List of all unique characters in a string?

Question

I want to append characters to a string, but want to make sure all the letters in the final list are unique.

Example: \"aaabcabccd\" →

Answer 1

char_seen = []
for char in string:
    if char not in char_seen:
        char_seen.append(char)
print(''.join(char_seen))

This will preserve the order in which alphabets are coming,

output will be

abcd

Answer 2

Store Unique characters in list

Method 1:

uniue_char = list(set('aaabcabccd'))
#['a', 'b', 'c', 'd']

Method 2: By Loop ( Complex )

uniue_char = []
for c in 'aaabcabccd':
    if not c in uniue_char:
        uniue_char.append(c)
print(uniue_char)
#['a', 'b', 'c', 'd']

Answer 3

For completeness sake, here's another recipe that sorts the letters as a byproduct of the way it works:

>>> from itertools import groupby
>>> ''.join(k for k, g in groupby(sorted("aaabcabccd")))
'abcd'

Answer 4

I have an idea. Why not use the ascii_lowercase constant?

For example, running the following code:

# string module, contains constant ascii_lowercase which is all the lowercase
# letters of the English alphabet
import string
# Example value of s, a string
s = 'aaabcabccd'
# Result variable to store the resulting string
result = ''
# Goes through each letter in the alphabet and checks how many times it appears.
# If a letter appears at least oce, then it is added to the result variable
for letter in string.ascii_letters:
    if s.count(letter) >= 1:
        result+=letter

# Optional three lines to convert result variable to a list for sorting
# and then back to a string
result = list(result)
result.sort()
result = ''.join(result)

print(result)

Will print 'abcd'

There you go, all duplicates removed and optionally sorted

Answer 5

if the result does not need to be order-preserving, then you can simply use a set

>>> ''.join(set( "aaabcabccd"))
'acbd'
>>>

Answer 6

Use an OrderedDict. This will ensure that the order is preserved

>>> ''.join(OrderedDict.fromkeys( "aaabcabccd").keys())
'abcd'

PS: I just timed both the OrderedDict and Set solution, and the later is faster. If order does not matter, set should be the natural solution, if Order Matter;s this is how you should do.

>>> from timeit import Timer
>>> t1 = Timer(stmt=stmt1, setup="from __main__ import data, OrderedDict")
>>> t2 = Timer(stmt=stmt2, setup="from __main__ import data")
>>> t1.timeit(number=1000)
1.2893918431815337
>>> t2.timeit(number=1000)
0.0632140599081196