Python for循环中的循环计数器 [英] Python loop counter in a for loop
问题描述
在我下面的示例代码中,计数器= 0是否真的需要,还是有一个更好,更多的Python,方法来访问循环计数器?我看到一些与循环计数器相关的PEP,但是它们要么被延迟,要么被拒绝( PEP 212 和 PEP 281 )。
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 (PEP 212 and PEP 281).
这是我的问题的简单例子。在我的真实应用中,这是用图形完成的,整个菜单必须重新绘制每一帧。但是,这以简单的文本方式演示,很容易重现。
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.
也许我还应该补充一点,我使用的是Python 2.5,尽管我仍然对
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.
# 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
运行时输出:
When run, it outputs:
[ ] Option 0
[ ] Option 1
[*] Option 2
[ ] Option 3
推荐答案
使用 enumerate()
就像这样:
def draw_menu(options, selected_index):
for counter, option in enumerate(options):
if counter == selected_index:
print " [*] %s" % option
else:
print " [ ] %s" % option
options = ['Option 0', 'Option 1', 'Option 2', 'Option 3']
draw_menu(options, 2)
注意:您可以选择在计数器,选项
之前放置括号,如(counter,option)
,如果你想,但它们是无关的,通常不包括在内。
Note: You can optionally put parenthesis around counter, option
, like (counter, option)
, if you want, but they're extraneous and not normally included.
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