python equivalent of R table

Question

I have a list

[[12, 6], [12, 0], [0, 6], [12, 0], [12, 0], [6, 0], [12, 6], [0, 6], [12, 0], [0, 6], [0, 6], [12, 0], [0, 6], [6, 0], [6, 0], [12, 0], [6, 0], [         


        

Answer 1

You can probably do a 1-dimensional count with list comprehension.

L = [[12, 6], [12, 0], [0, 6], [12, 0], [12, 0], [6, 0], [12, 6], [0, 6], [12, 0], [0, 6], [0, 6], [12, 0], [0, 6], [6, 0], [6, 0], [12, 0], [6, 0], [12, 0], [12, 0], [0, 6], [0, 6], [12, 6], [6, 0], [6, 0], [12, 6], [12, 0], [12, 0], [0, 6], [6, 0], [12, 6], [12, 6], [12, 6], [12, 0], [12, 0], [12, 0], [12, 0], [12, 6], [12, 0], [12, 0], [12, 6], [0, 6], [0, 6], [6, 0], [12, 6], [12, 6], [12, 6], [12, 6], [12, 6], [12, 0], [0, 6], [6, 0], [12, 0], [0, 6], [12, 6], [12, 6], [0, 6], [12, 0], [6, 0], [6, 0], [12, 6], [12, 0], [0, 6], [12, 0], [12, 0], [12, 0], [6, 0], [12, 6], [12, 6], [12, 6], [12, 6], [0, 6], [12, 0], [12, 6], [0, 6], [0, 6], [12, 0], [0, 6], [12, 6], [6, 0], [12, 6], [12, 6], [12, 0], [12, 0], [12, 6], [0, 6], [6, 0], [12, 0], [6, 0], [12, 0], [12, 0], [12, 6], [12, 0], [6, 0], [12, 6], [6, 0], [12, 0], [6, 0], [12, 0], [6, 0], [6, 0]]
countey = [tuple(x) for x in L]
freq = {x:countey.count(x) for x in set(countey)}

In [2]: %timeit {x:countey.count(x) for x in set(countey)}
        100000 loops, best of 3: 15.2 µs per loop   

In [4]: print(freq)
Out[4]: {(0, 6): 19, (6, 0): 20, (12, 0): 33, (12, 6): 28}

In [5]: print(freq[(12,6)])
Out[5]: 28