blocks|key|1200389|text|您的问题是将key和value放在引号中，使其成为字符串，即将aKey设置为包含字符串"key"，而不是变量key的值。此外，您不会清空temp列表，所以每次都要向其中添加内容，而不是只在其中添加两项。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1200390|要修复您的代码，请尝试如下所示：|1200391|for+key,+value+in+dict.iteritems():
++++temp+=+[key,value]
++++dictlist.append(temp)|code-block|syntax|javascript|1200392|在使用循环变量key和value之前，不需要将它们复制到另一个变量中，所以我把它们去掉了。类似地，您不需要使用append来构建列表，您只需在方括号中指定它，如上图所示。如果我们想要尽可能简短，我们可以做dictlist.append([key,value])。|1200393|或者按照建议的那样使用dict.items()。|1200394|entityMap^0|6|3|A|5|V|4|17|5|1I|3|1W|4|0|0|0|7|3|B|5|2U|S|0|B|C|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]|$9|Z|A|10|B|C]|$9|11|A|12|B|C]|$9|13|A|14|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|15|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|16|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|17|8|@$9|18|A|19|B|C]|$9|1A|A|1B|B|C]|$9|1C|A|1D|B|C]]|D|@]|E|$]]|$1|O|3|P|5|6|7|1E|8|@$9|1F|A|1G|B|C]]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|1H|8|@]|D|@]|E|$]]]|R|$]]

Your problem is that you have <code>key</code> and <code>value</code> in quotes making them strings, i.e. you're setting <code>aKey</code> to contain the string <code>"key"</code> and not the value of the variable <code>key</code>. Also, you're not clearing out the <code>temp</code> list, so you're adding to it each time, instead of just having two items in it.

To fix your code, try something like:

<pre><code>for key, value in dict.iteritems():
 temp = [key,value]
 dictlist.append(temp)
</code></pre>

You don't need to copy the loop variables <code>key</code> and <code>value</code> into another variable before using them so I dropped them out. Similarly, you don't need to use append to build up a list, you can just specify it between square brackets as shown above. And we could have done <code>dictlist.append([key,value])</code> if we wanted to be as brief as possible.

Or just use <code>dict.items()</code> as has been suggested.

blocks|key|1200310|text|dict.items()|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1200311|真的很管用。|unstyled|1200312|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>dict.items()
</code></pre>

Does the trick.

blocks|key|1415858|text|在Python语言中，可以轻松地将dict转换为list。三个例子：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1415859|>>+d+=+{'a':+'Arthur',+'b':+'Belling'}

>>+d.items()
[('a',+'Arthur'),+('b',+'Belling')]

>>+d.keys()
['a',+'b']

>>+d.values()
['Arthur',+'Belling']|code-block|syntax|javascript|1415860|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/library/stdtypes.html#mapping-types-dict^0|H|4|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

Converting from <a href="http://docs.python.org/library/stdtypes.html#mapping-types-dict">dict</a> to list is made easy in Python. Three examples:

<pre><code>&gt;&gt; d = {'a': 'Arthur', 'b': 'Belling'}

&gt;&gt; d.items()
[('a', 'Arthur'), ('b', 'Belling')]

&gt;&gt; d.keys()
['a', 'b']

&gt;&gt; d.values()
['Arthur', 'Belling']
</code></pre>

blocks|key|1415713|text|您应该使用dict.items()。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1415714|以下是您的问题的一个线性解决方案：|1415715|[(k,v)+for+k,v+in+dict.items()]|code-block|syntax|javascript|1415716|和结果：|1415717|[('Food',+'Fish&Chips'),+('2012',+'Olympics'),+('Capital',+'London')]|1415718|或者你可以这样做|1415719|l=[]
[l.extend([k,v])+for+k,v+in+dict.items()]|1415720|适用于：|1415721|['Food',+'Fish&Chips',+'2012',+'Olympics',+'Capital',+'London']|1415722|entityMap^0|5|C|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@$9|11|A|12|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|13|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|14|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|15|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|16|8|@]|D|@]|E|$K|L]]|$1|Q|3|R|5|6|7|17|8|@]|D|@]|E|$]]|$1|S|3|T|5|J|7|18|8|@]|D|@]|E|$K|L]]|$1|U|3|V|5|6|7|19|8|@]|D|@]|E|$]]|$1|W|3|X|5|J|7|1A|8|@]|D|@]|E|$K|L]]|$1|Y|3|-4|5|6|7|1B|8|@]|D|@]|E|$]]]|Z|$]]

You should use <code>dict.items()</code>.

Here is a one liner solution for your problem:

<pre><code>[(k,v) for k,v in dict.items()]
</code></pre>

and result:

<pre><code>[('Food', 'Fish&amp;Chips'), ('2012', 'Olympics'), ('Capital', 'London')]
</code></pre>

or you can do

<pre><code>l=[]
[l.extend([k,v]) for k,v in dict.items()]
</code></pre>

for:

<pre><code>['Food', 'Fish&amp;Chips', '2012', 'Olympics', 'Capital', 'London']
</code></pre>

blocks|key|1415539|text|+>>>+a+=+{'foo':+'bar',+'baz':+'quux',+'hello':+'world'}
+>>>+list(reduce(lambda+x,+y:+x+%2B+y,+a.items()))
+['foo',+'bar',+'baz',+'quux',+'hello',+'world']|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1415540|解释:+a.items()返回一个元组列表。将两个元组加在一起会形成一个包含所有元素的元组。因此，缩减创建了一个包含所有键和值的元组，然后是列表(...)列出了一份清单。|unstyled|1415541|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code> &gt;&gt;&gt; a = {'foo': 'bar', 'baz': 'quux', 'hello': 'world'}
 &gt;&gt;&gt; list(reduce(lambda x, y: x + y, a.items()))
 ['foo', 'bar', 'baz', 'quux', 'hello', 'world']
</code></pre>

