blocks|key|306225|text|可以通过以下方式实现：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|306226|t+=+('275',+'54000',+'0.0',+'5000.0',+'0.0')
lst+=+list(t)
lst[0]+=+'300'
t+=+tuple(lst)|code-block|syntax|javascript|306227|但是，如果您需要更改某些内容，最好将其保留为list|offset|length|style|CODE|306228|entityMap^0|0|0|M|4|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

It's possible via:
<pre><code>t = ('275', '54000', '0.0', '5000.0', '0.0')
lst = list(t)
lst[0] = '300'
t = tuple(lst)
</code></pre>
But if you're going to need to change things, you probably are better off keeping it as a <code>list</code>

blocks|key|306407|text|根据你的问题，切片可能是一个非常好的解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|306408|>>>+b+=+(1,+2,+3,+4,+5)
>>>+b[:2]+%2B+(8,9)+%2B+b[3:]
(1,+2,+8,+9,+4,+5)
>>>+b[:2]+%2B+(8,)+%2B+b[3:]
(1,+2,+8,+4,+5)|code-block|syntax|javascript|306409|这允许您添加多个元素或替换一些元素(特别是如果它们是“邻居”的话。在上面的例子中，转换为列表可能更合适，也更具可读性(尽管切片符号要短得多)。|306410|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|$]]

Depending on your problem slicing can be a really neat solution:

<pre><code>&gt;&gt;&gt; b = (1, 2, 3, 4, 5)
&gt;&gt;&gt; b[:2] + (8,9) + b[3:]
(1, 2, 8, 9, 4, 5)
&gt;&gt;&gt; b[:2] + (8,) + b[3:]
(1, 2, 8, 4, 5)
</code></pre>

This allows you to add multiple elements or also to replace a few elements (especially if they are "neighbours". In the above case casting to a list is probably more appropriate and readable (even though the slicing notation is much shorter).

blocks|key|306647|text|使用one+liner是可能的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|306648|values+=+('275',+'54000',+'0.0',+'5000.0',+'0.0')
values+=+('300',+*values[1:])|code-block|syntax|javascript|306649|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|$]]

It is possible with a one liner:

<pre><code>values = ('275', '54000', '0.0', '5000.0', '0.0')
values = ('300', *values[1:])
</code></pre>

blocks|key|523224|text|正如Hunter+McMillen提到的，元组是不可变的，您需要创建一个新的元组来实现这一点。例如：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|523225|>>>+tpl+=+('275',+'54000',+'0.0',+'5000.0',+'0.0')
>>>+change_value+=+200
>>>+tpl+=+(change_value,)+%2B+tpl[1:]
>>>+tpl
(200,+'54000',+'0.0',+'5000.0',+'0.0')|code-block|syntax|javascript|523226|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|$]]

As Hunter McMillen mentioned, tuples are immutable, you need to create a new tuple in order to achieve this. For instance:
<pre><code>&gt;&gt;&gt; tpl = ('275', '54000', '0.0', '5000.0', '0.0')
&gt;&gt;&gt; change_value = 200
&gt;&gt;&gt; tpl = (change_value,) + tpl[1:]
&gt;&gt;&gt; tpl
(200, '54000', '0.0', '5000.0', '0.0')
</code></pre>

blocks|key|306511|text|这并不是说这是优越的，但如果有人好奇的话，可以用一句话来完成：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|306512|tuple+=+tuple([200+if+i+==+0+else+_+for+i,+_+in+enumerate(tuple)])|code-block|syntax|javascript|306513|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|$]]

Not that this is superior, but if anyone is curious it can be done on one line with:

<pre><code>tuple = tuple([200 if i == 0 else _ for i, _ in enumerate(tuple)])
</code></pre>

blocks|key|306544|text|我相信这从技术上回答了这个问题，但不要在家里这样做。目前，所有答案都涉及创建一个新的元组，但您可以使用ctypes修改内存中的元组。依赖于64位系统上CPython的各种实现细节，一种方法如下：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|306545|def+modify_tuple(t,+idx,+new_value):
++++#+`id`+happens+to+give+the+memory+address+in+CPython;+you+may
++++#+want+to+use+`ctypes.addressof`+instead.
++++element_ptr+=+(ctypes.c_longlong).from_address(id(t)+%2B+(3+%2B+idx)*8)
++++element_ptr.value+=+id(new_value)
++++#+Manually+increment+the+reference+count+to+`new_value`+to+pretend+that
++++#+this+is+not+a+terrible+idea.
++++ref_count+=+(ctypes.c_longlong).from_address(id(new_value))
++++ref_count.value+%2B=+1

t+=+(10,+20,+30)
modify_tuple(t,+1,+50)+++#+t+is+now+(10,+50,+30)
modify_tuple(t,+-1,+50)++#+Will+probably+crash+your+Python+runtime|code-block|syntax|javascript|306546|entityMap^0|1F|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|$]]

