问题描述
我的代码：
directions = ["north", "south", "east", "west"]
def scan(sentence):
global sentence_list
sentence_list = []
sentence.split()
for i in sentence:
if i in directions:
a = ('direction', i)
sentence_list.append(a)
del a
return sentence_list
我试图拆分一个字符串并返回列表中元组中的单词，但是每当我使用返回空列表对其进行测试时。
这是我的输出：
PS C:\Users\dell 3521\lpythw\ex48> nosetests
F
======================================================================
FAIL: tests.lexicon_tests.test_directions
----------------------------------------------------------------------
Traceback (most recent call last):
File "C:\Users\dell 3521\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:\Users\dell 3521\lpythw\ex48\tests\lexicon_tests.py", line 5, in
test_directions
assert_equal(lexicon.scan("north"), [('direction', 'north')])
AssertionError: Lists differ: [] != [('direction', 'north')]
Second list contains 1 additional elements.
First extra element 0:
('direction', 'north')
- []
+ [('direction', 'north')]
----------------------------------------------------------------------
Ran 1 test in 0.021s
FAILED (failures=1)
提前致谢。
1楼
您必须将sentence重新分配给sentence.split()的返回值或直接迭代sentence.split() ，因为方法不会就地修改sentence ，而是返回一个列表。
你也不需要del a语句。
将您的代码更改为
directions = ["north", "south", "east", "west"]
def scan(sentence):
global sentence_list
sentence_list = []
for i in sentence.split():
if i in directions:
a = ('direction', i)
sentence_list.append(a)
return sentence_list
或者更短的方法是使用
directions = ["north", "south", "east", "west"]
def scan(sentence):
global sentence_list
sentence_list = [('direction', i) for i in sentence.split() if i in directions]
return sentence_list
输出是
>>> scan("north")
[('direction', 'north')]
并且您可能希望在代码中考虑使用global语句。
正如，为了代码的可读性和可维护性，您希望避免使用全局变量。
2楼
str.split()方法不会就地修改字符串。
您应该将str.split()的返回值分配给一个变量，或者在这种情况下，您可以简单地迭代它：
sentence_list = []
for i in sentence.split():