blocks|key|614393|text|如果顺序不是问题：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|614394|>>>+foo='mppmt'
>>>+''.join(set(foo))
'pmt'|code-block|syntax|javascript|614395|要保持顺序，请执行以下操作：|614396|>>>+foo='mppmt'
>>>+''.join([j+for+i,j+in+enumerate(foo)+if+j+not+in+foo[:i]])
'mpt'|614397|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

If order is not the matter:

<pre><code>&gt;&gt;&gt; foo='mppmt'
&gt;&gt;&gt; ''.join(set(foo))
'pmt'
</code></pre>

To keep the order:

<pre><code>&gt;&gt;&gt; foo='mppmt'
&gt;&gt;&gt; ''.join([j for i,j in enumerate(foo) if j not in foo[:i]])
'mpt'
</code></pre>

blocks|key|614455|text|如果顺序很重要，|type|unstyled|depth|inlineStyleRanges|entityRanges|data|614456|seen+=+set()
result+=+[]
for+c+in+foo:
++++if+c+not+in+seen:
++++++++result.append(c)
++++++++seen.add(c)
result+=+''.join(result)|code-block|syntax|javascript|614457|或者在没有集合的情况下完成：|614458|result+=+[]
for+c+in+foo:
++++if+c+not+in+result:
++++++++result.append(c)
result+=+''.join(result)|614459|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

If order is important, 

<pre><code>seen = set()
result = []
for c in foo:
 if c not in seen:
 result.append(c)
 seen.add(c)
result = ''.join(result)
</code></pre>

Or to do it without sets:

<pre><code>result = []
for c in foo:
 if c not in result:
 result.append(c)
result = ''.join(result)
</code></pre>

blocks|key|614548|text|正如前面提到的，"".join(set(foo))和collections.OrderedDict就可以了。增加了foo+=+foo.lower()，以防字符串包含大写和小写字符，并且您需要删除所有重复项，无论它们是大写字符还是小写字符。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|614549|from+collections+import+OrderedDict
foo+=+"EugeneEhGhsnaWW"
foo+=+foo.lower()
print+"".join(OrderedDict.fromkeys(foo))|code-block|syntax|javascript|614550|印花真锯片|614551|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

As was mentioned "".join(set(foo)) and collections.OrderedDict will do. 
A added foo = foo.lower() in case the string has upper and lower case characters and you need to remove ALL duplicates no matter if they're upper or lower characters. 

<pre><code>from collections import OrderedDict
foo = "EugeneEhGhsnaWW"
foo = foo.lower()
print "".join(OrderedDict.fromkeys(foo))
</code></pre>

prints eugnhsaw

blocks|key|398832|text|在Python中创建一个列表和一个不允许任何重复的set。Solution1：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|398833|def+fix(string):
++++s+=+set()
++++list+=+[]
++++for+ch+in+string:
++++++++if+ch+not+in+s:
++++++++++++s.add(ch)
++++++++++++list.append(ch)

++++return+''.join(list)++++++++

string+=+"Protiijaayiiii"
print(fix(string))|code-block|syntax|javascript|398834|方法二：|398835|s+=+"Protijayi"

aa+=+[+ch++for+i,+ch+in+enumerate(s)+if+ch+not+in+s[:i]]
print(''.join(aa))|398836|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Create a list in Python and also a set which doesn't allow any duplicates.
Solution1 :

<pre><code>def fix(string):
 s = set()
 list = []
 for ch in string:
 if ch not in s:
 s.add(ch)
 list.append(ch)

 return ''.join(list) 

string = "Protiijaayiiii"
print(fix(string))
</code></pre>

Method 2 :

<pre><code>s = "Protijayi"

aa = [ ch for i, ch in enumerate(s) if ch not in s[:i]]
print(''.join(aa))
</code></pre>

blocks|key|614876|text|def+dupe(str1):
++++s=set(str1)

++++return+"".join(s)
str1='geeksforgeeks'
a=dupe(str1)
print(a)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|614877|如果顺序不重要，则效果很好。|unstyled|614878|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>def dupe(str1):
 s=set(str1)

 return "".join(s)
str1='geeksforgeeks'
a=dupe(str1)
print(a)
</code></pre>

works well if order is not important.

blocks|key|398996|text|d+=+{}
s="YOUR_DESIRED_STRING"
res=[]
for+c+in+s:
++++if+c+not+in+d:
++++++res.append(c)
++++++d[c]=1
print+("".join(res))|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|398997|变量'c‘遍历for循环中的字符串's’，并检查c是否在集合d+(最初没有元素)中，如果c不在d中，则将c附加到字符数组'+res‘中，则集合d的索引c被更改为1。在退出循环后，即c完成遍历字符串以将唯一元素存储在集合d中之后，打印具有所有唯一字符的结果res。|unstyled|398998|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>d = {}
s=&quot;YOUR_DESIRED_STRING&quot;
res=[]
for c in s:
 if c not in d:
 res.append(c)
 d[c]=1
print (&quot;&quot;.join(res))
</code></pre>
variable 'c' traverses through String 's' in the for loop and is checked if c is in a set d (which initially has no element) and if c is not in d, c is appended to the character array 'res' then the index c of set d is changed to 1. after the loop is exited i.e c finishes traversing through the string to store unique elements in set d, the resultant res which has all unique characters is printed.

