List with duplicated values and suffix
37
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']
I think my solution is a bit over-complicated. Is there some better, more pythonic solution?
python list list-comprehension suffix
edited Jul 15 '17 at 20:46
Christian Dean
14.9k52655
asked Jul 15 '17 at 19:45
jezraeljezrael
329k23271350
add a comment |
37
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']
I think my solution is a bit over-complicated. Is there some better, more pythonic solution?
python list list-comprehension suffix
edited Jul 15 '17 at 20:46
Christian Dean
14.9k52655
asked Jul 15 '17 at 19:45
jezraeljezrael
329k23271350
6
For the record, there's nothing wrong with this solution.
– coldspeed
Jul 15 '17 at 20:09
add a comment |
37
37
37
9
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']
I think my solution is a bit over-complicated. Is there some better, more pythonic solution?
python list list-comprehension suffix
edited Jul 15 '17 at 20:46
Christian Dean
14.9k52655
asked Jul 15 '17 at 19:45
jezraeljezrael
329k23271350
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']
I think my solution is a bit over-complicated. Is there some better, more pythonic solution?
python list list-comprehension suffix
python list list-comprehension suffix
edited Jul 15 '17 at 20:46
Christian Dean
14.9k52655
asked Jul 15 '17 at 19:45
jezraeljezrael
329k23271350
edited Jul 15 '17 at 20:46
Christian Dean
14.9k52655
asked Jul 15 '17 at 19:45
jezraeljezrael
329k23271350
edited Jul 15 '17 at 20:46
Christian Dean
14.9k52655
edited Jul 15 '17 at 20:46
Christian Dean
14.9k52655
edited Jul 15 '17 at 20:46
Christian Dean
14.9k52655
14.9k52655
asked Jul 15 '17 at 19:45
jezraeljezrael
329k23271350
asked Jul 15 '17 at 19:45
jezraeljezrael
329k23271350
asked Jul 15 '17 at 19:45
jezraeljezrael
329k23271350
329k23271350
6
For the record, there's nothing wrong with this solution.
– coldspeed
Jul 15 '17 at 20:09
add a comment |
6
For the record, there's nothing wrong with this solution.
– coldspeed
Jul 15 '17 at 20:09
6
6
For the record, there's nothing wrong with this solution.
– coldspeed
Jul 15 '17 at 20:09
For the record, there's nothing wrong with this solution.
– coldspeed
Jul 15 '17 at 20:09
add a comment |
6 Answers
6
active
oldest
votes
52
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 2:11
coldspeed
127k23128214
answered Jul 15 '17 at 19:51
MSeifertMSeifert
74.8k17145178
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']
– coldspeed
Jul 15 '17 at 20:12
16
@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
– coldspeed
Jul 15 '17 at 20:16
3
@StefanPochmann It's py3k only. They added yield support for list comps to allow generators to be created.
– coldspeed
Jul 15 '17 at 20:19
3
@idjaw Oh wait, just saw the red underlines on my IDE too. xD
– coldspeed
Jul 15 '17 at 20:27
|
show 5 more comments
26
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
PsidomPsidom
123k1285127
add a comment |
14
Another alternative with splicing (Python2.x, 3.x):
In [642]: result = [None] * len(a) * 2
In [643]: result[::2], result[1::2] = a, map(lambda x: x + '_ind', a)
In [644]: result
Out[644]: ['a', 'a_ind', 'b', 'b_ind', 'c', 'c_ind']
answered Jul 15 '17 at 20:00
coldspeedcoldspeed
127k23128214
1
"Python 3.x only" - Actually it's only Python 3.3 and greater. See here
– Christian Dean
Jul 15 '17 at 21:18
2
@Bergi Yes, because a list comp with yield returns a generator.
– coldspeed
Jul 16 '17 at 7:10
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
1
@coldspeed - thanks.
– jezrael
Jan 8 at 10:34
|
show 3 more comments
5
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']
answered Jul 15 '17 at 19:58
Ajax1234Ajax1234
41.1k42853
4
I would usechain.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])) )orlist(itertools.chain(*zip(a, map(lambda x: x + '_ind', a))))
– coldspeed
Jul 15 '17 at 20:03
1
@cᴏʟᴅsᴘᴇᴇᴅ Agreed, but again,chain.from_iterablewould 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
add a comment |
3
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
LeonLeon
20.8k23270
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
add a comment |
3
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 8 '18 at 20:12
today
10.7k21837
answered Oct 6 '17 at 9:39
EladianEladian
492316
add a comment |
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
52
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 2:11
coldspeed
127k23128214
answered Jul 15 '17 at 19:51
MSeifertMSeifert
74.8k17145178
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']
– coldspeed
Jul 15 '17 at 20:12
16
@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
– coldspeed
Jul 15 '17 at 20:16
3
@StefanPochmann It's py3k only. They added yield support for list comps to allow generators to be created.
