Making my dictionary comprehension more efficient

user_17580381 0
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I have a list of strings. For example:

lst = ['aa bb cc', 'dd ee ff gg']

Each string in the list is known to contain 2 or more whitespace delimited tokens.

I want to build a dictionary keyed by the last token with the first token as its value.

The following dictionary comprehension achieves this:

d = {e.split()[-1]: e.split()[0] for e in lst}

This gives me:

{'cc': 'aa', 'gg': 'dd'}

...which is exactly what I want.

However, this means that the element e will have its split() function called twice per iteration over lst.

I can't help thinking that there must be a way to avoid this but I just can't figure it out.

Any ideas?

python
4 Answers

Using map:

d = {v[-1]: v[0] for v in map(str.split, lst)}
user_122990000

You can use walrus operator. But list comprehension is not meant to efficient, it's just shorter(I think)

lst = ['aa bb cc', 'dd ee ff gg']

d={j[-1]:j[0] for i in lst if (j:=i.split())}
print(d)
user_117550180

If you use a function it would be much cleaner as you are carrying out an additional instruction, not just creating a list. Since you are looking to create a single holder of split values you will need to ensure there is some way to access index 0 or -1.

But it can be done like so:

d = {e[0]: e[-1] for e in [item.split() for item in lst]}

Should work for you.

Updated answer for python 3.8 +

I just recalled a recent addition on this. You can also achieve this without the inner comprehension:

d = {parts[0]: parts[-1] for e in lst if (parts := e.split())}

https://www.python.org/dev/peps/pep-0572/

user_73984760

As an alternative to other answers, you can make use of unpacking to get rid of the indexing, IF the elements in your list are always going to have at least 2 strings separated by spaces (which as you say, they do), otherwise this won't work:

d = {last: first for first, *_, last in map(str.split, lst)}
print(d)
# {'cc': 'aa', 'gg': 'dd'}
user_36254770
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