I believe this technically answers the question, but don't do this at home. At the moment, all answers involve creating a new tuple, but you can use <code>ctypes</code> to modify a tuple in-memory. Relying on various implementation details of CPython on a 64-bit system, one way to do this is as follows:

<pre><code>def modify_tuple(t, idx, new_value):
 # `id` happens to give the memory address in CPython; you may
 # want to use `ctypes.addressof` instead.
 element_ptr = (ctypes.c_longlong).from_address(id(t) + (3 + idx)*8)
 element_ptr.value = id(new_value)
 # Manually increment the reference count to `new_value` to pretend that
 # this is not a terrible idea.
 ref_count = (ctypes.c_longlong).from_address(id(new_value))
 ref_count.value += 1

t = (10, 20, 30)
modify_tuple(t, 1, 50) # t is now (10, 50, 30)
modify_tuple(t, -1, 50) # Will probably crash your Python runtime
</code></pre>

blocks|key|523103|text|编辑:这还不适用于具有重复条目的元组！！|type|unstyled|depth|inlineStyleRanges|entityRanges|data|523104|基于Pooya's+idea|offset|length|523105|如果你打算经常这样做(你不应该这样做，因为元组是不变的)，你应该这样做：|523106|def+modTupByIndex(tup,+index,+ins):
++++return+tuple(tup[0:index])+%2B+(ins,)+%2B+tuple(tup[index%2B1:])

print+modTupByIndex((1,2,3),2,"a")|code-block|syntax|javascript|523107|或者基于Jon's+idea|523108|def+modTupByIndex(tup,+index,+ins):
++++lst+=+list(tup)
++++lst[index]+=+ins
++++return+tuple(lst)

print+modTupByIndex((1,2,3),1,"a")|523109|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/11458412/463065|1|https://stackoverflow.com/a/11458274/463065^0|0|2|C|0|0|0|0|4|A|1|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|11|8|@]|9|@$D|12|E|13|1|14]]|A|$]]|$1|F|3|G|5|6|7|15|8|@]|9|@]|A|$]]|$1|H|3|I|5|J|7|16|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|17|8|@]|9|@$D|18|E|19|1|1A]]|A|$]]|$1|O|3|P|5|J|7|1B|8|@]|9|@]|A|$K|L]]|$1|Q|3|-4|5|6|7|1C|8|@]|9|@]|A|$]]]|R|$S|$5|T|U|V|A|$W|X]]|Y|$5|T|U|V|A|$W|Z]]]]

EDIT: This doesn't work on tuples with duplicate entries yet!!

Based on <a href="https://stackoverflow.com/a/11458412/463065">Pooya's idea</a>:

If you are planning on doing this often (which you shouldn't since tuples are inmutable for a reason) you should do something like this:

<pre><code>def modTupByIndex(tup, index, ins):
 return tuple(tup[0:index]) + (ins,) + tuple(tup[index+1:])

print modTupByIndex((1,2,3),2,"a")
</code></pre>

Or based on <a href="https://stackoverflow.com/a/11458274/463065">Jon's idea</a>:

<pre><code>def modTupByIndex(tup, index, ins):
 lst = list(tup)
 lst[index] = ins
 return tuple(lst)

print modTupByIndex((1,2,3),1,"a")
</code></pre>

blocks|key|306802|text|您不能修改元组中的项，但可以修改元组中可变对象的属性(例如，如果这些对象是列表或实际的类对象)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|306803|例如|306804|my_list+=+[1,2]
tuple_of_lists+=+(my_list,'hello')
print(tuple_of_lists)+#+([1,+2],+'hello')
my_list[0]+=+0
print(tuple_of_lists)+#+([0,+2],+'hello')|code-block|syntax|javascript|306805|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

You can't modify items in tuple, but you can modify properties of mutable objects in tuples (for example if those objects are lists or actual class objects)
For example
<pre class="lang-py prettyprint-override"><code>my_list = [1,2]
tuple_of_lists = (my_list,'hello')
print(tuple_of_lists) # ([1, 2], 'hello')
my_list[0] = 0
print(tuple_of_lists) # ([0, 2], 'hello')
</code></pre>

blocks|key|523039|text|基于Jon的想法和亲爱的Trufa|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|523040|def+modifyTuple(tup,+oldval,+newval):
++++lst=list(tup)
++++for+i+in+range(tup.count(oldval)):
++++++++index+=+lst.index(oldval)
++++++++lst[index]=newval