– coldspeed
Jul 15 '17 at 20:19
3
@idjaw Oh wait, just saw the red underlines on my IDE too. xD
– coldspeed
Jul 15 '17 at 20:27
|
show 5 more comments
52
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 2:11
coldspeed
127k23128214
answered Jul 15 '17 at 19:51
MSeifertMSeifert
74.8k17145178
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']
– coldspeed
Jul 15 '17 at 20:12
16
@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
– coldspeed
Jul 15 '17 at 20:16
3
@StefanPochmann It's py3k only. They added yield support for list comps to allow generators to be created.
– coldspeed
Jul 15 '17 at 20:19
3
@idjaw Oh wait, just saw the red underlines on my IDE too. xD
– coldspeed
Jul 15 '17 at 20:27
|
show 5 more comments
52
52
52
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 2:11
coldspeed
127k23128214
answered Jul 15 '17 at 19:51
MSeifertMSeifert
74.8k17145178
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 2:11
coldspeed
127k23128214
answered Jul 15 '17 at 19:51
MSeifertMSeifert
74.8k17145178
edited Nov 14 '18 at 2:11
coldspeed
127k23128214
edited Nov 14 '18 at 2:11
coldspeed
127k23128214
edited Nov 14 '18 at 2:11
coldspeed
127k23128214
127k23128214
answered Jul 15 '17 at 19:51
MSeifertMSeifert
74.8k17145178
answered Jul 15 '17 at 19:51
MSeifertMSeifert
74.8k17145178
answered Jul 15 '17 at 19:51
MSeifertMSeifert
74.8k17145178
74.8k17145178
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']
– coldspeed
Jul 15 '17 at 20:12
16
@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
– coldspeed
Jul 15 '17 at 20:16
3
@StefanPochmann It's py3k only. They added yield support for list comps to allow generators to be created.
– coldspeed
Jul 15 '17 at 20:19
3
@idjaw Oh wait, just saw the red underlines on my IDE too. xD
– coldspeed
Jul 15 '17 at 20:27
|
show 5 more comments
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']
– coldspeed
Jul 15 '17 at 20:12
16
@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
– coldspeed
Jul 15 '17 at 20:16
3
@StefanPochmann It's py3k only. They added yield support for list comps to allow generators to be created.
– coldspeed
Jul 15 '17 at 20:19
3
@idjaw Oh wait, just saw the red underlines on my IDE too. xD
– coldspeed
Jul 15 '17 at 20:27
5
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']– coldspeed
Jul 15 '17 at 20:12
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']– coldspeed
Jul 15 '17 at 20:12
16
16
@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
@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
2
@StefanPochmann Killer. Want to post an answer with it? If not, I'll edit mine. :p
– coldspeed
Jul 15 '17 at 20:16
@StefanPochmann Killer. Want to post an answer with it? If not, I'll edit mine. :p
– coldspeed
Jul 15 '17 at 20:16
3
3
@StefanPochmann It's py3k only. They added yield support for list comps to allow generators to be created.
– coldspeed
Jul 15 '17 at 20:19
@StefanPochmann It's py3k only. They added yield support for list comps to allow generators to be created.
– coldspeed
Jul 15 '17 at 20:19
3
3
@idjaw Oh wait, just saw the red underlines on my IDE too. xD
– coldspeed
Jul 15 '17 at 20:27
@idjaw Oh wait, just saw the red underlines on my IDE too. xD
– coldspeed
Jul 15 '17 at 20:27
|
show 5 more comments
26
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
PsidomPsidom
123k1285127
add a comment |
26
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
PsidomPsidom
123k1285127
add a comment |
26
26
26
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
PsidomPsidom
123k1285127
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
PsidomPsidom
123k1285127
answered Jul 15 '17 at 19:50
PsidomPsidom
123k1285127
answered Jul 15 '17 at 19:50
PsidomPsidom
123k1285127
answered Jul 15 '17 at 19:50
PsidomPsidom
123k1285127
123k1285127
add a comment |
add a comment |
14
Another alternative with splicing (Python2.x, 3.x):
In [642]: result = [None] * len(a) * 2
In [643]: result[::2], result[1::2] = a, map(lambda x: x + '_ind', a)
In [644]: result
Out[644]: ['a', 'a_ind', 'b', 'b_ind', 'c', 'c_ind']
answered Jul 15 '17 at 20:00
coldspeedcoldspeed
127k23128214
1
"Python 3.x only" - Actually it's only Python 3.3 and greater. See here
– Christian Dean
Jul 15 '17 at 21:18
2
@Bergi Yes, because a list comp with yield returns a generator.
