Python 字符串中所有唯一字符的列表？
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List of all unique characters in a string?
提问by Ali
I want to append characters to a string, but want to make sure all the letters in the final list are unique.
我想将字符附加到字符串，但要确保最终列表中的所有字母都是唯一的。
Example: "aaabcabccd"→ "abcd"
示例："aaabcabccd"→"abcd"
Now of course I have two solutions in my mind. One is using a listthat will map the characters with their ASCII codes. So whenever I encounter a letter it will set the index to True. Afterwards I will scan the list and append all the ones that were set. It will have a time complexity of O(n).
现在我当然有两个解决方案。一种是使用list将字符与其 ASCII 代码进行映射。因此，每当我遇到一个字母时，它都会将索引设置为True. 之后我将扫描列表并附加所有设置的列表。它将具有O(n)的时间复杂度。
Another solution would be using a dictand following the same procedure. After mapping every char, I will do the operation for each key in the dictionary. This will have a linearrunning time as well.
另一种解决方案是使用 adict并遵循相同的程序。映射完每个字符后，我将对字典中的每个键进行操作。这也将具有线性运行时间。
Since I am a Python newbie, I was wondering which would be more space efficient. Which one could be implemented more efficiently?
由于我是 Python 新手，我想知道哪个更节省空间。哪一个可以更有效地实施？
PS: Order is notimportant while creating the list.
PS：创建列表时顺序并不重要。
采纳答案by NPE
The simplest solution is probably:
最简单的解决方案可能是：
In [10]: ''.join(set('aaabcabccd'))
Out[10]: 'acbd'
Note that this doesn't guarantee the order in which the letters appear in the output, even though the example might suggest otherwise.
请注意，这并不能保证字母在输出中出现的顺序，即使示例可能另有建议。
You refer to the output as a "list". If a list is what you really want, replace ''.joinwith list:
您将输出称为“列表”。如果列表是您真正想要的，请替换''.join为list：
In [1]: list(set('aaabcabccd'))
Out[1]: ['a', 'c', 'b', 'd']
As far as performance goes, worrying about it at this stage sounds like premature optimization.
就性能而言，现阶段担心它听起来像是过早的优化。
回答by gefei
if the result does not need to be order-preserving, then you can simply use a set
如果结果不需要保序，那么你可以简单地使用一个集合
>>> ''.join(set( "aaabcabccd"))
'acbd'
>>>
回答by Abhijit
Use an OrderedDict. This will ensure that the order is preserved
使用OrderedDict。这将确保订单被保留
>>> ''.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.
PS：我只是对 OrderedDict 和 Set 解决方案进行了计时，后者更快。如果顺序无关紧要， set 应该是自然的解决方案，如果 Order Matter;s 这就是你应该做的。
>>> 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
回答by martineau
For completeness sake, here's another recipe that sorts the letters as a byproduct of the way it works:
为了完整起见，这是另一个将字母作为其工作方式的副产品进行排序的方法：
>>> from itertools import groupby
>>> ''.join(k for k, g in groupby(sorted("aaabcabccd")))
'abcd'
回答by Brent Pappas
I have an idea. Why not use the ascii_lowercaseconstant?
我有个主意。为什么不使用ascii_lowercase常量？
For example, running the following code:
例如，运行以下代码：
# 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'
会打印 'abcd'
There you go, all duplicates removed and optionally sorted
好了，所有重复项都已删除并可选择排序
回答by dipenparmar12
Store Unique characters in list
在列表中存储唯一字符
Method 1:
方法一：
uniue_char = list(set('aaabcabccd'))
#['a', 'b', 'c', 'd']
Method 2: By Loop ( Complex )
方法 2：按循环（复杂）
uniue_char = []
for c in 'aaabcabccd':
if not c in uniue_char:
uniue_char.append(c)
print(uniue_char)
#['a', 'b', 'c', 'd']
回答by Amit Gupta
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,
这将保留字母出现的顺序，
output will be
输出将是
abcd
