy - which given your example seems like a good idea:

import numpy as np
yourlist = [
    [0, 0, 0],
    [0, 1, 1],
    [1, 0, 2]
]
a = np.array(yourlist)
print a[:,0]
# [0 0 1]
bc = np.bincount(a[:,0])
# array([2, 1])
count = bc[bc==1].size
# 1
# or... (I think it's probably better...)
count = np.count_nonzero(bc == 1)