如何从元组列表中提取第n个元素？

I’m trying to obtain the n-th elements from a list of tuples.

I have something like:

elements = [(1,1,1),(2,3,7),(3,5,10)]

I wish to extract only the second elements of each tuple into a list:

seconds = [1, 3, 5]

I know that it could be done with a for loop but I wanted to know if there’s another way since I have thousands 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.

map (lambda x:(x[1]),elements)

Using islice and chain.from_iterable:

>>> from itertools import chain, islice
>>> elements = [(1,1,1),(2,3,7),(3,5,10)]
>>> list(chain.from_iterable(islice(item, 1, 2) for item in elements))
[1, 3, 5]

This can be useful when you need more than one element:

>>> elements = [(0, 1, 2, 3, 4, 5), 
                (10, 11, 12, 13, 14, 15), 
                (20, 21, 22, 23, 24, 25)]
>>> list(chain.from_iterable(islice(tuple_, 2, 5) for tuple_ in elements))
[2, 3, 4, 12, 13, 14, 22, 23, 24]