blocks|key|43906|text|你可以只做(a,)。不需要使用函数。(请注意，逗号是必需的。)|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|43907|本质上，tuple(a)意味着创建a内容的元组，而不仅仅是由a本身组成的元组。字符串的“内容”(当您迭代它时得到的内容)是它的字符，这就是它被拆分成字符的原因。|43908|entityMap^0|5|4|0|4|8|H|1|U|1|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|H|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|I|$]]

You can just do <code>(a,)</code>. No need to use a function. (Note that the comma is necessary.)

Essentially, <code>tuple(a)</code> means to make a tuple of the contents of <code>a</code>, not a tuple consisting of just <code>a</code> itself. The "contents" of a string (what you get when you iterate over it) are its characters, which is why it is split into characters.

blocks|key|2729414|text|我使用此函数将字符串转换为元组|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2729415|import+ast

def+parse_tuple(string):
++++try:
++++++++s+=+ast.literal_eval(str(string))
++++++++if+type(s)+==+tuple:
++++++++++++return+s
++++++++return
++++except:
++++++++return|code-block|syntax|javascript|2729416|用法|2729417|parse_tuple('("A","B","C",)')++#+Result:+('A',+'B',+'C')|2729418|在您的情况下，您需要|2729419|value+=+parse_tuple("('%25s',)"+%25+a)|2729420|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

I use this function to convert string to tuple

<pre><code>import ast

def parse_tuple(string):
 try:
 s = ast.literal_eval(str(string))
 if type(s) == tuple:
 return s
 return
 except:
 return
</code></pre>

Usage

<pre><code>parse_tuple('("A","B","C",)') # Result: ('A', 'B', 'C')
</code></pre>

 
In your case, you do

<pre><code>value = parse_tuple("('%s',)" % a)
</code></pre>

blocks|key|2729372|text|以防有人想知道如何创建一个元组，将字符串"Quattro“和"TT”的每一部分赋给列表的一个元素，它就像这样的print+tuple(a.split())|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2729373|entityMap^0|1J|M|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

Just in case someone comes here trying to know how to create a tuple assigning each part of the string "Quattro" and "TT" to an element of the list, it would be like this 
<code>print tuple(a.split())</code>

blocks|key|2519713|text|您可以使用以下解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2519714|s="jack"

tup=tuple(s.split("+"))

output=('jack')|code-block|syntax|javascript|2519715|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|$]]

You can use the following solution:

<pre><code>s="jack"

tup=tuple(s.split(" "))

output=('jack')
</code></pre>

blocks|key|44206|text|您可以使用eval()|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|44207|>>>+a+=+['Quattro+TT']++
>>>+eval(str(a))
['Quattro+TT']|code-block|syntax|javascript|44208|entityMap^0|5|6|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

You can use <code>eval()</code>
<pre><code>&gt;&gt;&gt; a = ['Quattro TT'] 
&gt;&gt;&gt; eval(str(a))
['Quattro TT']
</code></pre>

blocks|key|43915|text|对元组进行子类化，其中一些子类实例可能需要是单字符串实例，这会抛出一些有趣的东西。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|43916|class+Sequence(+tuple+):
++++def+__init__(+self,+*args+):
++++++++#+initialisation...
++++++++self.instances+=+[]

++++def+__new__(+cls,+*args+):
++++++++for+arg+in+args:
++++++++++++assert+isinstance(+arg,+unicode+),+'#+arg+%25s+not+unicode'+%25+(+arg,+)
++++++++if+len(+args+)+==+1:
++++++++++++seq+=+super(+Sequence,+cls+).__new__(+cls,+(+args[+0+],+)+)
++++++++else:
++++++++++++seq+=+super(+Sequence,+cls+).__new__(+cls,+args+)
++++++++print(+'#+END+new+Sequence+len+%25d'+%25+(+len(+seq+),+))
++++++++return+seq|code-block|syntax|javascript|43917|注意，正如我从这个帖子中学到的，你必须把逗号放在args[+0+]后面。|offset|length|style|CODE|43918|打印行显示没有拆分单个字符串。|43919|注意:子类的构造函数中的逗号现在是可选的：|43920|Sequence(+u'silly'+)|43921|或|43922|Sequence(+u'silly',+)|43923|entityMap^0|0|0|O|9|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|10|8|@$I|11|J|12|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|13|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|14|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|15|8|@]|9|@]|A|$E|F]]|$1|S|3|T|5|6|7|16|8|@]|9|@]|A|$]]|$1|U|3|V|5|D|7|17|8|@]|9|@]|A|$E|F]]|$1|W|3|-4|5|6|7|18|8|@]|9|@]|A|$]]]|X|$]]

Subclassing tuple where some of these subclass instances may need to be one-string instances throws up something interesting.

