blocks|key|2930833|text|这正是random.sample()所做的。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2930834|>>>+random.sample(range(1,+16),+3)
[11,+10,+2]|code-block|syntax|javascript|2930835|编辑：我几乎可以肯定这不是你所要求的，但我被要求添加这样的评论:如果你想从其中获取样本的总体包含重复项，你必须首先删除它们：|BOLD|2930836|population+=+[1,+2,+3,+4,+5,+6,+5,+4,+3,+2,+1]
population+=+set(population)
samples+=+random.sample(population,+3)|2930837|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/library/random.html#random.sample^0|3|F|3|F|0|0|0|0|2|0|0^^$0|@$1|2|3|4|5|6|7|X|8|@$9|Y|A|Z|B|C]]|D|@$9|10|A|11|1|12]]|E|$]]|$1|F|3|G|5|H|7|13|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|14|8|@$9|15|A|16|B|M]]|D|@]|E|$]]|$1|N|3|O|5|H|7|17|8|@]|D|@]|E|$I|J]]|$1|P|3|-4|5|6|7|18|8|@]|D|@]|E|$]]]|Q|$R|$5|S|T|U|E|$V|W]]]]

That's exactly what <a href="http://docs.python.org/library/random.html#random.sample"><code>random.sample()</code></a> does.

<pre><code>&gt;&gt;&gt; random.sample(range(1, 16), 3)
[11, 10, 2]
</code></pre>

Edit: I'm almost certain this is not what you asked, but I was pushed to include this comment: If the population you want to take samples from contains duplicates, you have to remove them first:

<pre><code>population = [1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1]
population = set(population)
samples = random.sample(population, 3)
</code></pre>

blocks|key|5001446|text|如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5001447|all_data+=+[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]
from+random+import+shuffle
shuffle(all_data)
res+=+all_data[:3]#+or+any+other+number+of+items|code-block|syntax|javascript|5001448|或者：|5001449|from+random+import+sample
number_of_items+=+4
sample(all_data,+number_of_items)|5001450|如果all_data可能包含重复条目，请先修改代码以删除重复条目，然后使用shuffle或sample：|5001451|all_data+=+list(set(all_data))
shuffle(all_data)
res+=+all_data[:3]#+or+any+other+number+of+items|5001452|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|$]]

Something like this:

<pre><code>all_data = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]
from random import shuffle
shuffle(all_data)
res = all_data[:3]# or any other number of items
</code></pre>

OR:

<pre><code>from random import sample
number_of_items = 4
sample(all_data, number_of_items)
</code></pre>

If all_data could contains duplicate entries than modify your code to remove duplicates first and then use shuffle or sample:

<pre><code>all_data = list(set(all_data))
shuffle(all_data)
res = all_data[:3]# or any other number of items
</code></pre>

blocks|key|5001465|text|其他人则建议您使用random.sample。虽然这是一个有效的建议，但每个人都忽略了一个微妙之处：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|5001466|5001467|如果总体包含重复，则每次出现都是样本中可能的选择。|blockquote|5001468|5001469|5001470|因此，您需要将列表转换为一个集合，以避免重复值：|5001471|import+random
L+=+[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]
random.sample(set(L),+x)+#+where+x+is+the+number+of+samples+that+you+want|code-block|syntax|javascript|5001472|entityMap^0|9|D|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|X|8|@]|D|@]|E|$]]|$1|G|3|H|5|I|7|Y|8|@]|D|@]|E|$]]|$1|J|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]|$1|K|3|-4|5|6|7|10|8|@]|D|@]|E|$]]|$1|L|3|M|5|6|7|11|8|@]|D|@]|E|$]]|$1|N|3|O|5|P|7|12|8|@]|D|@]|E|$Q|R]]|$1|S|3|-4|5|6|7|13|8|@]|D|@]|E|$]]]|T|$]]

Others have suggested that you use <code>random.sample</code>. While this is a valid suggestion, there is one subtlety that everyone has ignored:

<blockquote>
 If the population contains repeats, 
 then each occurrence is a possible
 selection in the sample.
</blockquote>

Thus, you need to turn your list into a set, to avoid repeated values:

<pre><code>import random
L = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]
random.sample(set(L), x) # where x is the number of samples that you want
</code></pre>

blocks|key|3216564|text|另一种方式，当然，对于所有的解决方案，您必须确保原始列表中至少有3个唯一值。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3216565|all_data+=+[1,2,2,3,4,5,6,7,8,8,9,10,11,11,12,13,14,15,15]
choices+=+[]
while+len(choices)+<+3:
++++selection+=+random.choice(all_data)
++++if+selection+not+in+choices:
++++++++choices.append(selection)
print+choices+|code-block|syntax|javascript|3216566|​|3216567|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|$]]

Another way, of course with all the solutions you have to be sure that there are at least 3 unique values in the original list. <pre><code>all_data = [1,2,2,3,4,5,6,7,8,8,9,10,11,11,12,13,14,15,15]
choices = []
while len(choices) &lt; 3:
 selection = random.choice(all_data)
 if selection not in choices:
 choices.append(selection)
print choices </code></pre>

blocks|key|2930921|text|您还可以使用itertools.combinations和random.shuffle生成随机选择列表。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2930922|all_data+=+[1,2,2,3,4,5,6,7,8,8,9,10,11,11,12,13,14,15,15]

#+Remove+duplicates
unique_data+=+set(all_data)

#+Generate+a+list+of+combinations+of+three+elements
list_of_three+=+list(itertools.combinations(unique_data,+3))

#+Shuffle+the+list+of+combinations+of+three+elements
random.shuffle(list_of_three)|code-block|syntax|javascript|2930923|输出：|2930924|[(2,+5,+15),+(11,+13,+15),+(3,+10,+15),+(1,+6,+9),+(1,+7,+8),+...]|2930925|entityMap^0|6|M|T|E|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|$]]

You can also generate a list of random choices, using <code>itertools.combinations</code> and <code>random.shuffle</code>.

