Iterating to produce a unique list

Question

This is the initial code:

word_list = ['cat','dog','rabbit']
letter_list = [ ]
for a_word in word_list:
   for a_letter in a_word:
      letter_list.append(a_letter)
print(letter_list)

I need to modify it to produce a list of unique letters.

Could somebody please advise how to do this without using set()

The result should be like this

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

What is the problem with set? Is there any restriction? You can use [numpy unique](http://docs.scipy.org/doc/numpy-1.10.1/reference/generated/numpy.unique.html) — Quazi Marufur Rahman, Feb 25 '16 at 06:26
Without a set or something close to it, you get a complexity of O(n^2) instead of O(n), which is much worse. — Quentin Pradet, Feb 25 '16 at 06:35

Himanshu · Accepted Answer · 2016-02-25T06:35:44.910

2

Only problem that I can see is that you have not checked if the letter is already present in list or not. Try this:

>>> word_list= ['cat', 'dog', 'rabbit']
>>> letter_list= []
>>> for a_word in word_list:
    for a_letter in a_word:
        if a_letter not in letter_list:
            letter_list.append(a_letter)


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

edited Feb 25 '16 at 06:35

answered Feb 25 '16 at 06:27

Himanshu

1,002
9
22

how would i do this without using a function? – user1014691 Feb 25 '16 at 06:32
Edited my answer. Please check. :) – Himanshu Feb 25 '16 at 06:35

Eric · Answer 2 · 2016-02-25T07:12:55.153

1

You can do like this:

>>> word_list = ['cat', 'dog', 'rabbit']
>>> chars = [char for word in word_list for char in list(word)] # combine all chars
>>> letter_list = [ii for n, ii in enumerate(chars) if ii not in chars[:n]] # remove duplicated chars
>>>
>>> print letter_list
['c', 'a', 't', 'd', 'o', 'g', 'r', 'b', 'i']

Hope it helps.

edited Feb 25 '16 at 07:12

answered Feb 25 '16 at 06:42

Eric

2,636
21
25

score 1 · Answer 3 · answered Feb 25 '16 at 08:23

1

Just place this condition : if a_letter not in letter_list after the second for loop.

answered Feb 25 '16 at 08:23

score 0 · Answer 4 · answered Feb 25 '16 at 06:40

Use a dictionary, it is optimized for key based random look-ups. Keep the value as 1 if the key is encountered. Finally, extract all the keys at the end.

unique_chars = {}
word_list = ['cat','dog','rabbit']
for word in word_list:
    for alph in word:
        unique_chars[alph] = 1 #or any other value
letter_list = unique_chars.keys()

score 0 · Answer 5 · answered Feb 25 '16 at 08:17

All you have to do is add a condition :

if a_letter not in letter_list

And add the a_letter is not in the letter_list

The code is as follows :

word_list = ['cat','dog','rabbit']
letter_list = []

for a_word in word_list:
   for a_letter in a_word:
      if a_letter not in letter_list
        letter_list.append(a_letter)

print(letter_list)

The output for this would be :

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

score 0 · Answer 6 · answered Mar 03 '20 at 12:53

0

li = [char for word in word_list for char in word] # create list with individual characters
li_without_duplicates = list(dict.fromkeys(li)) #remove duplicates
print(li_without_duplicates)

answered Mar 03 '20 at 12:53

jay fegade

83
8

Iterating to produce a unique list

6 Answers6