blocks|key|486720|text|使用numpy数组和numpy.repeat函数：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|486721|import+numpy+as+np

x+=+np.array(["A",+"B"])
print+np.repeat(x,+[2,+3],+axis=0)

['A'+'A'+'B'+'B'+'B']|code-block|syntax|javascript|486722|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.scipy.org/doc/numpy/reference/generated/numpy.repeat.html^0|2|5|A|C|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@$9|V|A|W|1|X]]|E|$]]|$1|F|3|G|5|H|7|Y|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]]]

Use <code>numpy</code> arrays and the <a href="http://docs.scipy.org/doc/numpy/reference/generated/numpy.repeat.html">numpy.repeat</a> function:

<pre><code>import numpy as np

x = np.array(["A", "B"])
print np.repeat(x, [2, 3], axis=0)

['A' 'A' 'B' 'B' 'B']
</code></pre>

blocks|key|269691|text|既然你说的是"array“并提到R，你可能无论如何都想要使用numpy数组，然后使用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|269692|import+numpy+as+np
np.repeat(np.array([1,2]),+[2,3])|code-block|syntax|javascript|269693|编辑:既然你提到你也想重复行，我认为你应该使用numpy。np.repeat有一个轴参数可以做到这一点。|offset|length|style|CODE|269694|除此之外，也许：|269695|from+itertools+import+izip,+chain,+repeat
list(chain(*(repeat(a,b)+for+a,+b+in+izip([1,2],+[2,3]))))|269696|因为它没有假设你有一个要乘的列表或字符串。虽然我承认，将所有内容都作为参数传递到chain中可能并不完美，因此编写自己的迭代器可能会更好。|269697|entityMap^0|0|0|T|9|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@$I|X|J|Y|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|Z|8|@]|9|@]|A|$]]|$1|O|3|P|5|D|7|10|8|@]|9|@]|A|$E|F]]|$1|Q|3|R|5|6|7|11|8|@]|9|@]|A|$]]|$1|S|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|T|$]]

Since you say "array" and mention R. You may want to use numpy arrays anyways, and then use:

<pre><code>import numpy as np
np.repeat(np.array([1,2]), [2,3])
</code></pre>

EDIT: Since you mention you want to repeat rows as well, I think you should use numpy. <code>np.repeat</code> has an axis argument to do this.

Other then that, maybe:

<pre><code>from itertools import izip, chain, repeat
list(chain(*(repeat(a,b) for a, b in izip([1,2], [2,3]))))
</code></pre>

As it doesn't make the assumption you have a list or string to multiply. Though I admit, passing everything as argument into chain is maybe not perfect, so writing your own iterator may be better.

blocks|key|269601|text|l+=+['A','B']
n+=+[2,+4]|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|269602|您的示例使用的字符串已经是可迭代的。您可以生成类似于列表的结果字符串。|unstyled|269603|''.join([e+*+m+for+e,+m+in+zip(l,+n)])
'AABBBB'|269604|更新:这里不需要列表理解：|269605|''.join(e+*+m+for+e,+m+in+zip(l,+n))
'AABBBB'|269606|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|P|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$B|C]]|$1|I|3|J|5|F|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$B|C]]|$1|M|3|-4|5|F|7|T|8|@]|9|@]|A|$]]]|N|$]]

<pre><code>l = ['A','B']
n = [2, 4]
</code></pre>

Your example uses strings which are already iterables.
You can produce a result string which is similar to a list.

<pre><code>''.join([e * m for e, m in zip(l, n)])
'AABBBB'
</code></pre>

Update: the list comprehension is not required here:

<pre><code>''.join(e * m for e, m in zip(l, n))
'AABBBB'
</code></pre>

blocks|key|269731|text|你对这种方式有什么看法？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|269732|要重复某个值，请执行以下操作：|269733|>>>+repetitions=[]
>>>+torep=3
>>>+nrep=5
>>>+for+i+in+range(nrep):
>>>+++++i=torep
>>>+++++repetitions.append(i)
[3,+3,+3,+3,+3]|code-block|syntax|javascript|269734|重复序列的步骤：|269735|>>>+repetitions=[]
>>>+torep=[1,2,3,4]
>>>+nrep=+2
>>>+for+i+in+range(nrep):
>>>+++++repetitions=repetitions%2Btorep
>>>+print(repetitions)
[1,+2,+3,+4,+1,+2,+3,+4]|269736|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|$]]

What do you think about this way?

