blocks|key|205406|text|当您将包含20个元素的三个列表zip()在一起时，结果为20个元素。每个元素都是一个三元组。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|205407|你自己看一下：|205408|In+[1]:+a+=+b+=+c+=+range(20)

In+[2]:+zip(a,+b,+c)
Out[2]:+
[(0,+0,+0),
+(1,+1,+1),
+...
+(17,+17,+17),
+(18,+18,+18),
+(19,+19,+19)]|code-block|syntax|javascript|205409|要找出每个元组包含多少个元素，可以检查第一个元素的长度：|205410|In+[3]:+result+=+zip(a,+b,+c)

In+[4]:+len(result[0])
Out[4]:+3|205411|当然，如果列表一开始就是空的，那么这就不会起作用。|205412|entityMap^0|F|5|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|X|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|Y|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|Z|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|10|8|@]|D|@]|E|$K|L]]|$1|Q|3|R|5|6|7|11|8|@]|D|@]|E|$]]|$1|S|3|-4|5|6|7|12|8|@]|D|@]|E|$]]]|T|$]]

When you <code>zip()</code> together three lists containing 20 elements each, the result has twenty elements. Each element is a three-tuple.

See for yourself:

<pre><code>In [1]: a = b = c = range(20)

In [2]: zip(a, b, c)
Out[2]: 
[(0, 0, 0),
 (1, 1, 1),
 ...
 (17, 17, 17),
 (18, 18, 18),
 (19, 19, 19)]
</code></pre>

To find out how many elements each tuple contains, you could examine the length of the first element:

<pre><code>In [3]: result = zip(a, b, c)

In [4]: len(result[0])
Out[4]: 3
</code></pre>

Of course, this won't work if the lists were empty to start with.

blocks|key|423631|text|zip有一堆像这样的列表|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|423632|a:+a1+a2+a3+a4+a5+a6+a7...
b:+b1+b2+b3+b4+b5+b6+b7...
c:+c1+c2+c3+c4+c5+c6+c7...|code-block|syntax|javascript|423633|并将它们“压缩”到一个列表中，该列表的条目是3元组(ai,+bi,+ci)。想象一下从左到右水平绘制一条拉链。|423634|entityMap^0|0|3|0|0|P|C|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|$]]

<code>zip</code> takes a bunch of lists likes

<pre><code>a: a1 a2 a3 a4 a5 a6 a7...
b: b1 b2 b3 b4 b5 b6 b7...
c: c1 c2 c3 c4 c5 c6 c7...
</code></pre>

and "zips" them into one list whose entries are 3-tuples <code>(ai, bi, ci)</code>. Imagine drawing a zipper horizontally from left to right.

blocks|key|205564|text|在Python2.7中，这可能工作得很好：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|205565|>>>+a+=+b+=+c+=+range(20)
>>>+zip(a,+b,+c)|code-block|syntax|javascript|205566|但在Python3.4中应该是这样(否则，结果将类似于<zip+object+at+0x00000256124E7DC8>)：|offset|length|style|CODE|205567|>>>+a+=+b+=+c+=+range(20)
>>>+list(zip(a,+b,+c))|205568|entityMap^0|0|0|R|Y|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]]|$1|G|3|H|5|6|7|S|8|@$I|T|J|U|K|L]]|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|$]]

In Python 2.7 this might have worked fine:

<pre><code>&gt;&gt;&gt; a = b = c = range(20)
&gt;&gt;&gt; zip(a, b, c)
</code></pre>

But in Python 3.4 it should be (otherwise, the result will be something like <code>&lt;zip object at 0x00000256124E7DC8&gt;</code>):

<pre><code>&gt;&gt;&gt; a = b = c = range(20)
&gt;&gt;&gt; list(zip(a, b, c))
</code></pre>

blocks|key|205447|text|zip创建一个新列表，其中填充了包含可迭代参数中的元素的元组：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|205448|>>>+zip+([1,2],[3,4])
[(1,3),+(2,4)]|code-block|syntax|javascript|205449|我希望您尝试创建一个元组，其中每个元素都是一个列表。|205450|entityMap^0|0|3|0|0|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]]|$1|K|3|L|5|6|7|S|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|N|$]]

<code>zip</code> creates a new list, filled with tuples containing elements from the iterable arguments:

<pre><code>&gt;&gt;&gt; zip ([1,2],[3,4])
[(1,3), (2,4)]
</code></pre>

I expect what you try to so is create a tuple where each element is a list.

blocks|key|205523|text|基本上，zip函数适用于Python中的列表、元组和字典。如果您使用的是IPython，那么只需输入zip？并检查zip()是关于什么的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|205524|如果您没有使用IPython，那么只需安装它："pip+install+ipython“|205525|对于列表|205526|a+=+['a',+'b',+'c']
b+=+['p',+'q',+'r']
zip(a,+b)|code-block|syntax|javascript|205527|输出为[('a',+'p'),+('b',+'q'),+('c',+'r')|offset|length|style|CODE|205528|对于字典：|205529|c+=+{'gaurav':'waghs',+'nilesh':'kashid',+'ramesh':'sawant',+'anu':'raje'}
d+=+{'amit':'wagh',+'swapnil':'dalavi',+'anish':'mane',+'raghu':'rokda'}
zip(c,+d)|205530|输出为：|205531|[('gaurav',+'amit'),
+('nilesh',+'swapnil'),
+('ramesh',+'anish'),
+('anu',+'raghu')]|205532|entityMap^0|0|0|0|0|3|Z|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|11|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|12|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|13|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|14|8|@$M|15|N|16|O|P]]|9|@]|A|$]]|$1|Q|3|R|5|6|7|17|8|@]|9|@]|A|$]]|$1|S|3|T|5|H|7|18|8|@]|9|@]|A|$I|J]]|$1|U|3|V|5|6|7|19|8|@]|9|@]|A|$]]|$1|W|3|X|5|H|7|1A|8|@]|9|@]|A|$I|J]]|$1|Y|3|-4|5|6|7|1B|8|@]|9|@]|A|$]]]|Z|$]]

Basically the zip function works on lists, tuples and dictionaries in Python.
If you are using IPython then just type zip? And check what zip() is about.

