将字符串列表展平为字符，然后对新列表进行重复数据删除

Question

I have the following and have flattened the list via this documentation

>>> wordlist = ['cat','dog','rabbit']
>>> letterlist = [lt for wd in wordlist for lt in wd]
>>> print(letterlist)
['c', 'a', 't', 'd', 'o', 'g', 'r', 'a', 'b', 'b', 'i', 't']

Can the list comprehension be extended to remove duplicate characters. The desired result is the following (in any order):

['a', 'c', 'b', 'd', 'g', 'i', 'o', 'r', 't']

I can convert to a set and then back to a list but I'd prefer to keep it as a list.

Answer 1

Easiest is to use a set comprehension instead of a list comp:

letterlist = {lt for wd in wordlist for lt in wd}

All I did was replace the square brackets with curly braces. This works in Python 2.7 and up.

For Python 2.6 and earlier, you'd use the set() callable with a generator expression instead:

letterlist = set(lt for wd in wordlist for lt in wd)

Last, but not least, you can replace the comprehension syntax altogether by producing the letters from all sequences by chaining the strings together, treat them all like one long sequence, with itertools.chain.from_iterable() ; you give that a sequence of sequences and it'll give you back one long sequence:

from itertools import chain
letterlist = set(chain.from_iterable(wordlist))

Answer 2

Sets are an easy way to get unique elements from an iterable. To flatten a list of lists, itertools.chain provides a handy way to do that.

from itertools import chain

>>> set(chain.from_iterable(['cat','dog','rabbit'])
{'a', 'b', 'c', 'd', 'g', 'i', 'o', 'r', 't'}

Answer 3

I think set comprehension should be used

wordlist = ['cat','dog','rabbit']
letterlist = {lt for wd in wordlist for lt in wd}
print(letterlist)

this will work only in python 2.7 and higher for previous versions use set instead of {}

wordlist = ['cat','dog','rabbit']
letterlist = set(lt for wd in wordlist for lt in wd)
print(letterlist)