++++return+tuple(lst)

print+modTupByIndex((1,+1,+3),+1,+"a")|code-block|syntax|javascript|523041|它会改变所有出现的旧值|523042|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/users/1252759/jon-clements|1|https://stackoverflow.com/users/463065/trufa^0|2|3|0|C|5|1|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]|$A|Y|B|Z|1|10]]|C|$]]|$1|D|3|E|5|F|7|11|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|12|8|@]|9|@]|C|$]]|$1|K|3|-4|5|6|7|13|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]|S|$5|N|O|P|C|$Q|T]]]]

based on <a href="https://stackoverflow.com/users/1252759/jon-clements">Jon</a>'s Idea and dear <a href="https://stackoverflow.com/users/463065/trufa">Trufa</a>

<pre><code>def modifyTuple(tup, oldval, newval):
 lst=list(tup)
 for i in range(tup.count(oldval)):
 index = lst.index(oldval)
 lst[index]=newval

 return tuple(lst)

print modTupByIndex((1, 1, 3), 1, "a")
</code></pre>

it changes all of your old values occurrences

blocks|key|523003|text|你不能。如果你想改变它，你需要使用列表而不是元组。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|523004|请注意，您可以使用新值作为其第一个元素来创建一个新的元组。|523005|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

You can't. If you want to change it, you need to use a list instead of a tuple.

Note that you could instead make a new tuple that has the new value as its first element.

blocks|key|306731|text|tldr；“解决办法”是创建一个新的元组对象，而不是实际修改原始的元组对象|type|unstyled|depth|inlineStyleRanges|entityRanges|data|306732|虽然这是一个非常古老的问题，但有人告诉我关于Python变异元组的疯狂。这让我非常惊讶/好奇，我在谷歌上做了一些搜索，我在这里登陆(以及其他类似的在线示例)|306733|我做了一些测试来证明我的理论|306734|注意：==做了值相等，而is做了引用相等(+obj+a和obj+b是相同的实例吗)|offset|length|style|CODE|306735|a+=+("apple",+"canana",+"cherry")
b+=+tuple(["apple",+"canana",+"cherry"])
c+=+a

print("a:+"+%2B+str(a))
print("b:+"+%2B+str(b))
print("c:+"+%2B+str(c))
print("a+==+b+::+%25s"+%25+(a==b))
print("b+==+c+::+%25s"+%25+(b==c))
print("a+==+c+::+%25s"+%25+(a==c))
print("a+is+b+::+%25s"+%25+(a+is+b))
print("b+is+c+::+%25s"+%25+(b+is+c))
print("a+is+c+::+%25s"+%25+(a+is+c))

d+=+list(a)
d[1]+=+"kiwi"
a+=+tuple(d)

print("a:+"+%2B+str(a))
print("b:+"+%2B+str(b))
print("c:+"+%2B+str(c))
print("a+==+b+::+%25s"+%25+(a==b))
print("b+==+c+::+%25s"+%25+(b==c))
print("a+==+c+::+%25s"+%25+(a==c))
print("a+is+b+::+%25s"+%25+(a+is+b))
print("b+is+c+::+%25s"+%25+(b+is+c))
print("a+is+c+::+%25s"+%25+(a+is+c))|code-block|syntax|javascript|306736|收益率：|306737|a:+('apple',+'canana',+'cherry')
b:+('apple',+'canana',+'cherry')
c:+('apple',+'canana',+'cherry')
a+==+b+::+True
b+==+c+::+True
a+==+c+::+True
a+is+b+::+False
b+is+c+::+False
a+is+c+::+True
a:+('apple',+'kiwi',+'cherry')
b:+('apple',+'canana',+'cherry')
c:+('apple',+'canana',+'cherry')
a+==+b+::+False
b+==+c+::+True
a+==+c+::+False
a+is+b+::+False
b+is+c+::+False
a+is+c+::+False|306738|entityMap^0|0|0|0|3|2|C|2|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|X|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|Y|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|Z|8|@$H|10|I|11|J|K]|$H|12|I|13|J|K]]|9|@]|A|$]]|$1|L|3|M|5|N|7|14|8|@]|9|@]|A|$O|P]]|$1|Q|3|R|5|6|7|15|8|@]|9|@]|A|$]]|$1|S|3|T|5|N|7|16|8|@]|9|@]|A|$O|P]]|$1|U|3|-4|5|6|7|17|8|@]|9|@]|A|$]]]|V|$]]

tldr; the "workaround" is creating a new tuple object, not actually modifying the original

While this is a very old question, someone told me about this Python mutating tuples madness. Which I was very much surprised/intrigued, and doing some googling, I landed here (and other similar samples online)