To explain: a.items() returns a list of tuples. Adding two tuples together makes one tuple containing all elements. Thus the reduction creates one tuple containing all keys and values and then the list(...) makes a list from that.

blocks|key|1200544|text|你可能只想要这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1200545|dictList+=+dict.items()|code-block|syntax|javascript|1200546|你的方法有两个问题。对于一种方法，您可以在引号中使用key和value，它们是包含字母“键”和“值”的字符串，与这些名称的变量无关。此外，您还会不断向“临时”列表添加元素，并且永远不会删除先前迭代中已存在的旧元素。确保在每次迭代中都有一个新的空temp列表，并使用key和value变量：|offset|length|style|CODE|1200547|for+key,+value+in+dict.iteritems():
++++temp+=+[]
++++aKey+=+key
++++aValue+=+value
++++temp.append(aKey)
++++temp.append(aValue)
++++dictList.append(temp)|1200548|还要注意的是，在没有临时变量的情况下，这段代码可以写得更短(在Python3中，使用items()代替iteritems())：|1200549|for+key,+value+in+dict.items():
++++dictList.append([key,+value])|1200550|entityMap^0|0|0|Q|3|U|5|3E|4|3O|3|3S|5|0|0|16|7|1F|B|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]|$I|Z|J|10|K|L]|$I|11|J|12|K|L]|$I|13|J|14|K|L]|$I|15|J|16|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|17|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|18|8|@$I|19|J|1A|K|L]|$I|1B|J|1C|K|L]]|9|@]|A|$]]|$1|Q|3|R|5|D|7|1D|8|@]|9|@]|A|$E|F]]|$1|S|3|-4|5|6|7|1E|8|@]|9|@]|A|$]]]|T|$]]

Probably you just want this:

<pre><code>dictList = dict.items()
</code></pre>

Your approach has two problems. For one you use <code>key</code> and <code>value</code> in quotes, which are strings with the letters "key" and "value", not related to the variables of that names. Also you keep adding elements to the "temporary" list and never get rid of old elements that are already in it from previous iterations. Make sure you have a new and empty <code>temp</code> list in each iteration and use the <code>key</code> and <code>value</code> variables:

<pre><code>for key, value in dict.iteritems():
 temp = []
 aKey = key
 aValue = value
 temp.append(aKey)
 temp.append(aValue)
 dictList.append(temp)
</code></pre>

Also note that this could be written shorter without the temporary variables (and in Python 3 with <code>items()</code> instead of <code>iteritems()</code>):

<pre><code>for key, value in dict.items():
 dictList.append([key, value])
</code></pre>

blocks|key|1415788|text|如果您创建一个字典只是为了创建一个元组列表，因为创建像您这样的字典可能很痛苦，那么您可以考虑使用zip()|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1415789|如果你有一个标题和多行，它特别有用。例如，如果我假设您想要各个国家的奥运会统计数据：|1415790|headers+=+['Capital',+'Food',+'Year']
countries+=+[
++++['London',+'Fish+&+Chips',+'2012'],
++++['Beijing',+'Noodles',+'2008'],
]

for+olympics+in+countries:
++++print+zip(headers,+olympics)|code-block|syntax|javascript|1415791|给出|1415792|[('Capital',+'London'),+('Food',+'Fish+&+Chips'),+('Year',+'2012')]
[('Capital',+'Beijing'),+('Food',+'Noodles'),+('Year',+'2008')]|1415793|我不知道这是否是我的最终目标，也不知道我是否偏离了话题，但它可能是一些需要牢记的东西。|1415794|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|$]]

If you're making a dictionary only to make a list of tuples, as creating dicts like you are may be a pain, you might look into using zip()

Its especialy useful if you've got one heading, and multiple rows. For instance if I assume that you want Olympics stats for countries:

<pre><code>headers = ['Capital', 'Food', 'Year']
countries = [
 ['London', 'Fish &amp; Chips', '2012'],
 ['Beijing', 'Noodles', '2008'],
]

for olympics in countries:
 print zip(headers, olympics)
</code></pre>

gives

<pre><code>[('Capital', 'London'), ('Food', 'Fish &amp; Chips'), ('Year', '2012')]
[('Capital', 'Beijing'), ('Food', 'Noodles'), ('Year', '2008')]
</code></pre>

Don't know if thats the end goal, and my be off topic, but it could be something to keep in mind.

I'm trying to convert a Python dictionary into a Python list, in order to perform some calculations.

<pre><code>#My dictionary
dict = {}
dict['Capital']="London"
dict['Food']="Fish&amp;Chips"
dict['2012']="Olympics"

#lists
temp = []
dictList = []

#My attempt:
for key, value in dict.iteritems():
 aKey = key
 aValue = value
 temp.append(aKey)
 temp.append(aValue)
 dictList.append(temp) 
 aKey = ""
 aValue = ""
</code></pre>

That's my attempt at it... but I can't work out what's wrong?

Converting Dictionary to List?

Python

我正在尝试将Python字典转换为Python列表，以便执行一些计算。#My dictionarydict = {}dict['Capital']="London"dict['Food']="Fish&Chips"dict['2012']="Olympics"#liststemp = []dictList = []#My attempt:for key, value in dict.iterite

问是否将字典转换为列表？
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