I need to concatenate strings in a list sequentially to join the two parts of words that were separated by line breaks. Can someone help me?
list = ["a", "b", "c", "d"]
Required output:
"ab"
"cd"
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Are the variables string with line break in you want to remove?doctorlove– doctorlove2019-02-28 12:26:07 +00:00Commented Feb 28, 2019 at 12:26
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Is the List Fix contains elementravishankar chavare– ravishankar chavare2019-02-28 12:26:08 +00:00Commented Feb 28, 2019 at 12:26
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8 Answers
Assuming you want to join pairs of consecutive items:
>>> lst = ['a', 'b', 'c', 'd']
>>> list(map(''.join, zip(*([iter(lst)]*2))))
['ab', 'cd']
Here, zip(*([iter(lst)]*2)) is a common recipe for pairs of elements by ziping two instances of the same iterator on the list, but there are many other ways to do the same.
(Note: Renamed list to lst for not shadowing the builtin type)
1 Comment
Natalia Resende
Thank you so much!!
for i in range(0, len(l), 2):
''.join(l[i:i + 2])
2 Comments
Natalia Resende
Thank you again! Is this another method?
I159
It does the job as all the rest. I like it because it is not tricky but just a normal loop.
Using a shared iterator,
>>> [x + next(itr) for itr in [iter(lst)] for x in itr]
['ab', 'cd']
In the forthcoming Python 3.8 (ETA fall 2019), that can be more succinctly written (I believe) as
[x + next(itr) for x in (itr:=iter(lst))]
4 Comments
tobias_k
Nice. Only approach so far that will raise an exception if the list has an odd length instead of just omitting the last item, but a bit nasty to extend to more than two elements.
Natalia Resende
Thank you so much!
tobias_k
BTW, do you know why they don't use e.g.
for x in (iter(lst) as itr), reusing as from import, with and except instead of introducing new :=?
chepner
The decision to support some form of assignment expression was made quickly. The debate over the exact syntax to use dragged on... and on... and on.
as was eventually rejected to avoid adding another set of slightly different semantics on top of what existing uses of as already defined. See PEP 572 for all the gory details.Given a list,
L=['a', 'b', 'c', 'd']
(note, not hijacking the keyword list for a variable name)
you can walk through this in steps of two:
for i in range(0,len(L)-1,2):
print(L[i]+L[i+1])
Use rstrip to remove characters o the right, for example L[i].rstrip('\n')
Comments
You could define a method which groups for you the elements in the list:
def group_by(iterable, n = 2):
if len(iterable)%n != 0: raise Exception("Error: uneven grouping")
# if n < 2: n = 1
i, size = 0, len(iterable)
while i < size-n+1:
yield iterable[i:i+n]
i += n
So for example if your list is:
words = ["a", "b", "c", "d"]
group_by(words, 2) #=> [['a', 'b'], ['c', 'd']]
Or you can group by 3:
words = ["a", "b", "c", "d", "e", "f"]
cons = group_by(words, 3) #=> [['a', 'b', 'c'], ['d', 'e', 'f']]
Then you can use in this way:
res = [ "".join(pair) for pair in group_by(words, 2) ] # maybe you want to add "\n" to the joined string
#=> ['ab', 'cd', 'ef']
The method throws an error if number of elements can not be evenly grouped:
words = ["a", "b", "c", "d", "e"]
cons = group_by(words, 2) #=> Exception: Error: uneven grouping
Comments
I'm sorry that I use most of the time numpy. So a solution can be:
li = ['a', 'b', 'c', 'd']
L = np.array(li)
a1 = np.char.array([L[0], L[2]])
a2 = np.char.array([L[1], L[3]])
a1 + a2
chararray(['ab', 'cd'], dtype='<U2')
Comments
You can also starmap zipped lists to the operator concat:
from operator import concat
from itertools import starmap
l = ["a", "b", "c", "d"]
z = zip(l[::2], l[1::2])
list(starmap(concat, z))
# ['ab', 'cd']
Comments
You could slice your list into the even elements and odd elements.
Let's say your list is called l
Even elements - l[::2]
Odd elements - l[1::2]
Then you could use a list comprehension to concatenate them after zipping. So:
[x + y for x, y in zip(l[::2], l[1::2])]
