所以我正在制作一个程序来创建portmanteaus。我有我需要的代码和功能,我把它们放在一起。
以下是代码:
def portmanteauscore(start, mid, end):
totallen = len(start) + len(mid) + len(end)
return totallen - abs((len(start)/totallen) - len(start)) - abs((len(mid)/totallen) - len(mid)) - abs((len(end)/totallen) - len(end))
def portmanteaugenerator(word1, word2, words):
mid = longest_common_substring(word1, word2)
start = word1[:word1.index(mid)]
end = word2[len(mid):]
if start + mid in words and mid + end in words:
return start, mid, end
def natalie(words):
"Find the best Portmanteau word formed from any two of the list of words."
wordpermutations = list(itertools.permutations(words))
maxscore, bestnatalie = 0, ''
for perm in wordpermutations:
start, mid, end = portmanteaugenerator(perm[0], perm[1], words)
if portmanteauscore(start, mid, end) > maxscore:
bestnatalie, maxscore = start + mid + end, portmanteauscore(start, mid, end)
print bestnatalie
return bestnatalie
def longest_common_substring(s1, s2):
m = [[0] * (1 + len(s2)) for i in xrange(1 + len(s1))]
longest, x_longest = 0, 0
for x in xrange(1, 1 + len(s1)):
for y in xrange(1, 1 + len(s2)):
if s1[x - 1] == s2[y - 1]:
m[x][y] = m[x - 1][y - 1] + 1
if m[x][y] > longest:
longest = m[x][y]
x_longest = x
else:
m[x][y] = 0
return s1[x_longest - longest: x_longest]
但是当我运行代码时,我不断收到此错误消息,
Traceback (most recent call last):
File "vm_main.py", line 33, in <module>
import main
File "/tmp/vmuser_ijxrjleuxj/main.py", line 107, in <module>
print test_natalie()
File "/tmp/vmuser_ijxrjleuxj/main.py", line 87, in test_natalie
assert natalie(['adolescent', 'scented', 'centennial', 'always', 'ado']) in ('adolescented','adolescentennial')
File "/tmp/vmuser_ijxrjleuxj/main.py", line 67, in natalie
start,mid,end=portmanteaugenerator(perm[0],perm[1],words)
TypeError: 'NoneType' object is not iterable
当我返回portmanteau生成器的开始,中间和结束变量时会发生这种情况。当给出一个单词列表时,它应该从两个单词中返回一个portmanteau,这是根据portmanteau得分最好的。
但是由于某种原因我一直都会遇到这种类型错误。我尝试过开始,中间,结束列表,但仍然无法运行。你能帮我么?
答案 0 :(得分:6)
最有可能的是,start+mid in words and mid+end in words
返回False
,因此函数返回if-statement
而不是通过None
(因为如果某个函数没有返回某些内容,那么默认为None
)。
然后你要尝试做类似的事情:
start,mid,end = None
python尝试做的是在这三个变量中拆分None
。就像这样做:
one, two, three = (1, 2, 3)
但你不能,因为None
不是可迭代的。