Python list comprehension with multiple variables

Question

I am trying to get a list with a specific output from an index in another list, for example:

L = [(0, 1, 2, 3, 4, 5), (6, 7, 8, 9, 10,...etc), (...etc)]
multiple_index = [entry[0, 3, 4] for entry in L] 
#----> I know this specific code is wrong

I would love it if the above code could output:

[(0, 3, 4), (6, 9, 10), (...etc)]

I want the individual sub-indices from each index in the main list to be grouped as shown, if that is at all possible, and am wondering what code I could use to properly pull this off, thanks.

EDIT: Also, How could I format it to display as rows cleanly, I am outputting them to a text file using .writelines and a separate output line, thanks again!

Your edit makes this another question which already has plenty of answers on SO. — Jon Clements
– Jon Clements, Commented Jun 21, 2013 at 19:57
That's interesting, sorry I couldn't find the answer to the original question elsewhere, and the edit is just asking for an additional bit of help, that's all. — ImmortalxR
– ImmortalxR, Commented Jun 21, 2013 at 20:01

Martijn Pieters · Accepted Answer · 2013-06-21 19:44:13Z

9

Use operator.itemgetter():

from operator import itemgetter

multiple_index = map(itemgetter(0, 3, 4), L)

or in a list comprehension:

multiple_index = [itemgetter(0, 3, 4)(i) for i in L]

answered Jun 21, 2013 at 19:44

Martijn Pieters

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5 Comments

ImmortalxR Over a year ago

This works wonderfully, as a quick aside if you could answer this, how could I get the multiple_index to display the output in separate rows, I edited the original post with the new format :)

Martijn Pieters Over a year ago

You can use str.join() to join lists into strings: '\n'.join([',',join(i) for i in multiple_index]) for one string joining the outer lists by newlines, inner lists by comma.

ImmortalxR Over a year ago

It gives me an invalid syntax error when I apply the above code? Any idea why?

Martijn Pieters Over a year ago

@ImmortalxR: Because I accidentally typed a comma where a full-stop goes. You also have integer inputs, so you'd need to map to str first: '\n'.join([','.join(map(str, i)) for i in multiple_index])

ImmortalxR Over a year ago

Thanks! I had a feeling we might have had a typo in there haha

Andrew Clark · Accepted Answer · 2013-06-21 19:43:55Z

3

Here is one option:

L = [(0, 1, 2, 3, 4, 5), (6, 7, 8, 9, 10, 11), (11, 12, 13, 14, 15, 16)]
multiple_index = [(entry[0], entry[3], entry[4]) for entry in L]

Or using operator.itemgetter():

from operator import itemgetter
indices = itemgetter(0, 3, 4)
multiple_index = [indices(entry) for entry in L]

answered Jun 21, 2013 at 19:43

Andrew Clark

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Comments

akonsu · Accepted Answer · 2013-06-21 19:44:39Z

2

Are You interested in this?

L = [(0, 1, 2, 3, 4, 5), (6, 7, 8, 9, 10,...etc), (...etc)]
multiple_index = [(entry[0], entry[3], entry[4]) for entry in L] 
#----> I know this specific code is wrong

edited Jun 21, 2013 at 19:44

akonsu

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answered Jun 21, 2013 at 19:44

oleg

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Comments

JAB · Accepted Answer · 2013-06-21 20:03:01Z

from operator import itemgetter
get = itemgetter(0, 3, 4)
L = [(0, 1, 2, 3, 4, 5), (6, 7, 8, 9, 10,...etc), (...etc)]
multiple_index = [get(entry) for entry in L]

for a more functional style:

multiple_index = map(itemgetter(0, 3, 4), L)

Of course, if you're using numpy, you could do something like the following:

import numpy as np
L = np.array([(0, 1, 2, 3, 4, 5), (6, 7, 8, 9, 10, 11), (11, 12, 13, 14, 15, 16)])
multiple_index = L[:,(0, 3, 4)]

resulting in:

array([[ 0,  3,  4],
       [ 6,  9, 10],
       [11, 14, 15]])

Personally, I like the numpy version the best, but that requires you to have numpy installed. Here's some more on numpy indexing if you're interested: http://docs.scipy.org/doc/numpy/reference/arrays.indexing.html

Numpy also has some neat shortcuts/tricks for fancy slice and range building using np.s_, np.r_, and np.c_.

Jared · Accepted Answer · 2013-06-21 20:47:08Z

2

Just for some diversity, here's a way with itertools.compress,

>>> from itertools import compress, count
>>> indices = {0,3,4}
>>> items_at = lambda indices: (1 if n in indices else 0 for n in count())
>>> [tuple(compress(e, items_at(indices))) for e in L]
[(0, 3, 4), (6, 9, 10)]

edited Jun 21, 2013 at 20:47

answered Jun 21, 2013 at 19:55

Jared

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Comments

Richard · Accepted Answer · 2013-07-23 00:29:31Z

0

list tuple and dictioonary lookups are implemented using their getitem methods

myarray=[0,1,2]
print myarray[1]
#result:1
#equivalent to
print myarray.__getitem__(1)

you can map the indices you want through each list's getitem function. This returns a list containing the items at those indices for each list. Modifying your example code:

L = [(0, 1, 2, 3, 4, 5), (6, 7, 8, 9, 10,...etc), (...etc)]
multiple_index = [map(entry.__getitem__,[0, 3, 4]) for entry in L]

this produces the desired output.

for more on python's magic methods see this:

answered Jul 23, 2013 at 0:29

Richard

1

Comments

Stefan Gruenwald · Accepted Answer · 2014-07-04 06:17:11Z

0

Here is how I would do it:

    L=[tuple(range(0,6*1)),tuple(range(6*1,6*2)),tuple(range(6*2,6*3))]
    print [tuple(map(lambda i: entry[i],[0,3,4])) for entry in L]

answered Jul 4, 2014 at 6:17

Stefan Gruenwald

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