blocks|key|198183|text|首先将整数转换为字符串，然后使用map对其应用int：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|198184|>>>+num+=+132
>>>+map(int,+str(num))++++#note,+This+will+return+a+map+object+in+python+3.
[1,+3,+2]|code-block|syntax|javascript|198185|或者使用列表理解：|198186|>>>+[int(x)+for+x+in+str(num)]
[1,+3,+2]|198187|entityMap^0|G|3|N|3|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|Y|8|@]|D|@]|E|$]]]|P|$]]

Convert the integer to string first, and then use <code>map</code> to apply <code>int</code> on it:

<pre><code>&gt;&gt;&gt; num = 132
&gt;&gt;&gt; map(int, str(num)) #note, This will return a map object in python 3.
[1, 3, 2]
</code></pre>

or using a list comprehension:

<pre><code>&gt;&gt;&gt; [int(x) for x in str(num)]
[1, 3, 2]
</code></pre>

blocks|key|416455|text|最短最好的方法已经回答了，但我首先想到的是数学方法，所以它是这样的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|416456|def+intlist(n):
++++q+=+n
++++ret+=+[]
++++while+q+!=+0:
++++++++q,+r+=+divmod(q,+10)+#+Divide+by+10,+see+the+remainder
++++++++ret.insert(0,+r)+#+The+remainder+is+the+first+to+the+right+digit
++++return+ret

print+intlist(3)
print+'-'
print+intlist(10)
print+'--'
print+intlist(137)|code-block|syntax|javascript|416457|这只是另一种有趣的方法，你绝对不需要在实际用例中使用这样的东西。|416458|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|$]]

The shortest and best way is already answered, but the first thing I thought of was the mathematical way, so here it is:

<pre><code>def intlist(n):
 q = n
 ret = []
 while q != 0:
 q, r = divmod(q, 10) # Divide by 10, see the remainder
 ret.insert(0, r) # The remainder is the first to the right digit
 return ret

print intlist(3)
print '-'
print intlist(10)
print '--'
print intlist(137)
</code></pre>

It's just another interesting approach, you definitely don't have to use such a thing in practical use cases.

blocks|key|198271|text|n+=+int(raw_input("n=+"))

def+int_to_list(n):
++++l+=+[]
++++while+n+!=+0:
++++++++l+=+[n+%25+10]+%2B+l
++++++++n+=+n+//+10
++++return+l

print+int_to_list(n)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|198272|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>n = int(raw_input("n= "))

def int_to_list(n):
 l = []
 while n != 0:
 l = [n % 10] + l
 n = n // 10
 return l

print int_to_list(n)
</code></pre>

blocks|key|198349|text|>>>list(map(int,+str(number)))++#number+is+a+given+integer|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|198350|它返回number的所有数字的列表。|unstyled|198351|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>&gt;&gt;&gt;list(map(int, str(number))) #number is a given integer
</code></pre>

It returns a list of all digits of number.

blocks|key|416413|text|对转换为字符串的数字使用list：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|416414|In+[1]:+[int(x)+for+x+in+list(str(123))]
Out[2]:+[1,+2,+3]|code-block|syntax|javascript|416415|entityMap^0|C|4|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|$]]

Use <code>list</code> on a number converted to string:

<pre><code>In [1]: [int(x) for x in list(str(123))]
Out[2]: [1, 2, 3]
</code></pre>

blocks|key|416531|text|如果你有一个像这样的字符串：'123456‘，并且你想要一个像这样的整数列表:+1,2,3,4,5,6，使用这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|416532|>>>s+=+'123456'++++
>>>list1+=+[int(i)+for+i+in+list(s)]
>>>print(list1)

[1,2,3,4,5,6]|code-block|syntax|javascript|416533|或者，如果你想要像这样的字符串列表：'1'，'2'，'3'，'4'，'5'，'6'，使用这个：|416534|>>>s+=+'123456'++++
>>>list1+=+list(s)
>>>print(list1)

['1','2','3','4','5','6']|416535|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 you have a string like this: '123456'
and you want a list of integers like this: [1,2,3,4,5,6], use this:

<pre><code>&gt;&gt;&gt;s = '123456' 
&gt;&gt;&gt;list1 = [int(i) for i in list(s)]
&gt;&gt;&gt;print(list1)

[1,2,3,4,5,6]
</code></pre>

or if you want a list of strings like this: ['1','2','3','4','5','6'], use this:

<pre><code>&gt;&gt;&gt;s = '123456' 
&gt;&gt;&gt;list1 = list(s)
&gt;&gt;&gt;print(list1)

['1','2','3','4','5','6']
</code></pre>

blocks|key|416512|text|您可以使用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|416513|首先转换字符串中的值来迭代它，它们中的每个值都可以转换为整数value+=+12345|offset|length|style|CODE|416514|l+=+[+int(item)+for+item+in+str(value)+]|416515|entityMap^0|0|U|D|0|0|14|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|M|8|@$D|N|E|O|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|P|8|@$D|Q|E|R|F|G]]|9|@]|A|$]]|$1|J|3|-4|5|6|7|S|8|@]|9|@]|A|$]]]|K|$]]

<h3>you can use:</h3>

First convert the value in a string to iterate it, Them each value can be convert to a Integer <code>value = 12345</code>

