How to check the condition of a list during list comprehension?

Question

Say I have a list of three words and I want to generate a list containing the characters in the word.

This can be achieved easily using for-loops or list-comprehension:

wordList = ["one", "two", "three"]
letterList = []

# Solution 1:

for word in wordList:
    for letter in word:
        letterList.append(letter)

# Solution 2:

letterList = [letter for word in wordList for letter in word]

print(letterList) # Prints: ['o', 'n', 'e', 't', 'w', 'o', 't', 'h', 'r', 'e', 'e']

However, when this problem is made more difficult, I am unable to find a solution using a list-comprehension.

For example, assume I want to generate the previous list, but I also want any duplicate characters to be removed.

I can do this using for-loops:

wordList = ["one", "two", "three"]
letterList = []

for word in wordList:
    for letter in word:
        if letter not in letterList:
            letterList.append(letter)

print(letterList) # Prints: ['o', 'n', 'e', 't', 'w', 'h', 'r']

So, my question is: How do I check to see if an item is in the list being build during a list comprehension? Or put more simply: How do I do the above using list-comprehensions?

Answer 1

In Python 3.6 dictionaries are ordered. Therefore you can use a dict comprehension:

>>> list({letter:None for word in wordList for letter in word}.keys())
['o', 'n', 'e', 't', 'w', 'h', 'r']

Starting from Python 3.7 the ordering will be guaranteed:

python news: 😀 @gvanrossum just pronounced that dicts are now guaranteed to retain insertion order. This is the end of a long journey. source (2017-12-15)

Answer 2

if order is not important just use a set comprehension

letter_set = {letter for word in wordList for letter in word}

if order is important it gets a bit grosser

seen = set()
letter_list = [x for word in wordlist for x in word if x not in seen and not seen.add(s)]

or using something like ordered dict

from collections import OrderedDict
my_list = OrderedDict.fromkeys(letter for word in wordList for letter in word).keys()

Answer 3

You already got the set solution but here is one different solution :

Instead of using that list comprehension result , use that list comprehension for logic and append it to new list :

wordList = ["one", "two", "three"]
letterList = []
[letterList.append(letter) for word in wordList for letter in word if letter not in letterList]
print(letterList)

output:

['o', 'n', 'e', 't', 'w', 'h', 'r']