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

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
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回答 11

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