<pre><code>class Sequence( tuple ):
 def __init__( self, *args ):
 # initialisation...
 self.instances = []

 def __new__( cls, *args ):
 for arg in args:
 assert isinstance( arg, unicode ), '# arg %s not unicode' % ( arg, )
 if len( args ) == 1:
 seq = super( Sequence, cls ).__new__( cls, ( args[ 0 ], ) )
 else:
 seq = super( Sequence, cls ).__new__( cls, args )
 print( '# END new Sequence len %d' % ( len( seq ), ))
 return seq
</code></pre>

NB as I learnt from this thread, you have to put the comma after <code>args[ 0 ]</code>.

The print line shows that a single string does not get split up.

NB the comma in the constructor of the subclass now becomes optional :

<pre><code>Sequence( u'silly' )
</code></pre>

or 

<pre><code>Sequence( u'silly', )
</code></pre>

blocks|key|43944|text|请参见：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|43945|'Quattro+TT'|code-block|syntax|javascript|43946|是一个字符串。|43947|因为string是一个字符列表，所以它与|43948|['Q',+'u',+'a',+'t',+'t',+'r',+'o',+'+',+'T',+'T']|43949|现在..。|43950|['Quattro+TT']|43951|是在第一个位置有一个字符串的列表。|43952|还有..。|43953|a+=+'Quattro+TT'

list(a)|43954|是将转换为列表的字符串。|offset|length|style|BOLD|43955|同样，因为string是一个字符列表，所以没有太大的变化。|43956|另一个信息..。|43957|tuple(something)|43958|这会将一些东西转换成元组。|43959|理解了所有这一切，我想你可以得出结论，没有什么是失败的。|43960|entityMap^0|0|0|0|0|0|0|0|0|0|0|B|1|0|0|T|0|0|8|0|0|0|D|0|0|S|0^^$0|@$1|2|3|4|5|6|7|1E|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|1F|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|1G|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|1H|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|1I|8|@]|9|@]|A|$E|F]]|$1|M|3|N|5|6|7|1J|8|@]|9|@]|A|$]]|$1|O|3|P|5|D|7|1K|8|@]|9|@]|A|$E|F]]|$1|Q|3|R|5|6|7|1L|8|@]|9|@]|A|$]]|$1|S|3|T|5|6|7|1M|8|@]|9|@]|A|$]]|$1|U|3|V|5|D|7|1N|8|@]|9|@]|A|$E|F]]|$1|W|3|X|5|6|7|1O|8|@$Y|1P|Z|1Q|10|11]]|9|@]|A|$]]|$1|12|3|13|5|6|7|1R|8|@$Y|1S|Z|1T|10|11]]|9|@]|A|$]]|$1|14|3|15|5|6|7|1U|8|@$Y|1V|Z|1W|10|11]]|9|@]|A|$]]|$1|16|3|17|5|D|7|1X|8|@]|9|@]|A|$E|F]]|$1|18|3|19|5|6|7|1Y|8|@$Y|1Z|Z|20|10|11]]|9|@]|A|$]]|$1|1A|3|1B|5|6|7|21|8|@$Y|22|Z|23|10|11]]|9|@]|A|$]]|$1|1C|3|-4|5|6|7|24|8|@]|9|@]|A|$]]]|1D|$]]

See:

<pre><code>'Quattro TT'
</code></pre>

is a string.

Since string is a list of characters, this is the same as

<pre><code>['Q', 'u', 'a', 't', 't', 'r', 'o', ' ', 'T', 'T']
</code></pre>

Now...

<pre><code>['Quattro TT']
</code></pre>

is a list with a string in the first position.

Also...

<pre><code>a = 'Quattro TT'

list(a)
</code></pre>

is a string converted into a list.

Again, since string is a list of characters, there is not much change.

Another information...

<pre><code>tuple(something)
</code></pre>

This convert something into tuple.

Understanding all of this, I think you can conclude that nothing fails.