– coldspeed
Jul 16 '17 at 7:10
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
1
@coldspeed - thanks.
– jezrael
Jan 8 at 10:34
|
show 3 more comments
14
Another alternative with splicing (Python2.x, 3.x):
In [642]: result = [None] * len(a) * 2
In [643]: result[::2], result[1::2] = a, map(lambda x: x + '_ind', a)
In [644]: result
Out[644]: ['a', 'a_ind', 'b', 'b_ind', 'c', 'c_ind']
answered Jul 15 '17 at 20:00
coldspeedcoldspeed
127k23128214
1
"Python 3.x only" - Actually it's only Python 3.3 and greater. See here
– Christian Dean
Jul 15 '17 at 21:18
2
@Bergi Yes, because a list comp with yield returns a generator.
– coldspeed
Jul 16 '17 at 7:10
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
1
@coldspeed - thanks.
– jezrael
Jan 8 at 10:34
|
show 3 more comments
14
14
14
Another alternative with splicing (Python2.x, 3.x):
In [642]: result = [None] * len(a) * 2
In [643]: result[::2], result[1::2] = a, map(lambda x: x + '_ind', a)
In [644]: result
Out[644]: ['a', 'a_ind', 'b', 'b_ind', 'c', 'c_ind']
answered Jul 15 '17 at 20:00
coldspeedcoldspeed
127k23128214
Another alternative with splicing (Python2.x, 3.x):
In [642]: result = [None] * len(a) * 2
In [643]: result[::2], result[1::2] = a, map(lambda x: x + '_ind', a)
In [644]: result
Out[644]: ['a', 'a_ind', 'b', 'b_ind', 'c', 'c_ind']
answered Jul 15 '17 at 20:00
coldspeedcoldspeed
127k23128214
edited Oct 13 '17 at 7:16
answered Jul 15 '17 at 20:00
coldspeedcoldspeed
127k23128214
answered Jul 15 '17 at 20:00
coldspeedcoldspeed
127k23128214
answered Jul 15 '17 at 20:00
coldspeedcoldspeed
127k23128214
127k23128214
1
"Python 3.x only" - Actually it's only Python 3.3 and greater. See here
– Christian Dean
Jul 15 '17 at 21:18
2
@Bergi Yes, because a list comp with yield returns a generator.
– coldspeed
Jul 16 '17 at 7:10
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
1
@coldspeed - thanks.
– jezrael
Jan 8 at 10:34
|
show 3 more comments
1
"Python 3.x only" - Actually it's only Python 3.3 and greater. See here
– Christian Dean
Jul 15 '17 at 21:18
2
@Bergi Yes, because a list comp with yield returns a generator.
– coldspeed
Jul 16 '17 at 7:10
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
1
@coldspeed - thanks.
– jezrael
Jan 8 at 10:34
1
1
"Python 3.x only" - Actually it's only Python 3.3 and greater. See here
– Christian Dean
Jul 15 '17 at 21:18
"Python 3.x only" - Actually it's only Python 3.3 and greater. See here
– Christian Dean
Jul 15 '17 at 21:18
2
2
@Bergi Yes, because a list comp with yield returns a generator.
– coldspeed
Jul 16 '17 at 7:10
@Bergi Yes, because a list comp with yield returns a generator.
– coldspeed
Jul 16 '17 at 7:10
1
1
This option is most useful if you want to obfuscate your Python code.
– Sven Marnach
Aug 11 '17 at 15:01
This option is most useful if you want to obfuscate your Python code.
– Sven Marnach
Aug 11 '17 at 15:01
1
1
@cᴏʟᴅsᴘᴇᴇᴅ No, for code golf Psidom's answer wins. :)
– Sven Marnach
Aug 11 '17 at 19:17
@cᴏʟᴅsᴘᴇᴇᴅ No, for code golf Psidom's answer wins. :)
– Sven Marnach
Aug 11 '17 at 19:17
1
1
@coldspeed - thanks.
