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+--- Thread: duplication in list comprehension (/thread-1749.html)
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duplication in list comprehension - Skaperen - Jan-24-2017
if i have
x=['a','b','c','d','e','f']
how can i make duplicates in a list comprehension and get a result like:
['a','a','b','b','c','c','d','d','e','e','f','f']
i tried this and (as expected) it did not work:
Output:lt1/forums /home/forums 40> py3
Python 3.5.2 (default, Nov 17 2016, 17:05:23)
[GCC 5.4.0 20160609] on linux
Type "help", "copyright", "credits" or "license" for more information.
>>> x=['a','b','c','d','e','f']
>>> x
['a', 'b', 'c', 'd', 'e', 'f']
>>> [z z for z in x]
File "<stdin>", line 1
[z z for z in x]
^
SyntaxError: invalid syntax
>>> [z, z for z in x]
File "<stdin>", line 1
[z, z for z in x]
^
SyntaxError: invalid syntax
>>>
lt1/forums /home/forums 41>RE: duplication in list comprehension - Mekire - Jan-24-2017
Well, like this I guess:
x = ['a','b','c','d','e','f'] doubles = [item for pair in zip(x,x) for item in pair] print(doubles)But do you really think that is clearer than this:
doubles = [] for item in x: doubles.extend([item]*2) print(doubles)Matter of taste I guess. Honestly not sure which I prefer.
Also, this I guess:
import itertools x = ['a','b','c','d','e','f'] doubles = list(itertools.chain(*zip(x, x))) print(doubles)Basically it is a list flattening problem.
Lots of stuff here:
http://stackoverflow.com/questions/952914/making-a-flat-list-out-of-list-of-lists-in-python
RE: duplication in list comprehension - wavic - Jan-24-2017
In [11]: x = ['a','b','c','d','e','f'] In [12]: x = x*2 In [13]: x Out[13]: ['a', 'b', 'c', 'd', 'e', 'f', 'a', 'b', 'c', 'd', 'e', 'f'] In [14]: x.sort() In [15]: x Out[15]: ['a', 'a', 'b', 'b', 'c', 'c', 'd', 'd', 'e', 'e', 'f', 'f']
RE: duplication in list comprehension - Mekire - Jan-24-2017
That is definitely a way to do it, though if you have a sequence in which relative order matters but sorting would give a different order it might not be what you want.
For example:
>>> a = [5, 6, 2, 9, 13 ,2] >>> sorted(a*2) [2, 2, 2, 2, 5, 5, 6, 6, 9, 9, 13, 13] >>>Definitely something to consider though.
RE: duplication in list comprehension - wavic - Jan-24-2017
Yes, I am aware but that was not the case. Enyway, if there isn't a sequence in the list, this approach is not going to work
RE: duplication in list comprehension - wavic - Jan-30-2017
This should work in any case. Also, the list can be doubled, tripled, etcetera. The order doesn't matter.
In [1]: from itertools import repeat, tee In [2]: l = list(range(10)) In [3]: l Out[3]: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9] In [4]: it = tee(l, 2) In [5]: doubled = [] In [6]: for i, j in repeat(range(2), len(l)): ...: doubled.extend([next(it[i]), next(it[j])]) ...: In [7]: doubled Out[7]: [0, 0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8, 9, 9]
RE: duplication in list comprehension - Skaperen - Jan-31-2017
i wanted to do this in a comprehension.
but maybe i should not push comprehensions so much, considering this.
RE: duplication in list comprehension - wavic - Jan-31-2017
Well,
In [3]: [doubled.extend(list(i)) for i in zip(l, l)] Out[3]: [None, None, None, None, None, None, None, None, None, None] In [4]: doubled Out[4]: [0, 0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8, 9, 9]works but returns None after any iteration. It's basically the same as in the previous post.
I am not good with list comprehensions. Yet...
Here is a new one.
In [88]: [x for i in zip(l, l) if doubled.extend(list(i)) != None] Out[88]: [] In [89]: doubled Out[89]: [0, 0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8, 9, 9]
RE: duplication in list comprehension - micseydel - Jan-31-2017
Please don't use comprehensions for their side effects D=
RE: duplication in list comprehension - Skaperen - Feb-02-2017
what side-effects of comprehensions do you see being used here? is making duplicate items a side-effect?