blocks|key|2729431|text|这只涵盖了一个简单的案例：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2729432|a+=+‘Quattro+TT’
print+tuple(a)|code-block|syntax|javascript|2729433|如果你只使用像‘，’这样的分隔符，那么它就可以工作。|2729434|我使用了来自configparser的字符串，如下所示：|offset|length|style|CODE|2729435|list_users+=+(‘test1’,+‘test2’,+‘test3’)
and+the+i+get+from+file
tmp+=+config_ob.get(section_name,+option_name)
>>>”(‘test1’,+‘test2’,+‘test3’)”|2729436|在这种情况下，上述解决方案不起作用。但是，这是可行的：|2729437|def+fot_tuple(self,+some_str):
+++++#+(‘test1’,+‘test2’,+‘test3’)
+++++some_str+=+some_str.replace(‘(‘,+”)
+++++#+‘test1’,+‘test2’,+‘test3’)
+++++some_str+=+some_str.replace(‘)’,+”)
+++++#+‘test1’,+‘test2’,+‘test3’
+++++some_str+=+some_str.replace(“‘,+‘”,+‘,’)
+++++#+‘test1,test2,test3’
+++++some_str+=+some_str.replace(“‘”,+‘,’)
+++++#+test1,test2,test3
+++++#+and+now+i+could+convert+to+tuple
+++++return+tuple(item+for+item+in+some_str.split(‘,’)+if+item.strip())|2729438|entityMap^0|0|0|0|6|C|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|Z|8|@$K|10|L|11|M|N]]|9|@]|A|$]]|$1|O|3|P|5|D|7|12|8|@]|9|@]|A|$E|F]]|$1|Q|3|R|5|6|7|13|8|@]|9|@]|A|$]]|$1|S|3|T|5|D|7|14|8|@]|9|@]|A|$E|F]]|$1|U|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|V|$]]

This only covers a simple case:

<pre><code>a = ‘Quattro TT’
print tuple(a)
</code></pre>

If you use only delimiter like ‘,’, then it could work.

I used a string from <code>configparser</code> like so:

<pre><code>list_users = (‘test1’, ‘test2’, ‘test3’)
and the i get from file
tmp = config_ob.get(section_name, option_name)
&gt;&gt;&gt;”(‘test1’, ‘test2’, ‘test3’)”
</code></pre>

In this case the above solution does not work. However, this does work:

<pre><code>def fot_tuple(self, some_str):
 # (‘test1’, ‘test2’, ‘test3’)
 some_str = some_str.replace(‘(‘, ”)
 # ‘test1’, ‘test2’, ‘test3’)
 some_str = some_str.replace(‘)’, ”)
 # ‘test1’, ‘test2’, ‘test3’
 some_str = some_str.replace(“‘, ‘”, ‘,’)
 # ‘test1,test2,test3’
 some_str = some_str.replace(“‘”, ‘,’)
 # test1,test2,test3
 # and now i could convert to tuple
 return tuple(item for item in some_str.split(‘,’) if item.strip())
</code></pre>

blocks|key|2729472|text|val_1=input("").strip(")(").split(",")+#making+a+list+from+the+string

if+val_1[0].isdigit()==True:+#checks+if+your+contents+are+all+integers
++val_2+=+[]
++for+i+in+val_1:
++++val_2.append(int(i))
++val_2+=+tuple(val_2)
++print(val_2)
else:+#+if+the+contents+in+your+tuple+are+of+string,+tis+case+is+applied
++val_2=[]
++for+i+in+range(0,len(val_1)):
++++++if+i==0:
+++++++++val_2.append(val_1[i][1:-1])
++++++else:
+++++++++val_2.append(val_1[i][2:-1])

++val_2=tuple(val_2)
++print(val_2)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2729473|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>val_1=input(&quot;&quot;).strip(&quot;)(&quot;).split(&quot;,&quot;) #making a list from the string

if val_1[0].isdigit()==True: #checks if your contents are all integers
 val_2 = []
 for i in val_1:
 val_2.append(int(i))
 val_2 = tuple(val_2)
 print(val_2)
else: # if the contents in your tuple are of string, tis case is applied
 val_2=[]
 for i in range(0,len(val_1)):
 if i==0:
 val_2.append(val_1[i][1:-1])
 else:
 val_2.append(val_1[i][2:-1])

 val_2=tuple(val_2)
 print(val_2)
</code></pre>

I am striving to convert a string to a tuple without splitting the characters of the string in the process. Can somebody suggest an easy method to do this. Need a one liner.

Fails 

<pre><code> a = 'Quattro TT'
 print tuple(a)
</code></pre>

Works 

<pre><code> a = ['Quattro TT'] 
 print tuple(a)
</code></pre>

Since my input is a string, I tried the code below by converting the string to a list, which again splits the string into characters .. 

Fails 

<pre><code>a = 'Quattro TT'
print tuple(list(a))
</code></pre>

Expected Output: 

<pre><code>('Quattro TT')
</code></pre>

Generated Output: 

<pre><code>('Q', 'u', 'a', 't', 't', 'r', 'o', ' ', 'T', 'T')
</code></pre>

Converting string to tuple without splitting characters

有没有人能建议一个简单的方法来做这件事。我需要一条直线。出现故障   a = 'Quattro TT'   print tuple(a)Works  a = ['Quattro TT']    print tuple(a)因为我的输入是一个字符串，所以我尝试了下面的代码，将字符串转换为列表，这将再次将字符串拆分为字符。出现故障a = 'Quattro TT'print tuple(list(a))

问在不分割字符的情况下将字符串转换为元组
EN

回答 9

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