blocks|key|344996|text|这里有一种方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|344997|>>>+mylist+=+['a',+'b',+'c',+'d']
>>>+[item+for+item+in+enumerate(mylist)]
[(0,+'a'),+(1,+'b'),+(2,+'c'),+(3,+'d')]|code-block|syntax|javascript|344998|或者，您可以执行以下操作：|344999|>>>+[(i,+j)+for+i,+j+in+enumerate(mylist)]
[(0,+'a'),+(1,+'b'),+(2,+'c'),+(3,+'d')]|345000|你得到一个错误的原因是你错过了i和j周围的()来使它成为一个元组。|offset|length|style|CODE|345001|entityMap^0|0|0|0|0|F|1|H|1|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|W|8|@$M|X|N|Y|O|P]|$M|Z|N|10|O|P]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|R|$]]

Here's a way to do it:

<pre><code>&gt;&gt;&gt; mylist = ['a', 'b', 'c', 'd']
&gt;&gt;&gt; [item for item in enumerate(mylist)]
[(0, 'a'), (1, 'b'), (2, 'c'), (3, 'd')]
</code></pre>

Alternatively, you can do:

<pre><code>&gt;&gt;&gt; [(i, j) for i, j in enumerate(mylist)]
[(0, 'a'), (1, 'b'), (2, 'c'), (3, 'd')]
</code></pre>

The reason you got an error was that you were missing the () around <code>i</code> and <code>j</code> to make it a tuple.

blocks|key|560939|text|对元组要明确。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|560940|[(i,+j)+for+(i,+j)+in+enumerate(mylist)]|code-block|syntax|javascript|560941|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|$]]

Be explicit about the tuples.

<pre><code>[(i, j) for (i, j) in enumerate(mylist)]
</code></pre>

blocks|key|560974|text|或者，如果您不坚持使用列表理解：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|560975|>>>+mylist+=+["a","b","c","d"]
>>>+list(enumerate(mylist))
[(0,+'a'),+(1,+'b'),+(2,+'c'),+(3,+'d')]|code-block|syntax|javascript|560976|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|$]]

Or, if you don't insist on using a list comprehension:

<pre><code>&gt;&gt;&gt; mylist = ["a","b","c","d"]
&gt;&gt;&gt; list(enumerate(mylist))
[(0, 'a'), (1, 'b'), (2, 'c'), (3, 'd')]
</code></pre>

blocks|key|561013|text|需要明确的是，这与enumerate无关，而是与列表理解语法有关。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|561014|这个列表理解返回一个元组列表：|561015|[(i,j)+for+i+in+range(3)+for+j+in+'abc']|code-block|syntax|javascript|561016|这是一个字典列表：|561017|[{i:j}+for+i+in+range(3)+for+j+in+'abc']|561018|列表列表：|561019|[[i,j]+for+i+in+range(3)+for+j+in+'abc']|561020|语法错误：|561021|[i,j+for+i+in+range(3)+for+j+in+'abc']|561022|这与字典理解语法不一致(IMHO)且令人困惑：|561023|>>>+{i:j+for+i,j+in+enumerate('abcdef')}
{0:+'a',+1:+'b',+2:+'c',+3:+'d',+4:+'e',+5:+'f'}|561024|和一组元组：|561025|>>>+{(i,j)+for+i,j+in+enumerate('abcdef')}
set([(0,+'a'),+(4,+'e'),+(1,+'b'),+(2,+'c'),+(5,+'f'),+(3,+'d')])|561026|正如斯卡普·洛佩斯所说，你可以直接传递枚举元组：|561027|>>>+[t+for+t+in+enumerate('abcdef')+]+
[(0,+'a'),+(1,+'b'),+(2,+'c'),+(3,+'d'),+(4,+'e'),+(5,+'f')]|561028|entityMap^0|9|9|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1C|8|@$9|1D|A|1E|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|1F|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|1G|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|1H|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|1I|8|@]|D|@]|E|$K|L]]|$1|Q|3|R|5|6|7|1J|8|@]|D|@]|E|$]]|$1|S|3|T|5|J|7|1K|8|@]|D|@]|E|$K|L]]|$1|U|3|V|5|6|7|1L|8|@]|D|@]|E|$]]|$1|W|3|X|5|J|7|1M|8|@]|D|@]|E|$K|L]]|$1|Y|3|Z|5|6|7|1N|8|@]|D|@]|E|$]]|$1|10|3|11|5|J|7|1O|8|@]|D|@]|E|$K|L]]|$1|12|3|13|5|6|7|1P|8|@]|D|@]|E|$]]|$1|14|3|15|5|J|7|1Q|8|@]|D|@]|E|$K|L]]|$1|16|3|17|5|6|7|1R|8|@]|D|@]|E|$]]|$1|18|3|19|5|J|7|1S|8|@]|D|@]|E|$K|L]]|$1|1A|3|-4|5|6|7|1T|8|@]|D|@]|E|$]]]|1B|$]]

Just to be really clear, this has nothing to do with <code>enumerate</code> and everything to do with list comprehension syntax.

