blocks|key|1272266|text|你可以给自己写一个小函数来做你想做的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1272267|import+random
def+randomDigits(digits):
++++lower+=+10**(digits-1)
++++upper+=+10**digits+-+1
++++return+random.randint(lower,+upper)|code-block|syntax|javascript|1272268|基本上，10**(digits-1)给出的是最小的{-digit}数字，而10**digits+-+1给出的是最大的{数字}-digit数字(恰好是最小的{digit%2B1}-digit数字减1!)。然后我们只需要从这个范围中随机取一个整数。|offset|length|style|CODE|1272269|entityMap^0|0|0|4|E|11|E|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]|$I|T|J|U|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|V|8|@]|9|@]|A|$]]]|N|$]]

You could write yourself a little function to do what you want:

<pre><code>import random
def randomDigits(digits):
 lower = 10**(digits-1)
 upper = 10**digits - 1
 return random.randint(lower, upper)
</code></pre>

Basically, <code>10**(digits-1)</code> gives you the smallest {digit}-digit number, and <code>10**digits - 1</code> gives you the largest {digit}-digit number (which happens to be the smallest {digit+1}-digit number minus 1!). Then we just take a random integer from that range.

blocks|key|1272284|text|如果您希望它是一个字符串(例如，一个10位数字的电话号码)，您可以使用以下命令：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1272285|n+=+10
''.join(["{}".format(randint(0,+9))+for+num+in+range(0,+n)])|code-block|syntax|javascript|1272286|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|$]]

If you want it as a string (for example, a 10-digit phone number) you can use this:

<pre><code>n = 10
''.join(["{}".format(randint(0, 9)) for num in range(0, n)])
</code></pre>

blocks|key|1272334|text|我真的很喜欢RichieHindle的答案，但是我喜欢这个问题作为练习。下面是一个使用字符串的暴力实现：)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1272335|import+random
first+=+random.randint(1,9)
first+=+str(first)
n+=+5

nrs+=+[str(random.randrange(10))+for+i+in+range(n-1)]
for+i+in+range(len(nrs))++++:
++++first+%2B=+str(nrs[i])

print+str(first)|code-block|syntax|javascript|1272336|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|$]]

I really liked the answer of RichieHindle, however I liked the question as an exercise. Here's a brute force implementation using strings:)

<pre><code>import random
first = random.randint(1,9)
first = str(first)
n = 5

nrs = [str(random.randrange(10)) for i in range(n-1)]
for i in range(len(nrs)) :
 first += str(nrs[i])

print str(first)
</code></pre>

blocks|key|1272390|text|从official+documentation上看，sample()方法不适合用于此目的吗？|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1272391|import+random

def+random_digits(n):
++++num+=+range(0,+10)
++++lst+=+random.sample(num,+n)
++++print+str(lst).strip('[]')|code-block|syntax|javascript|1272392|输出：|1272393|>>>random_digits(5)
2,+5,+1,+0,+4|1272394|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3.1/library/random.html^0|1|M|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|Z|8|@]|9|@]|C|$]]|$1|K|3|L|5|F|7|10|8|@]|9|@]|C|$G|H]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]]]

From the <a href="https://docs.python.org/3.1/library/random.html" rel="nofollow noreferrer">official documentation</a>, does it not seem that the sample() method is appropriate for this purpose?

<pre><code>import random

def random_digits(n):
 num = range(0, 10)
 lst = random.sample(num, n)
 print str(lst).strip('[]')
</code></pre>

Output:

<pre><code>&gt;&gt;&gt;random_digits(5)
2, 5, 1, 0, 4
</code></pre>

blocks|key|1057688|text|您可以创建一个使用int列表的函数，将字符串转换为连接，然后再次转换为do+int，如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1057689|import+random

def+generate_random_number(length):
++++return+int(''.join([str(random.randint(0,10))+for+_+in+range(length)]))|code-block|syntax|javascript|1057690|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 could create a function who consumes an list of int, transforms in string to concatenate and cast do int again, something like this:

<pre><code>import random

def generate_random_number(length):
 return int(''.join([str(random.randint(0,10)) for _ in range(length)]))
</code></pre>

Let's say I need a 3-digit number, so it would be something like:
<pre><code>&gt;&gt;&gt; random(3)
563

or

&gt;&gt;&gt; random(5)
26748
&gt;&gt; random(2)
56
</code></pre>

How to generate a random number with a specific amount of digits?

假设我需要一个3位数的数字，那么它应该是这样的：>>> random(3)563or>>> random(5)26748>> random(2)56

问如何生成具有特定位数的随机数？
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