当想到任何问题时,通过名单理解自然会发生。
如何最好地思考这一问题?
Regards Ashish
对我来说,一些例子发挥了作用。 在你利用清单解决问题之前,清单核对表可能不会自然发生,或者说,如果在<<>代码>上填写<>代码/代码>,则通常不会解决。
某些想法(其他可能有更好的实例):
在这方面,我是如何通过名单理解来思考的。
1) I need to output a list
(2) Im, 先从清单/备份。
3) 我要么需要就所有要素采取行动,要么选择原始清单中的具体内容。
这导致我进行以下施工:
<代码>output = 选择(投入)]中的x[千兆(x)]
男性是改变一个要素的一些功能。 例如,我可以使用<代码>x.down(>)来降低要素。
我总是使用<条码>x作为代号。 只是保持一切一致(我从来不把它当作“<>条/代码>的序号”的传导器。
选择权()是产出<代码>True或<代码>False的职能。 如果说的话,通常会采取某种方式。 我主要利用这一试验进行生存,特别是如果I mmangling产出。 例如,如果输入,则输入中的x<代码>。
清单核对表确实是巨大的。 我认为,它们确实提高了可读性,而且其方式超过了一种 ne。 但是,他们记得,他们只是一线 lo。
尝试为假肢撰写文件并试图将其转化为理解清单,可能最容易。
如果你刚刚从清单中抽出,那么我怎么走。 第一封信给你正常的住宿:
results = []
# get all keys that aren t test strings and add to results list
for a in blah.keys():
if a not in ( test , foo ):
results.append(a)
现在要制定一份清单,让人们了解情况,如果发言顺序相同,并放在同一行文上,最终结果代码就在开头:
results = [ results.append(a) for a in blah.keys() if a not in ( test , foo ) ]
我们几乎在那里! 我们不需要人工阅读(a),因为我们在该部分的表述中将自动附在名单之后,因此最后的表述是:
# get all keys that aren t test strings and add to results list
results = [ a for a in blah.keys() if a not in ( test , foo ) ]
HTH!
清单核对表的概念与关于住所的思路相同,但只是按一线计算,你可以将其在线用作理由。
myFunc([s.strip() for s in mystrings])
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