Antonio Caminero Garcia writes:
> On Saturday, March 26, 2016 at 11:12:58 PM UTC+1, beli...@aol.com wrote:
>> I can create a list that has repeated elements of another list as
>> follows:
>>
>> xx = ["a","b"]
>> nrep = 3
>> print xx
>> yy = []
>> for aa in xx:
>> for i in range(nrep):
>> yy.append(aa)
>> print yy
>>
>> output:
>> ['a', 'b']
>> ['a', 'a', 'a', 'b', 'b', 'b']
>>
>> Is there a one-liner to create a list with repeated elements?
>
> What about this?
>
> def rep_elements(sequence, nrep):
> #return [ritem for item in sequence for ritem in [item]*nrep]
> return list(chain.from_iterable(([item]*nrep for item in sequence)))
>
> sequence = ['h','o','l','a']
> print(rep_elements(sequence, 3))
A thing to know about the comprehension-syntaxes is that they correspond
precisely to nested loops (and conditions, but conditions don't appear
in the present example) with an .append inside.
xx = "ab"
nrep = 3
print([ aa for aa in xx for i in range(nrep) ])
(This has been posted in this thread a few times already, but I think
the systematic correspondence to the original loops was left unstated.
Apologies in advance if I missed something.)
The resulting list has some hidden name. The original loops should be
re-mentalized for an even closer correspondence as follows.
g47 = []
for aa in xx: # Loopy ...
for i in range(nrep): # ... do!
g47.append(aa) # <-- _This_ aa is one of the result items.
yy = g47
A thing about range objects is that they can be reused, so the present
example could also reuse just one.
xx = "ab"
reps = range(3)
print([ aa for aa in xx for i in reps ])
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