This list comprehension returns a list of tuples:

<pre><code>[(i,j) for i in range(3) for j in 'abc']
</code></pre>

this a list of dicts:

<pre><code>[{i:j} for i in range(3) for j in 'abc']
</code></pre>

a list of lists:

<pre><code>[[i,j] for i in range(3) for j in 'abc']
</code></pre>

a syntax error:

<pre><code>[i,j for i in range(3) for j in 'abc']
</code></pre>

Which is inconsistent (IMHO) and confusing with dictionary comprehensions syntax:

<pre><code>&gt;&gt;&gt; {i:j for i,j in enumerate('abcdef')}
{0: 'a', 1: 'b', 2: 'c', 3: 'd', 4: 'e', 5: 'f'}
</code></pre>

And a set of tuples:

<pre><code>&gt;&gt;&gt; {(i,j) for i,j in enumerate('abcdef')}
set([(0, 'a'), (4, 'e'), (1, 'b'), (2, 'c'), (5, 'f'), (3, 'd')])
</code></pre>

As Óscar López stated, you can just pass the enumerate tuple directly:

<pre><code>&gt;&gt;&gt; [t for t in enumerate('abcdef') ] 
[(0, 'a'), (1, 'b'), (2, 'c'), (3, 'd'), (4, 'e'), (5, 'f')]
</code></pre>

blocks|key|561051|text|如果你使用的是长列表，那么列表的理解速度会更快，更不用说可读性了。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|561052|~$+python+-mtimeit+-s"mylist+=+['a','b','c','d']"+"list(enumerate(mylist))"
1000000+loops,+best+of+3:+1.61+usec+per+loop
~$+python+-mtimeit+-s"mylist+=+['a','b','c','d']"+"[(i,+j)+for+i,+j+in+enumerate(mylist)]"
1000000+loops,+best+of+3:+0.978+usec+per+loop
~$+python+-mtimeit+-s"mylist+=+['a','b','c','d']"+"[t+for+t+in+enumerate(mylist)]"
1000000+loops,+best+of+3:+0.767+usec+per+loop|code-block|syntax|javascript|561053|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're using long lists, it appears the list comprehension's faster, not to mention more readable.

<pre><code>~$ python -mtimeit -s"mylist = ['a','b','c','d']" "list(enumerate(mylist))"
1000000 loops, best of 3: 1.61 usec per loop
~$ python -mtimeit -s"mylist = ['a','b','c','d']" "[(i, j) for i, j in enumerate(mylist)]"
1000000 loops, best of 3: 0.978 usec per loop
~$ python -mtimeit -s"mylist = ['a','b','c','d']" "[t for t in enumerate(mylist)]"
1000000 loops, best of 3: 0.767 usec per loop
</code></pre>

blocks|key|561077|text|他们都回答得很好。我知道这里的问题是特定于枚举的，但是这样如何，只是另一种观点|type|unstyled|depth|inlineStyleRanges|entityRanges|data|561078|from+itertools+import+izip,+count
a+=+["5",+"6",+"1",+"2"]
tupleList+=+list(+izip(+count(),+a+)+)
print(tupleList)|code-block|syntax|javascript|561079|在性能方面，如果必须并行迭代多个列表，它将变得更加强大。只是一个想法|561080|a+=+["5",+"6",+"1",+"2"]
b+=+["a",+"b",+"c",+"d"]
tupleList+=+list(+izip(+count(),+a,+b+)+)
print(tupleList)|561081|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

All great answer guys. I know the question here is specific to enumeration but how about something like this, just another perspective

<pre><code>from itertools import izip, count
a = ["5", "6", "1", "2"]
tupleList = list( izip( count(), a ) )
print(tupleList)
</code></pre>

It becomes more powerful, if one has to iterate multiple lists in parallel in terms of performance. Just a thought

<pre><code>a = ["5", "6", "1", "2"]
b = ["a", "b", "c", "d"]
tupleList = list( izip( count(), a, b ) )
print(tupleList)
</code></pre>

Lets suppose I have a list like this:

<pre><code>mylist = ["a","b","c","d"]
</code></pre>

To get the values printed along with their index I can use Python's <code>enumerate</code> function like this

<pre><code>&gt;&gt;&gt; for i,j in enumerate(mylist):
... print i,j
...
0 a
1 b
2 c
3 d
&gt;&gt;&gt;
</code></pre>

Now, when I try to use it inside a <code>list comprehension</code> it gives me this error

<pre><code>&gt;&gt;&gt; [i,j for i,j in enumerate(mylist)]
 File "&lt;stdin&gt;", line 1
 [i,j for i,j in enumerate(mylist)]
 ^
SyntaxError: invalid syntax
</code></pre>

So, my question is: what is the correct way of using enumerate inside list comprehension?

Python using enumerate inside list comprehension

Python

假设我有一个这样的列表：mylist = ["a","b","c","d"]为了将这些值连同它们的索引一起打印出来，我可以使用Python的enumerate函数，如下所示>>> for i,j in enumerate(mylist):...     print i,j...0 a1 b2 c3 d>>>现在，当我试...

问Python在列表理解中使用枚举
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