It seems there have been a number of questions surrounding this exercise (48) from Learn Python the Hard Way, yet none answer my question. I have successfully completed the exercise in it's entirety and now wish to refactor it.
似乎围绕这个练习(48)有一些问题来自Learn Python the Hard Way,但没有人回答我的问题。我已经成功完成了整个练习,现在希望重构它。
Before I paste any code let me give you a brief explanation of what the exercise is for. The book provides you with a file (lexicon_tests.py) filled with various test functions and assert_equal statements. The objective is to create another file that contains the code that will allow the various tests to pass using nosetests command from the nose project.
在我粘贴任何代码之前,让我简要解释一下这个练习的内容。本书为您提供了一个文件(lexicon_tests.py),其中包含各种测试函数和assert_equal语句。目标是创建另一个文件,其中包含允许使用nose项目中的nosetests命令传递各种测试的代码。
I've created a dictionary with the necessary key-value pairs as well as a scan function that will check for int, word, or error. And all my tests pass.
我已经创建了一个包含必要键值对的字典以及一个将检查int,word或error的扫描函数。我所有的测试都通过了。
Now I would like to shorten up the code. Below is what I've attempted so far:
现在我想缩短代码。以下是我到目前为止所做的尝试:
# from collections import defaultdict
# What I'm trying to do
lexicon = {
"direction": ['north', 'south', 'east', 'west', 'down', 'up', 'left', 'right', 'back']
}
''' *What I've already done*
lexicon = {
"north": 'direction',
"south": 'direction',
"east": 'direction',
"west": 'direction',
"down": 'down',
"up": 'direction',
"left": 'direction',
"right": 'direction',
"back": 'direction',
"go": 'verb',
"stop": 'verb',
"kill": 'verb',
"eat": 'verb',
"the": 'stop',
"in": 'stop',
"of": 'stop',
"from": 'stop',
"at": 'stop',
"it": 'stop',
"bear": 'noun',
"door": 'noun',
"princess": 'noun',
"cabinet": 'noun',
"0": 'number',
"1": 'number',
"2": 'number',
"3": 'number',
"4": 'number',
"5": 'number',
"6": 'number',
"7": 'number',
"8": 'number',
"9": 'number',
}
'''
''' *Not sure how to incorporate this with my code below or if it will work*
lexicon = defaultdict(list)
for k, v in 1:
d1[k].append(v)
d = dict((k, tuple(v)) for k, v in d1.iteritems())
'''
# Scan Method - This works fine
def scan(sentence):
results = []
words = sentence.split()
for word in words:
try:
temp = int(word)
word_type = "number"
results.append((word_type, temp))
except ValueError:
word_type = lexicon.get(word)
if word_type == None:
results.append(('error', word))
else:
results.append((word_type, word))
return results
if __name__ == '__main__':
print(scan("ASDFADFASDF"))
Here is the error I'm receiving currently:
这是我目前收到的错误:
======================================================================
FAIL: tests.lexicon_tests.test_directions
----------------------------------------------------------------------
Traceback (most recent call last):
File "C:\Users\blalonde\AppData\Local\Programs\Python\Python36-32\lib\site-packages\nose-1.3.7-py3.6.egg\nose\case.py", line 198, in runTest
self.test(*self.arg)
File "C:\lpthw\projects\ex48\skeleton\tests\lexicon_tests.py", line 6, in test_directions
assert_equal(lexicon.scan("north"), [('direction', 'north')])
AssertionError: Lists differ: [('error', 'north')] != [('direction', 'north')]
First differing element 0:
('error', 'north')
('direction', 'north')
- [('error', 'north')]
? ^^ ^
+ [('direction', 'north')]
? +++ ^^^ ^
----------------------------------------------------------------------
Ran 1 test in 0.032s
FAILED (failures=1)
PS C:\lpthw\projects\ex48\skeleton>
Instead of having multiple lines of code for each key:value entry, is there a way to assign multiple values to a single key entry? The above example 'lexicon = { "direction": ['north', 'south', 'east', 'west', 'down', 'up', 'left', 'right', 'back'] }' does not work. And by all accounts I have been unable to find a clear answer on how to solve this problem.
不是每个键都有多行代码:值输入,有没有办法为单个键条目分配多个值?上面的例子'lexicon = {“direction”:['north','south','east','west','down','up','left','right','back']}'不起作用。从各方面来看,我一直无法找到解决这个问题的明确答案。
Any help is greatly appreciated. Thanks!
任何帮助是极大的赞赏。谢谢!
1 个解决方案
#1
0
"is there a way to assign multiple values to a single key entry?"
“有没有办法为单个键条目分配多个值?”
You can assign a list, like you are doing in your first lexicon. To quickly search if a word is in that list, you can use word in list like so:
您可以分配一个列表,就像您在第一个词典中所做的那样。要快速搜索单词是否在该列表中,您可以使用列表中的单词,如下所示:
for word in words:
if word in lexicon["direction"]:
word_type = "direction"
# You can also use get()
elif word in lexicon.get("number"):
word_type = "number"
else:
word_type = "err"
results.append((word_type, word))
return results
#1
0
"is there a way to assign multiple values to a single key entry?"
“有没有办法为单个键条目分配多个值?”
You can assign a list, like you are doing in your first lexicon. To quickly search if a word is in that list, you can use word in list like so:
您可以分配一个列表,就像您在第一个词典中所做的那样。要快速搜索单词是否在该列表中,您可以使用列表中的单词,如下所示:
for word in words:
if word in lexicon["direction"]:
word_type = "direction"
# You can also use get()
elif word in lexicon.get("number"):
word_type = "number"
else:
word_type = "err"
results.append((word_type, word))
return results