<pre><code>all_data = [1,2,2,3,4,5,6,7,8,8,9,10,11,11,12,13,14,15,15]

# Remove duplicates
unique_data = set(all_data)

# Generate a list of combinations of three elements
list_of_three = list(itertools.combinations(unique_data, 3))

# Shuffle the list of combinations of three elements
random.shuffle(list_of_three)
</code></pre>

Output:

<pre><code>[(2, 5, 15), (11, 13, 15), (3, 10, 15), (1, 6, 9), (1, 7, 8), ...]
</code></pre>

blocks|key|5001524|text|import+random
fruits_in_store+=+['apple','mango','orange','pineapple','fig','grapes','guava','litchi','almond']+
print('items+available+in+store+:')
print(fruits_in_store)
my_cart+=+[]
for+i+in+range(4):
++++#selecting+a+random+index
++++temp+=+int(random.random()*len(fruits_in_store))
++++#+adding+element+at+random+index+to+new+list
++++my_cart.append(fruits_in_store[temp])
++++#+removing+the+add+element+from+original+list
++++fruits_in_store.pop(temp)++
print('items+successfully+added+to+cart:')
print(my_cart)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|5001525|输出：|unstyled|5001526|items+available+in+store+:
['apple',+'mango',+'orange',+'pineapple',+'fig',+'grapes',+'guava',+'litchi',+'almond']
items+successfully+added+to+cart:
['orange',+'pineapple',+'mango',+'almond']|5001527|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>import random
fruits_in_store = ['apple','mango','orange','pineapple','fig','grapes','guava','litchi','almond'] 
print('items available in store :')
print(fruits_in_store)
my_cart = []
for i in range(4):
 #selecting a random index
 temp = int(random.random()*len(fruits_in_store))
 # adding element at random index to new list
 my_cart.append(fruits_in_store[temp])
 # removing the add element from original list
 fruits_in_store.pop(temp) 
print('items successfully added to cart:')
print(my_cart)
</code></pre>

Output:

<pre><code>items available in store :
['apple', 'mango', 'orange', 'pineapple', 'fig', 'grapes', 'guava', 'litchi', 'almond']
items successfully added to cart:
['orange', 'pineapple', 'mango', 'almond']
</code></pre>

blocks|key|5001555|text|如果重复的数据意味着我们更有可能绘制特定的数据，我们不能立即将其转换为集合(因为这样做会丢失该信息)。为此，我们需要一个接一个地挑选样本，并验证我们生成的集合的大小是否达到x(我们想要的样本数量)。类似于：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5001556|data=[0,+1,+2,+3,+4,+4,+4,+4,+5,+5,+6,+6]
x=3
res=set()
while(len(res)<x):
++++res.add(np.random.choice(data))
print(res)|code-block|syntax|javascript|5001557|一些输出：|5001558|{3,+4,+5}
{3,+5,+6}
{0,+4,+5}
{2,+4,+5}|5001559|正如我们所看到的，4或5出现的频率更高(我知道4个例子是不够好的统计数据)。|5001560|entityMap^0|0|0|0|0|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]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

If the data being repeated implies that we are more likely to draw that particular data, we can't turn it into a set right away (since we would loose that information by doing so). For this, we need to pick samples one by one and verify the size of the set that we generate has reached x (the number of samples that we want). Something like:
<pre><code>data=[0, 1, 2, 3, 4, 4, 4, 4, 5, 5, 6, 6]
x=3
res=set()
while(len(res)&lt;x):
 res.add(np.random.choice(data))
print(res)
</code></pre>
some outputs :
<pre><code>{3, 4, 5}
{3, 5, 6}
{0, 4, 5}
{2, 4, 5}
</code></pre>
As we can see 4 or 5 appear more frequently (I know 4 examples is not good enough statistics).

I need to pick out "x" number of non-repeating, random numbers out of a list. For example:

<pre><code>all_data = [1, 2, 2, 3, 4, 5, 6, 7, 8, 8, 9, 10, 11, 11, 12, 13, 14, 15, 15]
</code></pre>

How do I pick out a list like <code>[2, 11, 15]</code> and not <code>[3, 8, 8]</code>?

How do you pick "x" number of unique numbers from a list in Python?

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。

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我需要从列表中挑选出"x“个不重复的随机数。例如：all_data = [1, 2, 2, 3, 4, 5, 6, 7, 8, 8, 9, 10, 11, 11, 12, 13, 14, 15, 15]如何选择像[2, 11, 15]而不是[3, 8, 8]这样的列表

问如何在Python中从列表中选择"x“个数的唯一数字？
EN

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问如何在Python中从列表中选择"x“个数的唯一数字？EN