List with duplicated values and suffix

Question

I have a list, a:

a = ['a','b','c']

and need to duplicate some values with the suffix _ind added this way (order is important):

['a', 'a_ind', 'b', 'b_ind', 'c', 'c_ind']

I tried:

b = [[x, x + '_ind'] for x in a]
c = [item for sublist in b for item in sublist]
print (c)
['a', 'a_ind', 'b', 'b_ind', 'c', 'c_ind']

Is there some better, more pythonic solution?

For the record, there's nothing wrong with this solution. – cs95 Jul 15 '17 at 20:09 — cs95, Jul 15 '17 at 20:09

score 66 · Accepted Answer · edited Nov 14 '18 at 02:11

66

You could make it a generator:

def mygen(lst):
    for item in lst:
        yield item
        yield item + '_ind'

>>> a = ['a','b','c']
>>> list(mygen(a))
['a', 'a_ind', 'b', 'b_ind', 'c', 'c_ind']

You could also do it with itertools.product, itertools.starmap or itertools.chain or nested comprehensions but in most cases I would prefer a simple to understand, custom generator-function.

With python3.3, you can also use yield from—generator delegation—to make this elegant solution just a bit more concise:

def mygen(lst):
    for item in lst:
        yield from (item, item + '_ind')

edited Nov 14 '18 at 02:11

cs95

379,657
97
704
746

answered Jul 15 '17 at 19:51

MSeifert

145,886
38
333
352

5

Might I suggest a list comprehension version? `list([((yield x), (yield (x + '_ind'))) for x in a]) ; ['a', 'a_ind', 'b', 'b_ind', 'c', 'c_ind']` – cs95 Jul 15 '17 at 20:12
18

@cᴏʟᴅsᴘᴇᴇᴅ Wtf... never seen such "inline" yield. Played with it some more, this works as well: `list([(yield from (x, x + '_ind')) for x in a])`. – Stefan Pochmann Jul 15 '17 at 20:15
2

@StefanPochmann Killer. Want to post an answer with it? If not, I'll edit mine. :p – cs95 Jul 15 '17 at 20:16
@cᴏʟᴅsᴘᴇᴇᴅ You do it. I don't even know how that works :-) – Stefan Pochmann Jul 15 '17 at 20:16
At first I thought there was weird side effect shenanigans happening here, but no...I don't think so...damn..this is nice. – idjaw Jul 15 '17 at 20:18
3

@StefanPochmann It's py3k only. They added yield support for list comps to allow generators to be created. – cs95 Jul 15 '17 at 20:19
1

@idjaw Yeah, python is full of easter eggs like this. I didn't even know yields could exist inside list comprehensions, but lo and behold they can, and they result in generators, not lists :) – cs95 Jul 15 '17 at 20:20
@cᴏʟᴅsᴘᴇᴇᴅ lol. PyCharm is reporting this as a syntax error (but still runs fine) even with a py3 interpreter attached. I think I'll see if there is a bug reported for it already. – idjaw Jul 15 '17 at 20:22
3

@idjaw Oh wait, just saw the red underlines on my IDE too. xD – cs95 Jul 15 '17 at 20:27
This seems to be by far the best way: simple, readable and efficient, no need for any other approaches – Chris_Rands Jul 17 '17 at 08:04

score 29 · Answer 2 · answered Jul 15 '17 at 19:50

29

It can be shortened a little bit by moving the options to the inner for loop in the list comprehension:

a = ['a','b','c']

[item for x in a for item in (x, x + '_ind')]
# ['a', 'a_ind', 'b', 'b_ind', 'c', 'c_ind']

answered Jul 15 '17 at 19:50

Psidom

209,562
33
339
356

cs95 · Answer 3 · 2020-03-05T17:00:20.470

15

Another alternative with splicing (Python2.x, 3.x):

result = [None] * len(a) * 2
result[::2], result[1::2] = a, map(lambda x: x + '_ind', a)

result
# ['a', 'a_ind', 'b', 'b_ind', 'c', 'c_ind']

edited Mar 05 '20 at 17:00

answered Jul 15 '17 at 20:00

cs95

379,657
97
704
746

1

_"Python 3.x only"_ - Actually it's only Python 3.3 and greater. See [here](https://docs.python.org/3/whatsnew/3.3.html#pep-380) – Christian Dean Jul 15 '17 at 21:18
@ChristianDean Thanks. Added the link. – cs95 Jul 15 '17 at 21:20
Do you need both the `list` call and the (list comprehension?) brackets? – Bergi Jul 16 '17 at 05:12
2

@Bergi Yes, because a list comp with yield returns a generator. – cs95 Jul 16 '17 at 07:10
1

This option is most useful if you want to obfuscate your Python code. – Sven Marnach Aug 11 '17 at 15:01
1

@cᴏʟᴅsᴘᴇᴇᴅ No, for code golf Psidom's answer wins. :) – Sven Marnach Aug 11 '17 at 19:17

Ajax1234 · Answer 4 · 2017-07-16T14:05:38.623

6

You can use itertools.chain():

import itertools

l = ['a','b','c']

new_list = list(itertools.chain.from_iterable([[i, i+"_ind"] for i in l]))

print new_list

Output:

['a', 'a_ind', 'b', 'b_ind', 'c', 'c_ind']

edited Jul 16 '17 at 14:05

answered Jul 15 '17 at 19:58

Ajax1234

69,937
8
61
102

4

I would use [`chain.from_iterable`](https://docs.python.org/library/itertools.html#itertools.chain.from_iterable). That way you don't need the unpacking with `*` – MSeifert Jul 15 '17 at 19:59
1

Alt: `list(itertools.chain(*zip(a, [x + '_ind' for x in a])) )` or `list(itertools.chain(*zip(a, map(lambda x: x + '_ind', a))))` – cs95 Jul 15 '17 at 20:03
1

@cᴏʟᴅsᴘᴇᴇᴅ Agreed, but again, `chain.from_iterable` would IMHO be a bit cleaner :) `list(chain.from_iterable(zip(a, [i+'_ind' for i in a])))`. Not that it is particularly important. – miradulo Jul 15 '17 at 20:08

score 5 · Answer 5 · answered Jul 17 '17 at 16:30

5

Before list comprehensions and generators were invented/became widespread, people used to think much simpler¹:

>>> a = ['a', 'b', 'c']
>>> b = []
>>> for x in a: b.extend([x, x+'_ind'])
... 
>>> b
['a', 'a_ind', 'b', 'b_ind', 'c', 'c_ind']

^* I don't mean that those constructs/tools are evil, just wanted to point out that there is a simple solution.

answered Jul 17 '17 at 16:30

Leon

31,443
4
72
97

As far as I know were comprehensions and generators invented (implemented) to make these "constructs and tools" simpler (and definitely a lot faster). But I guess that's a quite opinion-based statement so YMMV. – MSeifert Jul 17 '17 at 16:56

score 4 · Answer 6 · edited Sep 08 '18 at 20:12

4

Since you asked for "simple", I thought I'd throw this in (albeit, maybe not the pythonic way):

for i in mylist: 
    mylist1.append(i);
    mylist1.append(i + '_ind');

edited Sep 08 '18 at 20:12

today

32,602
8
95
115

answered Oct 06 '17 at 09:39

Eladian

958
10
29

List with duplicated values and suffix

6 Answers6

Linked