quickest way to to convert list of tuples to a series

Question

Consider a list of tuples lst

lst = [(\'a\', 10), (\'b\', 20)]

question
What is the quickest way to

Answer 1

Two possible downsides to @Divakar's np.asarray(lst) - it converts everything to string, requiring Pandas to convert them back. And speed - making arrays is relatively expensive.

An alternative is to use the zip(*) idiom to 'transpose' the list:

In [65]: lst = [('a', 10), ('b', 20), ('j',1000)]
In [66]: zlst = list(zip(*lst))
In [67]: zlst
Out[67]: [('a', 'b', 'j'), (10, 20, 1000)]
In [68]: out = pd.Series(zlst[1], index = zlst[0])
In [69]: out
Out[69]: 
a      10
b      20
j    1000
dtype: int32

Note that my dtype is int, not object.

In [79]: out.values
Out[79]: array(['10', '20', '1000'], dtype=object)

So in the array case, Pandas doesn't convert the values back to integer; it leaves them as strings.

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My guess about timings is off - I don't have any feel for pandas Series creation times. Also the sample is too small to do meaningful timings:

In [71]: %%timeit
    ...: out=pd.Series(dict(lst))
1000 loops, best of 3: 305 µs per loop
In [72]: %%timeit
    ...: arr=np.array(lst)
    ...: out = pd.Series(arr[:,1], index=arr[:,0])
10000 loops, best of 3: 198 µs per loop
In [73]: %%timeit
    ...: zlst = list(zip(*lst))
    ...: out = pd.Series(zlst[1], index=zlst[0])
    ...: 
1000 loops, best of 3: 275 µs per loop

Or forcing the integer interpretation

In [85]: %%timeit
    ...: arr=np.array(lst)
    ...: out = pd.Series(arr[:,1], index=arr[:,0], dtype=int)
    ...: 
    ...: 
1000 loops, best of 3: 253 µs per loop