blocks|key|1279328|text|您还可以执行以下操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1279329|+for+option+in+options:
++++++if+option+==+options[selected_index]:
+++++++++++#print
++++++else:
+++++++++++#print|code-block|syntax|javascript|1279330|尽管如果有重复的选项，您会遇到问题。|1279331|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|$]]

You could also do:

<pre><code> for option in options:
 if option == options[selected_index]:
 #print
 else:
 #print
</code></pre>

Although you'd run into issues if there are duplicate options.

blocks|key|1494179|text|我有时会这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1494180|def+draw_menu(options,+selected_index):
++++for+i+in+range(len(options)):
++++++++if+i+==+selected_index:
++++++++++++print+"+[*]+%25s"+%25+options[i]
++++++++else:
++++++++++++print+"+[+]+%25s"+%25+options[i]|code-block|syntax|javascript|1494181|虽然我倾向于避免这一点，如果这意味着我会多次说options[i]。|offset|length|style|CODE|1494182|entityMap^0|0|0|N|A|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

I'll sometimes do this:

<pre><code>def draw_menu(options, selected_index):
 for i in range(len(options)):
 if i == selected_index:
 print " [*] %s" % options[i]
 else:
 print " [ ] %s" % options[i]
</code></pre>

Though I tend to avoid this if it means I'll be saying <code>options[i]</code> more than a couple of times.

blocks|key|1494230|text|enumerate就是你要找的东西。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1494231|您可能还会对unpacking感兴趣|1494232|#+The+pattern
x,+y,+z+=+[1,+2,+3]

#+also+works+in+loops:
l+=+[(28,+'M'),+(4,+'a'),+(1990,+'r')]
for+x,+y+in+l:
++++print(x)++#+prints+the+numbers+28,+4,+1990

#+and+also
for+index,+(x,+y)+in+enumerate(l):
++++print(x)++#+prints+the+numbers+28,+4,+1990|code-block|syntax|javascript|1494233|此外，还有itertools.count()，因此您可以执行以下操作|1494234|import+itertools

for+index,+el+in+zip(itertools.count(),+[28,+4,+1990]):
++++print(el)++#+prints+the+numbers+28,+4,+1990|1494235|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/2/library/functions.html#enumerate|1|https://docs.python.org/2/tutorial/controlflow.html#unpacking-argument-lists|2|https://docs.python.org/2/library/itertools.html#itertools.count^0|0|9|0|9|0|0|6|9|1|0|0|5|H|5|H|2|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@$9|13|A|14|B|C]]|D|@$9|15|A|16|1|17]]|E|$]]|$1|F|3|G|5|6|7|18|8|@]|D|@$9|19|A|1A|1|1B]]|E|$]]|$1|H|3|I|5|J|7|1C|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|1D|8|@$9|1E|A|1F|B|C]]|D|@$9|1G|A|1H|1|1I]]|E|$]]|$1|O|3|P|5|J|7|1J|8|@]|D|@]|E|$K|L]]|$1|Q|3|-4|5|6|7|1K|8|@]|D|@]|E|$]]]|R|$S|$5|T|U|V|E|$W|X]]|Y|$5|T|U|V|E|$W|Z]]|10|$5|T|U|V|E|$W|11]]]]

<a href="https://docs.python.org/2/library/functions.html#enumerate" rel="nofollow noreferrer"><code>enumerate</code></a> is what you are looking for.

You might also be interested in <a href="https://docs.python.org/2/tutorial/controlflow.html#unpacking-argument-lists" rel="nofollow noreferrer">unpacking</a>:

<pre><code># The pattern
x, y, z = [1, 2, 3]

# also works in loops:
l = [(28, 'M'), (4, 'a'), (1990, 'r')]
for x, y in l:
 print(x) # prints the numbers 28, 4, 1990

# and also
for index, (x, y) in enumerate(l):
 print(x) # prints the numbers 28, 4, 1990
</code></pre>

Also, there is <a href="https://docs.python.org/2/library/itertools.html#itertools.count" rel="nofollow noreferrer"><code>itertools.count()</code></a> so you could do something like

<pre><code>import itertools

for index, el in zip(itertools.count(), [28, 4, 1990]):
 print(el) # prints the numbers 28, 4, 1990
</code></pre>

In my example code below, is the counter = 0 really required, or is there a better, more Python, way to get access to a loop counter? I saw a few PEPs related to loop counters, but they were either deferred or rejected (<a href="http://www.python.org/dev/peps/pep-0212/" rel="noreferrer">PEP 212</a> and <a href="http://www.python.org/dev/peps/pep-0281/" rel="noreferrer">PEP 281</a>).

This is a simplified example of my problem. In my real application this is done with graphics and the whole menu has to be repainted each frame. But this demonstrates it in a simple text way that is easy to reproduce.

Maybe I should also add that I'm using Python 2.5, although I'm still interested if there is a way specific to 2.6 or higher.

<pre><code># Draw all the options, but highlight the selected index
def draw_menu(options, selected_index):
 counter = 0
 for option in options:
 if counter == selected_index:
 print " [*] %s" % option
 else:
 print " [ ] %s" % option
 counter += 1


options = ['Option 0', 'Option 1', 'Option 2', 'Option 3']

draw_menu(option, 2) # Draw menu with "Option2" selected
</code></pre>

When run, it outputs:

<pre><code> [ ] Option 0
 [ ] Option 1
 [*] Option 2
 [ ] Option 3
</code></pre>

Python loop counter in a for loop

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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