blocks|key|399021|text|def+remove_duplicates(value):
++++var=""
++++for+i+in+value:
++++++++if+i+in+value:
++++++++++++if+i+in+var:
++++++++++++++++pass
++++++++++++else:
++++++++++++++++var=var%2Bi
++++return+var

print(remove_duplicates("11223445566666ababzzz@@@123#*#*"))|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|399022|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>def remove_duplicates(value):
 var=""
 for i in value:
 if i in value:
 if i in var:
 pass
 else:
 var=var+i
 return var

print(remove_duplicates("11223445566666ababzzz@@@123#*#*"))
</code></pre>

blocks|key|399068|text|由于字符串是一个字符列表，因此将其转换为字典将删除所有重复项并保留顺序。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|399069|"".join(list(dict.fromkeys(foo)))|offset|length|style|CODE|399070|entityMap^0|0|0|X|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|K|8|@$D|L|E|M|F|G]]|9|@]|A|$]]|$1|H|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|I|$]]

As string is a list of characters, converting it to dictionary will remove all duplicates and will retain the order.

<code>"".join(list(dict.fromkeys(foo)))</code>

blocks|key|399074|text|from+collections+import+OrderedDict
def+remove_duplicates(value):
++++++++m=list(OrderedDict.fromkeys(value))
++++++++s=''
++++++++for+i+in+m:
++++++++++++s%2B=i
++++++++return+s
print(remove_duplicates("11223445566666ababzzz@@@123#*#*"))|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|399075|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre class="lang-py prettyprint-override"><code>from collections import OrderedDict
def remove_duplicates(value):
 m=list(OrderedDict.fromkeys(value))
 s=''
 for i in m:
 s+=i
 return s
print(remove_duplicates("11223445566666ababzzz@@@123#*#*"))

</code></pre>

blocks|key|399079|text|+mylist=["ABA",+"CAA",+"ADA"]
+results=[]
+for+item+in+mylist:
+++++buffer=[]
+++++for+char+in+item:
+++++++++if+char+not+in+buffer:
+++++++++++++buffer.append(char)
+++++results.append("".join(buffer))
++++
+print(results)

+output
+ABA
+CAA
+ADA
+['AB',+'CA',+'AD']|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|399080|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code> mylist=[&quot;ABA&quot;, &quot;CAA&quot;, &quot;ADA&quot;]
 results=[]
 for item in mylist:
 buffer=[]
 for char in item:
 if char not in buffer:
 buffer.append(char)
 results.append(&quot;&quot;.join(buffer))
 
 print(results)

 output
 ABA
 CAA
 ADA
 ['AB', 'CA', 'AD']
</code></pre>

blocks|key|615106|text|保持顺序的函数式编程风格：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|615107|import+functools

def+get_unique_char(a,+b):
++++if+b+not+in+a:
++++++++return+a+%2B+b
++++else:
++++++++return+a

if+__name__+==+'__main__':
++++foo+=+'mppmt'

++++gen+=+functools.reduce(get_unique_char,+foo)
++++print(''.join(list(gen)))|code-block|syntax|javascript|615108|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Functional programming style while keeping order:
<pre><code>import functools

def get_unique_char(a, b):
 if b not in a:
 return a + b
 else:
 return a

if __name__ == '__main__':
 foo = 'mppmt'

 gen = functools.reduce(get_unique_char, foo)
 print(''.join(list(gen)))
</code></pre>

blocks|key|615114|text|使用正则表达式：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|615115|import+re
pattern+=+r'(.)\1%2B'+#+(.)+any+character+repeated+(\%2B)+more+than
repl+=+r'\1'++++++++#+replace+it+once
text+=+'shhhhh!!!
re.sub(pattern,repl,text)|code-block|syntax|javascript|615116|输出：|615117|sh!|615118|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Using regular expressions:
<pre><code>import re
pattern = r'(.)\1+' # (.) any character repeated (\+) more than
repl = r'\1' # replace it once
text = 'shhhhh!!!
re.sub(pattern,repl,text)
</code></pre>
output:
<pre><code>sh!
</code></pre>

How can I remove duplicate characters from a string using Python? For example, let's say I have a string:

<pre><code>foo = 'mppmt'
</code></pre>

How can I make the string:

<pre><code>foo = 'mpt'
</code></pre>

NOTE: Order is not important

Removing duplicate characters from a string

Python

如何使用Python删除字符串中的重复字符？例如，假设我有一个字符串：foo = 'mppmt'我怎样才能使字符串：foo = 'mpt'注意:顺序并不重要

问从字符串中删除重复字符
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