– jezrael
Jan 8 at 10:34
@coldspeed - thanks.
– jezrael
Jan 8 at 10:34
|
show 3 more comments
5
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']
answered Jul 15 '17 at 19:58
Ajax1234Ajax1234
41.1k42853
4
I would usechain.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])) )orlist(itertools.chain(*zip(a, map(lambda x: x + '_ind', a))))
– coldspeed
Jul 15 '17 at 20:03
1
@cᴏʟᴅsᴘᴇᴇᴅ Agreed, but again,chain.from_iterablewould 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
add a comment |
5
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']
answered Jul 15 '17 at 19:58
Ajax1234Ajax1234
41.1k42853
4
I would usechain.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])) )orlist(itertools.chain(*zip(a, map(lambda x: x + '_ind', a))))
– coldspeed
Jul 15 '17 at 20:03
1
@cᴏʟᴅsᴘᴇᴇᴅ Agreed, but again,chain.from_iterablewould 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
add a comment |
5
5
5
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']
answered Jul 15 '17 at 19:58
Ajax1234Ajax1234
41.1k42853
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']
answered Jul 15 '17 at 19:58
Ajax1234Ajax1234
41.1k42853
edited Jul 16 '17 at 14:05
answered Jul 15 '17 at 19:58
Ajax1234Ajax1234
41.1k42853
answered Jul 15 '17 at 19:58
Ajax1234Ajax1234
41.1k42853
answered Jul 15 '17 at 19:58
Ajax1234Ajax1234
41.1k42853
41.1k42853
4
I would usechain.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])) )orlist(itertools.chain(*zip(a, map(lambda x: x + '_ind', a))))
– coldspeed
Jul 15 '17 at 20:03
1
@cᴏʟᴅsᴘᴇᴇᴅ Agreed, but again,chain.from_iterablewould 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
add a comment |
4
I would usechain.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])) )orlist(itertools.chain(*zip(a, map(lambda x: x + '_ind', a))))
– coldspeed
Jul 15 '17 at 20:03
1
@cᴏʟᴅsᴘᴇᴇᴅ Agreed, but again,chain.from_iterablewould 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
4
4
I would use
chain.from_iterable. That way you don't need the unpacking with *– MSeifert
Jul 15 '17 at 19:59
I would use
chain.from_iterable. That way you don't need the unpacking with *– MSeifert
Jul 15 '17 at 19:59
1
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))))– coldspeed
Jul 15 '17 at 20:03
Alt:
list(itertools.chain(*zip(a, [x + '_ind' for x in a])) ) or list(itertools.chain(*zip(a, map(lambda x: x + '_ind', a))))– coldspeed
Jul 15 '17 at 20:03
1
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
@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
add a comment |
3
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
LeonLeon
20.8k23270
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
add a comment |
3
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
LeonLeon
20.8k23270
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
add a comment |
3
3
3
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
LeonLeon
20.8k23270
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
LeonLeon
20.8k23270
answered Jul 17 '17 at 16:30
LeonLeon
20.8k23270
answered Jul 17 '17 at 16:30
LeonLeon
20.8k23270
answered Jul 17 '17 at 16:30
LeonLeon
20.8k23270
20.8k23270
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
add a comment |
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
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
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
add a comment |
3
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 8 '18 at 20:12
today
10.7k21837
answered Oct 6 '17 at 9:39
EladianEladian
492316
add a comment |
3
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 8 '18 at 20:12
today
10.7k21837
answered Oct 6 '17 at 9:39
EladianEladian
492316
add a comment |
3
3
3
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 8 '18 at 20:12
today
10.7k21837
answered Oct 6 '17 at 9:39
EladianEladian
492316
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 8 '18 at 20:12
today
10.7k21837
answered Oct 6 '17 at 9:39
EladianEladian
492316
edited Sep 8 '18 at 20:12
today
10.7k21837
edited Sep 8 '18 at 20:12
today
10.7k21837
edited Sep 8 '18 at 20:12
today
10.7k21837
10.7k21837
answered Oct 6 '17 at 9:39
EladianEladian
492316
answered Oct 6 '17 at 9:39
EladianEladian
492316
answered Oct 6 '17 at 9:39
EladianEladian
492316
492316
add a comment |
add a comment |
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6
For the record, there's nothing wrong with this solution.
– coldspeed
Jul 15 '17 at 20:09