How to extract the n-th elements from a list of tuples
n = 1 # N. . .
[x[n] for x in elements]
This also works:
zip(*elements)[1]
(I am mainly posting this, to prove to myself that I have groked zip
...)
See it in action:
>>> help(zip)
Help on built-in function zip in module builtin:
zip(...)
zip(seq1 [, seq2 [...]]) -> [(seq1[0], seq2[0] ...), (...)]
Return a list of tuples, where each tuple contains the i-th element from each of the argument sequences. The returned list is truncated in length to the length of the shortest argument sequence.
>>> elements = [(1,1,1),(2,3,7),(3,5,10)]
>>> zip(*elements)
[(1, 2, 3), (1, 3, 5), (1, 7, 10)]
>>> zip(*elements)[1]
(1, 3, 5)
>>>
Neat thing I learned today: Use *list
in arguments to create a parameter list for a function...
Note: In Python3, zip
returns an iterator, so instead use list(zip(*elements))
to return a list of tuples.
I know that it could be done with a FOR but I wanted to know if there's another way
There is another way. You can also do it with map and itemgetter:
>>> from operator import itemgetter
>>> map(itemgetter(1), elements)
This still performs a loop internally though and it is slightly slower than the list comprehension:
setup = 'elements = [(1,1,1) for _ in range(100000)];from operator import itemgetter'
method1 = '[x[1] for x in elements]'
method2 = 'map(itemgetter(1), elements)'
import timeit
t = timeit.Timer(method1, setup)
print('Method 1: ' + str(t.timeit(100)))
t = timeit.Timer(method2, setup)
print('Method 2: ' + str(t.timeit(100)))
Results:
Method 1: 1.25699996948 Method 2: 1.46600008011
If you need to iterate over a list then using a for
is fine.
Found this as I was searching for which way is fastest to pull the second element of a 2-tuple list. Not what I wanted but ran same test as shown with a 3rd method plus test the zip method
setup = 'elements = [(1,1) for _ in range(100000)];from operator import itemgetter'
method1 = '[x[1] for x in elements]'
method2 = 'map(itemgetter(1), elements)'
method3 = 'dict(elements).values()'
method4 = 'zip(*elements)[1]'
import timeit
t = timeit.Timer(method1, setup)
print('Method 1: ' + str(t.timeit(100)))
t = timeit.Timer(method2, setup)
print('Method 2: ' + str(t.timeit(100)))
t = timeit.Timer(method3, setup)
print('Method 3: ' + str(t.timeit(100)))
t = timeit.Timer(method4, setup)
print('Method 4: ' + str(t.timeit(100)))
Method 1: 0.618785858154
Method 2: 0.711684942245
Method 3: 0.298138141632
Method 4: 1.32586884499
So over twice as fast if you have a 2 tuple pair to just convert to a dict and take the values.
Timings for Python 3.6 for extracting the second element from a 2-tuple list.
Also, added numpy
array method, which is simpler to read (but arguably simpler than the list comprehension).
from operator import itemgetter
elements = [(1,1) for _ in range(100000)]
%timeit second = [x[1] for x in elements]
%timeit second = list(map(itemgetter(1), elements))
%timeit second = dict(elements).values()
%timeit second = list(zip(*elements))[1]
%timeit second = np.array(elements)[:,1]
and the timings:
list comprehension: 4.73 ms ± 206 µs per loop
list(map): 5.3 ms ± 167 µs per loop
dict: 2.25 ms ± 103 µs per loop
list(zip) 5.2 ms ± 252 µs per loop
numpy array: 28.7 ms ± 1.88 ms per loop
Note that map()
and zip()
do not return a list anymore, hence the explicit conversion.