Question

In my code I have the following:

if all(requiredField in submittedFields for requiredField in requiredFields):
    # all required fields in submittedFields exist
else:
    # error handling

目的是核对在<代码>上填报的外地的示意图清单是否都出现在<代码>提交的外地上。

在“条码”中,该作品是一份长篇大体的插图;但是,如果你有类似的话,则会处以罚款。

requiredFields = ( single element )

那么,选择权会超越每一种特性,而不是扼杀本身。

因此,我的问题是,除了处理这种问题外,还有更多的老化手段。

try: 
    requiredFields.sort()
    # requiredFields is a list of strings
except AttributeError:
    # requiredFields is a string list whose length == 1

Answer 1

使用假装会提高效率:

submitted_fields = set([ length ,  width ,  color ,  single element ])
required_fields = set([ width ,  length ])
if submitted_fields >= required_fields:
    # all required fields in submittedFields exist
else:
    # error handling

几个优化使这一点迅速:

The hash table implementation of sets ensures a high likelihood of a match before doing a character by character equality test.
If two strings are identical (the same object in memory), the identity check will bypass a character-by-character equality check.

注这与你原来的问题一样,是tu。 Steven Rumbalski对此作了很好的阐述。当然,如果你使用成套设备,就成为一个不问题。

得到你的实地验证的好兆头:-

Answer 2

括号里的扼杀不是直观——它是一种扼杀。 a. 做一个项目,需要一个拖车:

>>> ( single element ) # this is not a tuple, it s a string
 single element 
>>> ( single element ,) # note the trailing comma
( single element ,)

欲了解更多信息,请参看Wiki on Tupleyntax Stack Overflow question Zar tuple comma syntax Rule。

Answer 3

首先,如果你想写上<代码>需要的领域=(单一要素)——如果你想写上<代码>需要填写的外地=(单一要素:)。

但是,假设你不掌握你的意见,检验的最初步方法是:

if isinstance(requiredFields, basestring):
    # requiredFields is a string
else:
    # requiredFields is iterable

Answer 4

为什么不要让你做这样的事情:

required = [ name ,  password ]
submitted = [ name ,  email ]

for field in required:
    if field not in submitted:
        print field +   is required

Answer 5

这是因为定义

requiredFields = ( single element )

实际等于

requiredFields =  single element

编制单一要素清单确实如此。

requiredFields = [ single element ]

a 仅作一个部分,确实如此。

requiredFields = ( single element ,)

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