<code>l = [ int(item) for item in str(value) ]</code>

blocks|key|198395|text|通过循环，它可以通过以下方式完成:)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|198396|num1=+int(input('Enter+the+number'))
sum1+=+num1+#making+a+alt+int+to+store+the+value+of+the+orginal+so+it+wont+be+affected
y+=+[]+#making+a+list+
while+True:
++++if(sum1==0):#checking+if+the+number+is+not+zero+so+it+can+break+if+it+is
++++++++break
++++d+=+sum1%2510+#last+number+of+your+integer+is+saved+in+d
++++sum1+=+int(sum1/10)+#integer+is+now+with+out+the+last+number+ie.4320/10+become+432
++++y.append(d)+#+appending+the+last+number+in+the+first+place

y.reverse()#as+last+is+in+first+,+reversing+the+number+to+orginal+form
print(y)|code-block|syntax|javascript|198397|答案变成了|198398|Enter+the+number2342
[2,+3,+4,+2]|198399|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|$]]

By looping it can be done the following way :)

<pre><code>num1= int(input('Enter the number'))
sum1 = num1 #making a alt int to store the value of the orginal so it wont be affected
y = [] #making a list 
while True:
 if(sum1==0):#checking if the number is not zero so it can break if it is
 break
 d = sum1%10 #last number of your integer is saved in d
 sum1 = int(sum1/10) #integer is now with out the last number ie.4320/10 become 432
 y.append(d) # appending the last number in the first place

y.reverse()#as last is in first , reversing the number to orginal form
print(y)
</code></pre>

Answer becomes 

<pre><code>Enter the number2342
[2, 3, 4, 2]
</code></pre>

blocks|key|416502|text|num+=+list(str(100))
index+=+len(num)
while+index+>+0:
++++index+-=+1
++++num[index]+=+int(num[index])
print(num)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|416503|它打印[1,+0,+0]对象。|unstyled|offset|length|style|CODE|416504|entityMap^0|0|3|9|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|N|8|@$G|O|H|P|I|J]]|9|@]|A|$]]|$1|K|3|-4|5|F|7|Q|8|@]|9|@]|A|$]]]|L|$]]

<pre><code>num = list(str(100))
index = len(num)
while index &gt; 0:
 index -= 1
 num[index] = int(num[index])
print(num)
</code></pre>

It prints <code>[1, 0, 0]</code> object.

blocks|key|416635|text|接受整数作为输入，并将其转换为数字列表。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|416636|代码：|416637|num+=+int(input())
print(list(str(num)))|code-block|syntax|javascript|416638|使用156789的输出：|416639|>>>+['1',+'5',+'6',+'7',+'8',+'9']|416640|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Q|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Takes an integer as input and converts it into list of digits.
code:
<pre class="lang-py prettyprint-override"><code>num = int(input())
print(list(str(num)))
</code></pre>
output using 156789:
<pre class="lang-py prettyprint-override"><code>&gt;&gt;&gt; ['1', '5', '6', '7', '8', '9']
</code></pre>

blocks|key|416652|text|num+=+123
print(num)
num+=+list(str(num))
num+=+[int(i)+for+i+in+num]
print(num)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|416653|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-python prettyprint-override"><code>num = 123
print(num)
num = list(str(num))
num = [int(i) for i in num]
print(num)
</code></pre>

What is the quickest and cleanest way to convert an <code>integer</code> into a <code>list</code>? 

For example, change <code>132</code> into <code>[1,3,2]</code> and <code>23</code> into <code>[2,3]</code>. I have a variable which is an <code>int</code>, and I want to be able to compare the individual digits so I thought making it into a list would be best, since I can just do <code>int(number[0])</code>, <code>int(number[1])</code> to easily convert the list element back into int for digit operations.

Converting integer to digit list

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

文章

问答

视频

学习中心

腾讯云实验室

直播

竞赛

腾讯云代码分析专区

腾讯iOA零信任安全管理系统专区

腾讯云架构师技术同盟交流圈

腾讯云数据库专区

腾讯云顾问专区

腾讯云原生专区

腾讯混元专区

腾讯云TCE专区

腾讯云Lighthouse专区

腾讯云HAI专区

腾讯云Edgeone专区

腾讯云存储专区

腾讯云智能专区

腾讯轻联专区 

腾讯云开发专区

TAPD专区

腾讯轻量云游戏服专区

腾讯云最具价值专家

腾讯云架构师技术同盟

腾讯云创作之星

腾讯云开发者先锋

腾讯云代码助手

云原生构建

TAPD 敏捷项目管理

Cloud Studio

SDK中心

API中心

命令行工具

涵盖代码开发、场景应用、自动测试全流程，助你从零构建专属AI助手

一站式MCP教程库，解锁AI应用新玩法

将integer转换为list的最快、最干净的方法是什么？例如，将132更改为[1,3,2]，将23更改为[2,3]。我有一个变量，它是一个整数，我想能够比较各个数字，所以我认为把它变成一个列表是最好的，因为我可以做int(number[0])，int(number[1])，来轻松地将列表元素转换回int，进行数字操作。

问将整数转换为数字列表
EN

回答 11

Stack Overflow用户

Stack Overflow用户

Stack Overflow用户

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问将整数转换为数字列表EN

回答 11

Stack Overflow用户

Stack Overflow用户

Stack Overflow用户

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问将整数转换为数字列表
EN