Python Join List Pairs

Given a list of strings. Join the first with the second string, the second with the third, and so on. The one-liner [lst[i] + lst[i+1] for i in range(0, len(lst), 2)] solves the problem by using the range function to iterate over every other index i=0, 2, 4, 5, ... to concatenate the i-th and the i+1-th elements in a list comprehension expression with lst[i] + lst[i+1].

You may already know the normal join function in Python:

Intro: Python Join

Problem: Given a list of elements. How to join the elements by concatenating all elements in the list?

Example: You want to convert list ['learn ', 'python ', 'fast'] to the string 'learn python fast'.

Quick Solution: to convert a list of strings to a string, do the following.

  • Call the ''.join(list) method on the empty string '' that glues together all strings in the list and returns a new string.
  • The string on which you call the join method is used as a delimiter between the list elements.
  • If you don’t need a delimiter, just use the empty string ''.

Code: Let’s have a look at the code.

lst = ['learn ', 'python ', 'fast']
print(''.join(lst))

The output is:

learn python fast

However, what if you want to do something different. Rather than joining all strings in the list to a single string, you want to join the strings in the list in pairs.

Problem: Python Join List Pairs

Problem: Given a list of strings. Join the first with the second string, the second with the third, and so on.

Example: Let’s consider the following minimal example:

['x', 'y', 'v', 'w']

Is there any simple way to pair the first with the second and the third with the fourth string to obtain the following output?

['xy', 'vw']

Note that the length of the strings in the list is variable so the following would be a perfectly acceptable input:

['aaaa', 'b', 'cc', 'dddd', 'eee', 'fff'] 

You can play with all three methods before diving into each of them:

Exercise: What’s the most Pythonic method?

Method 1: Zip() + List Comprehension

You can use the following smart one-liner solution

lst = ['aaaa', 'b', 'cc', 'dddd', 'eee', 'fff']
out = [x + y for x,y in zip(lst[::2], lst[1::2])]
print(out)
# ['aaaab', 'ccdddd', 'eeefff']

The one-liner uses the following strategy:

  • Obtain two slices lst[::2] and lst[1::2] of the original list over every other element starting from the first and the second elements, respectively. If you need to refresh your slicing skills, check out my detailed blog article.
  • Zip the two slices to a sequence of tuples using the zip(...) function. This aligns the first with the second elements from the original list, the third with the forth, and so on. To refresh your zip() skills, check out my blog tutorial here.
  • Use list comprehension to iterate over each pair of values x,y and concatenate them using list concatenation x+y. For a refresher on list comprehension, check out this free tutorial—and for a refresher on list concatenation, check out this one.

Method 2: Iterator + List Comprehension

You can also use an iterator to accomplish this:

lst = ['aaaa', 'b', 'cc', 'dddd', 'eee', 'fff']
it = iter(lst)
out = [x + next(it, '') for x in it] 
print(out)
# ['aaaab', 'ccdddd', 'eeefff']

Here’s the idea:

  • Create an iterator object it using the built-in function iter().
  • Use list comprehension to go over each element in the iterator.
  • Concatenate each element with the return value of calling the next() function on the iterator. This ensures that the iterator moves one position further iterating over the list. So, the next element x won’t be a duplicate.

Method 3: Use List Comprehension with Indexing

This method is the most straightforward one for Python beginners:

lst = ['aaaa', 'b', 'cc', 'dddd', 'eee', 'fff']
out = [lst[i] + lst[i+1] for i in range(0, len(lst), 2)]
print(out)
# ['aaaab', 'ccdddd', 'eeefff']

The idea is simply to use the range function to iterate over every other index i=0, 2, 4, 5, ... to access the i-th and the i+1-th elements at the same time in the expression statement of list comprehension (to concatenate those with lst[i] + lst[i+1]).

Where to Go From Here?

Enough theory, let’s get some practice!

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