To repeat a value:

<pre><code>&gt;&gt;&gt; repetitions=[]
&gt;&gt;&gt; torep=3
&gt;&gt;&gt; nrep=5
&gt;&gt;&gt; for i in range(nrep):
&gt;&gt;&gt; i=torep
&gt;&gt;&gt; repetitions.append(i)
[3, 3, 3, 3, 3]
</code></pre>

To repeat a sequence:

<pre><code>&gt;&gt;&gt; repetitions=[]
&gt;&gt;&gt; torep=[1,2,3,4]
&gt;&gt;&gt; nrep= 2
&gt;&gt;&gt; for i in range(nrep):
&gt;&gt;&gt; repetitions=repetitions+torep
&gt;&gt;&gt; print(repetitions)
[1, 2, 3, 4, 1, 2, 3, 4]
</code></pre>

blocks|key|486835|text|以下内容可能适用于您：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|486836|>>>[['a','b'],['A','B']]*5


[['a',+'b'],+['A',+'B'],+['a',+'b'],+['A',+'B'],+['a',+'b'],+['A',+'B'],+['a',+'b'],+['A',+'B'],+['a',+'b'],+['A',+'B']]|code-block|syntax|javascript|486837|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|$]]

The following might work for you:
<pre><code>&gt;&gt;&gt;[['a','b'],['A','B']]*5


[['a', 'b'], ['A', 'B'], ['a', 'b'], ['A', 'B'], ['a', 'b'], ['A', 'B'], ['a', 'b'], ['A', 'B'], ['a', 'b'], ['A', 'B']]
</code></pre>

blocks|key|269805|text|前面已经提到了numpy.repeat，它显然等同于您想要的东西。但是为了完整起见，还有来自itertools标准库的repeat。然而，这通常是针对可迭代的，所以它不允许按索引重复(因为可迭代一般没有定义索引)。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|269806|我们可以使用那里给出的代码作为粗略的等价物|269807|def+repeat(object,+times=None):
++++#+repeat(10,+3)+-->+10+10+10
++++if+times+is+None:
++++++++while+True:
++++++++++++yield+object
++++else:
++++++++for+i+in+xrange(times):
++++++++++++yield+object|code-block|syntax|javascript|269808|要定义我们自己的泛化重复：|269809|def+repeat_generalised(object,+times=None):
++++#+repeat(10,+3)+-->+10+10+10
++++if+times+is+None:
++++++++while+True:
++++++++++++yield+object
++++else:
++++++++for+reps,+elem+in+zip(times,+object):
++++++++++++for+i+in+xrange(reps):+
++++++++++++++++yield+elem|269810|当然，问题是您必须定义许多可能的边缘情况(如果对象和时间具有不同数量的元素会发生什么？)，这将取决于您个人的用例。|269811|entityMap^0|7|C|1A|9|1N|6|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|11|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|12|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|13|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|14|8|@]|D|@]|E|$K|L]]|$1|Q|3|R|5|6|7|15|8|@]|D|@]|E|$]]|$1|S|3|-4|5|6|7|16|8|@]|D|@]|E|$]]]|T|$]]

The <code>numpy.repeat</code> has been mentioned, and that's clearly the equivalent to what you want. But for completenes' sake, there's also <code>repeat</code> from the <code>itertools</code> standard library. However, this is intended for iterables in general, so it doesn't allow repetions by index (because iterables in general do not have an index defined).
We can use the code given there as a rough equivalent
<pre><code>def repeat(object, times=None):
 # repeat(10, 3) --&gt; 10 10 10
 if times is None:
 while True:
 yield object
 else:
 for i in xrange(times):
 yield object
</code></pre>
to define our own generalised repeat:
<pre><code>def repeat_generalised(object, times=None):
 # repeat(10, 3) --&gt; 10 10 10
 if times is None:
 while True:
 yield object
 else:
 for reps, elem in zip(times, object):
 for i in xrange(reps): 
 yield elem
</code></pre>
The problem of course is that there's a lot of possible edge cases you have to define (What should happen if object and times have a different number of elements?), and that would depend on you individual use case.

The R function rep() replicates each element of a vector:

<pre><code>&gt; rep(c("A","B"), times=2)
[1] "A" "B" "A" "B"
</code></pre>

This is like the list multiplication in Python:

<pre><code>&gt;&gt;&gt; ["A","B"]*2
['A', 'B', 'A', 'B']
</code></pre>

But with the rep() R function it is also possible to specifiy the number of repeats for each element of the vector:

<pre><code>&gt; rep(c("A","B"), times=c(2,3))
[1] "A" "A" "B" "B" "B"
</code></pre>

Is there such a function availbale in Python ? Otherwise how could one define it ? By the way I'm also interested in such a function for duplicating rows of an array.

R function rep() in Python (replicates elements of a list/vector)

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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R函数rep()复制向量的每个元素：> rep(c("A","B"), times=2)[1] "A" "B" "A" "B"这类似于Python中的列表乘法：>>> ["A","B"]*2['A', 'B', 'A', 'B']但是使用rep() R函数也可以指定向量的每个元素的重复次数：> rep(c("A","B"), times=c(2,3))[1] "A" "A" "B" "B" "B"

问Python中的r函数rep() (复制列表/向量的元素)
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回答 6

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