[英]python: flatten list while preserving nested structure for certain indexes
[英]Returning a certain list structure in Python
我在使用列表(和列表列表,以及lt ...的列表列表)结构时遇到麻烦。
使用+ =代替.append()运算符。 并使用第二个列表临时
new_list = []
for sublist in lines:
temp = []
for word in sublist:
temp+=key_to_phonemes[word]
new_list.append(temp)
编辑:
new_list = []
for n,sublist in enumerate(lines):
new_list.append([])
for word in sublist:
new_list[n]+=key_to_phonemes[word]
更好,因为它可以节省您的工作时间
EndEdit中
这里的行为差异如下。
[1,2,3] .append([4,5])
[ 1 , 2 , 3 , [ 4 , 5 ] ]
比。
[1,2,3] + [4,5]
[ 1 , 2 , 3 , 4 , 5 ]
您需要在循环中添加新列表并折叠key_to_phonemes:
new_list = []
for sublist in lines:
sub = []
for word in sublist:
for phoneme in key_to_phonemes[word]:
sub.append(phoneme)
new_list.append(sub)
return new_list
您还可以通过列表理解来做到这一点:
return [[p for word in sublist for p in key_to_phonemes[word]] for sublist in lines]
输出:
>>> convert_to_phonemes([['THE', 'FIRST'], ['WITH', 'A']], key_to_phonemes)
[['DH', 'AH0', 'F', 'ER1', 'S', 'T'], ['W', 'IH1', 'DH', 'AH0']]
for sublist in lines:
new_list.append([])
for word in sublist:
x = set(key_to_phonemes[word])
y = set(new_list[-1])
y = list(x.union(y))
new_list[-1] = y
return new_list
对于每个子列表,您都将创建一个新列表。然后使用集合概念,我们将不同的key_to_phonemes添加到列表的最后一个元素。
您可以使用list comprehension
来做到这一点
lines = [['THE', 'FIRST'], ['WITH', 'A']]
key_to_phonemes = {'WITH': ['W', 'IH1', 'DH'], 'THE': ['DH', 'AH0'], 'A': ['AH0'], 'FIRST': ['F', 'ER1', 'S', 'T']}
def convert_to_phonemes(lines, word_to_phonemes):
return [[k for j in i for k in key_to_phonemes[j]] for i in lines]
>>>convert_to_phonemes(lines, word_to_phonemes)
[['DH', 'AH0', 'F', 'ER1', 'S', 'T'], ['W', 'IH1', 'DH', 'AH0']]
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.