blocks|key|183827|text|>>>+A+=+np.random.randint(5,+size=(10,3))
>>>+A
array([[1,+3,+0],
+++++++[3,+2,+0],
+++++++[0,+2,+1],
+++++++[1,+1,+4],
+++++++[3,+2,+2],
+++++++[0,+1,+0],
+++++++[1,+3,+1],
+++++++[0,+4,+1],
+++++++[2,+4,+2],
+++++++[3,+3,+1]])
>>>+idx+=+np.random.randint(10,+size=2)
>>>+idx
array([7,+6])
>>>+A[idx,:]
array([[0,+4,+1],
+++++++[1,+3,+1]])|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|183828|把它们放在一起，形成一个一般的情况：|unstyled|183829|A[np.random.randint(A.shape[0],+size=2),+:]|183830|对于非替换(numpy+1.7.0%2B)：|183831|A[np.random.choice(A.shape[0],+2,+replace=False),+:]|183832|在1.7版本之前，我不相信有一个好的方法来生成随机列表而不进行替换。也许您可以设置一个较小的定义，以确保两个值不相同。|183833|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|R|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$B|C]]|$1|I|3|J|5|F|7|T|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$B|C]]|$1|M|3|N|5|F|7|V|8|@]|9|@]|A|$]]|$1|O|3|-4|5|F|7|W|8|@]|9|@]|A|$]]]|P|$]]

<pre><code>&gt;&gt;&gt; A = np.random.randint(5, size=(10,3))
&gt;&gt;&gt; A
array([[1, 3, 0],
 [3, 2, 0],
 [0, 2, 1],
 [1, 1, 4],
 [3, 2, 2],
 [0, 1, 0],
 [1, 3, 1],
 [0, 4, 1],
 [2, 4, 2],
 [3, 3, 1]])
&gt;&gt;&gt; idx = np.random.randint(10, size=2)
&gt;&gt;&gt; idx
array([7, 6])
&gt;&gt;&gt; A[idx,:]
array([[0, 4, 1],
 [1, 3, 1]])
</code></pre>

Putting it together for a general case:

<pre><code>A[np.random.randint(A.shape[0], size=2), :]
</code></pre>

For non replacement (numpy 1.7.0+):

<pre><code>A[np.random.choice(A.shape[0], 2, replace=False), :]
</code></pre>

I do not believe there is a good way to generate random list without replacement before 1.7. Perhaps you can setup a small definition that ensures the two values are not the same.

blocks|key|401928|text|这是一个老帖子，但这是最适合我的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|401929|A[np.random.choice(A.shape[0],+num_rows_2_sample,+replace=False)]|code-block|syntax|javascript|401930|将replace=False更改为True以获得相同的结果，但需要替换。|401931|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|$]]

This is an old post, but this is what works best for me:

<pre><code>A[np.random.choice(A.shape[0], num_rows_2_sample, replace=False)]
</code></pre>

change the replace=False to True to get the same thing, but with replacement.

blocks|key|183931|text|另一种选择是创建一个随机掩码，如果您只想按某个因子对数据进行下采样。假设我想将原始数据集的下采样降至25%25，该数据集当前保存在数组data_arr中|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|183932|#+generate+random+boolean+mask+the+length+of+data
#+use+p+0.75+for+False+and+0.25+for+True
mask+=+numpy.random.choice([False,+True],+len(data_arr),+p=[0.75,+0.25])|code-block|syntax|javascript|183933|现在，您可以调用data_arr[mask]并返回大约25%25的随机抽样的行。|183934|entityMap^0|1T|8|0|0|8|E|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|V|8|@]|D|@]|E|$]]]|N|$]]

Another option is to create a random mask if you just want to down-sample your data by a certain factor. Say I want to down-sample to 25% of my original data set, which is currently held in the array <code>data_arr</code>:

<pre><code># generate random boolean mask the length of data
# use p 0.75 for False and 0.25 for True
mask = numpy.random.choice([False, True], len(data_arr), p=[0.75, 0.25])
</code></pre>

Now you can call <code>data_arr[mask]</code> and return ~25% of the rows, randomly sampled.

blocks|key|402031|text|如果您需要相同的行，但只需要一个随机样本，|type|unstyled|depth|inlineStyleRanges|entityRanges|data|402032|import+random
new_array+=+random.sample(old_array,x)|code-block|syntax|javascript|402033|这里的x必须是一个'int‘，它定义了你想随机挑选的行数。|402034|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|$]]

