我整天都试图解决书籍Exercise 48: Advanced User Input的“Learn Python The Hard Way”中的test_errors()功能。
assert_equal()，测试中的一个函数按顺序要求我输入元组，但我无法按照这种方式编写代码。
我的循环总是首先返回名词并且最后返回错误元组，我不知道如何打破循环以便它再次启动但是使用正确的值继续或者按照它们应该的顺序对这些元组进行排序的任何必要条件是。
以下是代码：
class Lexicon(object):
def scan(self, stringo):
vocabulary = [[('direction', 'north'), ('direction', 'south'), ('direction', 'east'), ('direction', 'west')],
[('verb', 'go'), ('verb', 'kill'), ('verb', 'eat')],
[('stop', 'the'), ('stop', 'in'), ('stop', 'of')],
[('noun', 'bear'), ('noun', 'princess')], # Remember numbers
[('error', 'ASDFADFASDF'), ('error', 'IAS')],
[('number', '1234'), ('number','3'), ('number', '91234')]]
self.stringo = stringo
got_word = ''
value = []
rompe = self.stringo.split() #split rompe en los espacios
for asigna in vocabulary:
for encuentra in asigna:
if encuentra[1] in rompe:
value.append(encuentra)
return value
eLexicon = Lexicon()
from nose.tools import *
from ex48.ex48 import eLexicon
def test_directions():
assert_equal(eLexicon.scan("north"), [('direction', 'north')])
result = eLexicon.scan("north south east")
assert_equal(result, [('direction', 'north'),
('direction', 'south'),
('direction', 'east')])
def test_verbs():
assert_equal(eLexicon.scan("go"), [('verb', 'go')])
result = eLexicon.scan("go kill eat")
assert_equal(result, [('verb', 'go'),
('verb', 'kill'),
('verb', 'eat')])
def test_stops():
assert_equal(eLexicon.scan("the"), [('stop', 'the')])
result = eLexicon.scan("the in of")
assert_equal(result, [('stop', 'the'),
('stop', 'in'),
('stop', 'of')])
def test_nouns():
assert_equal(eLexicon.scan("bear"), [('noun', 'bear')])
result = eLexicon.scan("bear princess")
assert_equal(result, [('noun', 'bear'),
('noun', 'princess')])
#def test_numbers():
# assert_equal(lexicon.scan("1234"), [('number', 1234)])
# result = lexicon.scan("3 91234")
# assert_equal(result, [('number', 3),
# ('number', 91234)])
def test_errors():
assert_equal(eLexicon.scan("ASDFADFASDF"), [('error', 'ASDFADFASDF')])
result = eLexicon.scan("bear IAS princess")
assert_equal(result, [('noun', 'bear'),
('error', 'IAS'),
('noun', 'princess')])
======================================================================
FAIL: tests.ex48_tests.test_errors
----------------------------------------------------------------------
Traceback (most recent call last):
File "/usr/local/lib/python2.7/dist-packages/nose/case.py", line 197, in runTest
self.test(*self.arg)
File "/home/totoro/Desktop/Python/projects/ex48/tests/ex48_tests.py", line 43, in test_errors
('noun', 'princess')])
AssertionError: Lists differ: [('noun', 'bear'), ('noun', 'p... != [('noun', 'bear'), ('error', '...
First differing element 1:
('noun', 'princess')
('error', 'IAS')
- [('noun', 'bear'), ('noun', 'princess'), ('error', 'IAS')]
+ [('noun', 'bear'), ('error', 'IAS'), ('noun', 'princess')]
----------------------------------------------------------------------
Ran 5 tests in 0.006s
非常感谢您抽出时间。
答案 0 :(得分：1)
这对我来说很好，它是更短的代码。这本书不久之前，谢天谢地我还有文件......
def check(word):
lexicon = {
'direction': ['north', 'south', 'east', 'west'],
'verb': ['go', 'kill', 'eat'],
'stop': ['the', 'in', 'of'],
'noun': ['bear', 'princess'],
'error': ['ASDFADFASDF', 'IAS']
}
word = str(word)
for key in lexicon:
if word in lexicon[key]:
return (key, word)
elif word.isdigit():
return ('number', int(word))
def scan(words):
words = words.split()
to_return = []
for i in words:
to_return.append(check(i))
return to_return
这出现了：
......
----------------------------------------------------------------------
Ran 6 tests in 0.008s
OK
告诉我此代码是否有任何错误。请在下面评论：D。
答案 1 :(得分：0)
测试中的单词与出现时的顺序相同。因此，您需要重新排序for - 循环以首先迭代输入：
value = []
for rompe in stringo.split():
for asigna in vocabulary:
for encuentra in asigna:
if encuentra[1] == rompe:
value.append(encuentra)
这将以正确的顺序返回encuentra。
注1 ：您不应该对数字或错误进行硬编码。
注意2 ：您可以使用一两个字典来大幅降低此算法的复杂性。
vocabulary = {
'direction': 'north east south west up down left right back'.split(),
'noun': 'bear princess door cabinet'.split(),
'stop': 'the in of from at it'.split(),
'verb': 'go kill eat stop'.split(),
}
'''
This creates a lookup using a dictionary-comprehension:
{'at': 'stop',
# [...]
'up': 'direction',
'west': 'direction'}
'''
classifications = {i: k for k, v in vocabulary.iteritems() for i in v}
def classify(word):
try:
return 'number', int(word)
except ValueError:
return classifications.get(word, 'error'), word
def scan(words):
return [classify(word) for word in words.split()]
答案 2 :(得分：0)
for word in self.stringo.split():
for pair in vocabulary:
if pair[0][1] == word:
value.append(pair[0])
elif pair[1][1] == word:
value.append(pair[1])
elif pair[2][1] == word:
value.append(pair[2])
elif pair[3][1] == word:
value.append(pair[3])
答案 3 :(得分：0)
我刚刚完成了这个练习，希望这给你们一些新的想法。 这是我的解决方案：
#Set up datastructure
direction = ["north", "east", "south", "west", "up", "right", "down", "left", "back"]
verb = ["go", "stop", "kill", "eat"]
stop = ["the", "in", "of", "from", "at", "it"]
noun = ["door", "bear", "princess", "cabinet"]
vocabulary = [(direction, 'direction'), (verb, 'verb'), (stop, 'stop'), (noun, 'noun')]
def scan(sentence):
#searches the words in the datastructure, if not found checks if it is an integer, if not returns error.
results = []
words = sentence.split()
for word in words:
found = False
for category in vocabulary:
if word.lower() in category[0]:
results.append((category[1], word))
found = True
else:
pass
if found is False and isInt_str(word) is True:
results.append(('number', int(word)))
elif found is False and isInt_str(word) is False:
results.append(('error', word))
elif found is True:
pass
else:
print("I'm terribly sorry, but something you entered is neither a word nor a number.")
return results
def isInt_str(string):
#returns True or False if string equals an integer. (i.e. 2 = True, 2*-2 = True 2**0,5 = False)
string = str(string).strip()
return string=='0' or (string if string.find('..') > -1 else string.lstrip('-+').rstrip('0').rstrip('.')).isdigit()