If you are not using IPython then just install it: "pip install ipython"

For lists

<pre><code>a = ['a', 'b', 'c']
b = ['p', 'q', 'r']
zip(a, b)
</code></pre>

The output is <code>[('a', 'p'), ('b', 'q'), ('c', 'r')</code>

For dictionary:

<pre><code>c = {'gaurav':'waghs', 'nilesh':'kashid', 'ramesh':'sawant', 'anu':'raje'}
d = {'amit':'wagh', 'swapnil':'dalavi', 'anish':'mane', 'raghu':'rokda'}
zip(c, d)
</code></pre>

The output is:

<pre><code>[('gaurav', 'amit'),
 ('nilesh', 'swapnil'),
 ('ramesh', 'anish'),
 ('anu', 'raghu')]
</code></pre>

blocks|key|423708|text|为了完整起见。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|423709|当压缩列表的长度不相等时。结果列表的长度将成为最短的列表，不会出现任何错误|423710|>>>+a+=+[1]
>>>+b+=+["2",+3]
>>>+zip(a,b)
[(1,+'2')]|code-block|syntax|javascript|423711|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

For the completeness's sake.

When zipped lists' lengths are not equal.
The result list's length will become the shortest one without any error occurred

<pre><code>&gt;&gt;&gt; a = [1]
&gt;&gt;&gt; b = ["2", 3]
&gt;&gt;&gt; zip(a,b)
[(1, '2')]
</code></pre>

blocks|key|205662|text|这是一个值得在这里添加的问题，因为它是一个在zip上排名很高的问题。zip是一个很棒的、惯用的Python+--但它根本不能很好地扩展到大型列表。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|205663|而不是：|205664|books+=+['AAAAAAA',+'BAAAAAAA',+...+,+'ZZZZZZZ']
words+=+[345,+567,+...+,+672]

for+book,+word+in+zip(books,+words):
+++print('{}:+{}'.format(book,+word))|code-block|syntax|javascript|205665|使用izip。对于现代处理，它将其存储在L1缓存中，并且对于较大的列表性能要高得多。像添加i一样简单地使用它|205666|for+book,+word+in+izip(books,+words):
+++print('{}:+{}'.format(book,+word))|205667|entityMap^0|Y|3|0|0|0|2|4|19|1|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|V|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|W|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|X|8|@$9|Y|A|Z|B|C]|$9|10|A|11|B|C]]|D|@]|E|$]]|$1|O|3|P|5|J|7|12|8|@]|D|@]|E|$K|L]]|$1|Q|3|-4|5|6|7|13|8|@]|D|@]|E|$]]]|R|$]]

It's worth adding here as it is such a highly ranking question on zip. <code>zip</code> is great, idiomatic Python - but it doesn't scale very well at all for large lists.

Instead of:

<pre><code>books = ['AAAAAAA', 'BAAAAAAA', ... , 'ZZZZZZZ']
words = [345, 567, ... , 672]

for book, word in zip(books, words):
 print('{}: {}'.format(book, word))
</code></pre>

Use <code>izip</code>. For modern processing, it stores it in L1 Cache memory and is far more performant for larger lists. Use it as simply as adding an <code>i</code>:

<pre><code>for book, word in izip(books, words):
 print('{}: {}'.format(book, word))
</code></pre>

blocks|key|423722|text|我不认为zip会返回列表。zip返回一个生成器。您必须执行list(zip(a,+b))才能获得元组列表。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|423723|x+=+[1,+2,+3]
y+=+[4,+5,+6]
zipped+=+zip(x,+y)
list(zipped)|code-block|syntax|javascript|423724|entityMap^0|4|3|D|3|T|F|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|U|8|@]|D|@]|E|$]]]|L|$]]

I don't think <code>zip</code> returns a list. <code>zip</code> returns a generator. You have got to do <code>list(zip(a, b))</code> to get a list of tuples.

<pre><code>x = [1, 2, 3]
y = [4, 5, 6]
zipped = zip(x, y)
list(zipped)
</code></pre>

I am trying to learn how to "zip" lists. To this end, I have a program, where at a particular point, I do the following:

<pre><code>x1, x2, x3 = stuff.calculations(withdataa)
</code></pre>

This gives me three lists, <code>x1</code>, <code>x2</code>, and <code>x3</code>, each of, say, size 20.

Now, I do:

<pre><code>zipall = zip(x1, x2, x3)
</code></pre>

However, when I do:

<pre><code>print "len of zipall %s" % len(zipall)
</code></pre>

I get 20, which is not what I expected. I expected three. I think I am doing something fundamentally wrong.

Zip lists in Python

我正在努力学习如何“压缩”列表。为此，我有一个程序，其中在特定的点上，我执行以下操作：x1, x2, x3 = stuff.calculations(withdataa)这给了我三个列表，x1、x2和x3，每个列表的大小都是20。现在，我做到了：zipall = zip(x1, x2, x3)然而，当我这样做的时候：print "len of zipall %s" % len(zipall)我得到

问Python中的Zip列表
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