If you need the same rows but just a random sample then,

<pre><code>import random
new_array = random.sample(old_array,x)
</code></pre>

Here x, has to be an 'int' defining the number of rows you want to randomly pick.

blocks|key|402079|text|我看到已经有人提出了排列。事实上，它可以写成一行：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|402080|>>>+A+=+np.random.randint(5,+size=(10,3))
>>>+np.random.permutation(A)[:2]

array([[0,+3,+0],
+++++++[3,+1,+2]])|code-block|syntax|javascript|402081|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 see permutation has been suggested. In fact it can be made into one line:

<pre><code>&gt;&gt;&gt; A = np.random.randint(5, size=(10,3))
&gt;&gt;&gt; np.random.permutation(A)[:2]

array([[0, 3, 0],
 [3, 1, 2]])
</code></pre>

blocks|key|184067|text|如果您想要生成多个随机的行子集，例如，如果您正在执行RANSAC。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|184068|num_pop+=+10
num_samples+=+2
pop_in_sample+=+3
rows_to_sample+=+np.random.random([num_pop,+5])
random_numbers+=+np.random.random([num_samples,+num_pop])
samples+=+np.argsort(random_numbers,+axis=1)[:,+:pop_in_sample]
#+will+be+shape+[num_samples,+pop_in_sample,+5]
row_subsets+=+rows_to_sample[samples,+:]|code-block|syntax|javascript|184069|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 to generate multiple random subsets of rows, for example if your doing RANSAC.

<pre><code>num_pop = 10
num_samples = 2
pop_in_sample = 3
rows_to_sample = np.random.random([num_pop, 5])
random_numbers = np.random.random([num_samples, num_pop])
samples = np.argsort(random_numbers, axis=1)[:, :pop_in_sample]
# will be shape [num_samples, pop_in_sample, 5]
row_subsets = rows_to_sample[samples, :]
</code></pre>

blocks|key|402164|text|这是与Hezi+Rasheff提供的答案类似的答案，但经过简化，以便较新的python用户了解发生了什么(我注意到许多datascience的新学生以最奇怪的方式获取随机样本，因为他们不知道自己在python中做什么)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|402165|您可以通过使用以下命令从数组中获取大量随机索引：|402166|indices+=+np.random.choice(A.shape[0],+number_of_samples,+replace=False)|code-block|syntax|javascript|402167|然后，您可以对numpy数组使用奇特的索引，以获得这些索引处的样本：|402168|A[indices]|402169|这将从您的数据中获取指定数量的随机样本。|402170|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|T|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|U|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|V|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

This is a similar answer to the one Hezi Rasheff provided, but simplified so newer python users understand what's going on (I noticed many new datascience students fetch random samples in the weirdest ways because they don't know what they are doing in python).
You can get a number of random indices from your array by using:
<pre><code>indices = np.random.choice(A.shape[0], number_of_samples, replace=False)
</code></pre>
You can then use fancy indexing with your numpy array to get the samples at those indices:
<pre><code>A[indices]
</code></pre>
This will get you the specified number of random samples from your data.

blocks|key|184143|text|可以使用随机数生成器从给定的数组中生成随机样本：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|184144|rng+=+np.random.default_rng()
b+=+rng.choice(a,+2,+replace=False)
b
>>>+[[a4,++b4,++c4],
++++[a99,+b99,+c99]]|code-block|syntax|javascript|184145|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|$]]

One can generates a random sample from a given array with a random number generator:
<pre class="lang-py prettyprint-override"><code>rng = np.random.default_rng()
b = rng.choice(a, 2, replace=False)
b
&gt;&gt;&gt; [[a4, b4, c4],
 [a99, b99, c99]]
</code></pre>

I have a very large 2D array which looks something like this:

<pre><code>a=
[[a1, b1, c1],
 [a2, b2, c2],
 ...,
 [an, bn, cn]]
</code></pre>

Using numpy, is there an easy way to get a new 2D array with, e.g., 2 random rows from the initial array <code>a</code> (without replacement)?

e.g.

<pre><code>b=
[[a4, b4, c4],
 [a99, b99, c99]]
</code></pre>

Numpy: Get random set of rows from 2D array

我有一个非常大的2D数组，它看起来像这样：a=[[a1, b1, c1], [a2, b2, c2], ..., [an, bn, cn]]使用a，有没有一种简单的方法来获得一个新的2D数组，例如，从初始数组numpy中随机获得2行(不需要替换)？例如：b=[[a4,  b4,  c4], [a99, b99, c99...

问Numpy:从2D数组中获取随机行集合
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