concatenate strings sequentially in python

Question

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"   

Answer 1

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 zip ing 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)

Answer 2

for i in range(0, len(l), 2):
    ''.join(l[i:i + 2])

Answer 3

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))]

Answer 4

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')

Answer 5

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')

Answer 6

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 

Answer 7

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']

Answer 8

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])]