I ran some test to prove my theory

Note <code>==</code> does value equality while <code>is</code> does referential equality (is obj a the same instance as obj b)

<pre><code>a = ("apple", "canana", "cherry")
b = tuple(["apple", "canana", "cherry"])
c = a

print("a: " + str(a))
print("b: " + str(b))
print("c: " + str(c))
print("a == b :: %s" % (a==b))
print("b == c :: %s" % (b==c))
print("a == c :: %s" % (a==c))
print("a is b :: %s" % (a is b))
print("b is c :: %s" % (b is c))
print("a is c :: %s" % (a is c))

d = list(a)
d[1] = "kiwi"
a = tuple(d)

print("a: " + str(a))
print("b: " + str(b))
print("c: " + str(c))
print("a == b :: %s" % (a==b))
print("b == c :: %s" % (b==c))
print("a == c :: %s" % (a==c))
print("a is b :: %s" % (a is b))
print("b is c :: %s" % (b is c))
print("a is c :: %s" % (a is c))
</code></pre>

Yields:

<pre><code>a: ('apple', 'canana', 'cherry')
b: ('apple', 'canana', 'cherry')
c: ('apple', 'canana', 'cherry')
a == b :: True
b == c :: True
a == c :: True
a is b :: False
b is c :: False
a is c :: True
a: ('apple', 'kiwi', 'cherry')
b: ('apple', 'canana', 'cherry')
c: ('apple', 'canana', 'cherry')
a == b :: False
b == c :: True
a == c :: False
a is b :: False
b is c :: False
a is c :: False
</code></pre>

blocks|key|523170|text|我这样做了：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|523171|list+=+[1,2,3,4,5]
tuple+=+(list)|code-block|syntax|javascript|523172|想要改变，只需要做|523173|list[0]=6|523174|你可以改变一个元组:D|523175|这是从IDLE复制的|523176|>>>+list=[1,2,3,4,5,6,7,8,9]

>>>+tuple=(list)

>>>+print(tuple)

[1,+2,+3,+4,+5,+6,+7,+8,+9]

>>>+list[0]=6

>>>+print(tuple)

[6,+2,+3,+4,+5,+6,+7,+8,+9]|523177|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|W|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|X|8|@]|9|@]|A|$]]|$1|O|3|P|5|D|7|Y|8|@]|9|@]|A|$E|F]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

i did this:

<pre><code>list = [1,2,3,4,5]
tuple = (list)
</code></pre>

and to change, just do

<pre><code>list[0]=6
</code></pre>

and u can change a tuple :D

here is it copied exactly from IDLE

<pre><code>&gt;&gt;&gt; list=[1,2,3,4,5,6,7,8,9]

&gt;&gt;&gt; tuple=(list)

&gt;&gt;&gt; print(tuple)

[1, 2, 3, 4, 5, 6, 7, 8, 9]

&gt;&gt;&gt; list[0]=6

&gt;&gt;&gt; print(tuple)

[6, 2, 3, 4, 5, 6, 7, 8, 9]
</code></pre>

blocks|key|523337|text|您可以使用按引用复制来更改tuple的值|type|unstyled|depth|inlineStyleRanges|entityRanges|data|523338|>>>+tuple1=[20,30,40]

>>>+tuple2=tuple1

>>>+tuple2
++++[20,+30,+40]

>>>+tuple2[1]=10

>>>+print(tuple2)
++++[20,+10,+40]

>>>+print(tuple1)
++++[20,+10,+40]|code-block|syntax|javascript|523339|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 change the value of tuple using copy by reference

<pre><code>&gt;&gt;&gt; tuple1=[20,30,40]

&gt;&gt;&gt; tuple2=tuple1

&gt;&gt;&gt; tuple2
 [20, 30, 40]

&gt;&gt;&gt; tuple2[1]=10

&gt;&gt;&gt; print(tuple2)
 [20, 10, 40]

&gt;&gt;&gt; print(tuple1)
 [20, 10, 40]
</code></pre>

I have a tuple called <code>values</code> which contains the following:
<pre><code>('275', '54000', '0.0', '5000.0', '0.0')
</code></pre>
I want to change the first value (i.e., <code>275</code>) in this tuple but I understand that tuples are immutable so <code>values[0] = 200</code> will not work. How can I achieve this?

How to change values in a tuple?

我有一个名为values的元组，它包含以下内容：('275', '54000', '0.0', '5000.0', '0.0')我想更改这个元组中的第一个值(即275)，但我知道元组是不可变的，所以values[0] = 200将不起作用。我如何才能做到这一点？

问如何更改元组中的值？
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问如何更改元组中的值？
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