Question

我使用以下代码来标记字符串，从stdin读取。

d=[]
cur = ''
for i in sys.stdin.readline():
    if i in ' .':
        if cur not in d and (cur != ''):
            d.append(cur)
            cur = ''
    else:
        cur = cur + i.lower()

这给了我一系列不重复的单词。但是，在我的输出中，有些单词不会被分割。

我的输入是

Dan went to the north pole to lead an expedition during summer.

输出数组d是

['dan'，'去'，'到'，'''，'北'，'极'，'宽容'，'一个'，'远征'，'''，'夏天']

为什么tolead在一起？

Answer 1

试试这个

d=[]
cur = ''
for i in sys.stdin.readline():
    if i in ' .':
        if cur not in d and (cur != ''):
            d.append(cur)
        cur = '' # note the different indentation
    else:
        cur = cur + i.lower()

Answer 2

试试这个：

for line in sys.stdin.readline():
    res = set(word.lower() for word in line[:-1].split(" "))
    print res

示例：

line = "Dan went to the north pole to lead an expedition during summer."
res = set(word.lower() for word in line[:-1].split(" "))
print res

set(['north', 'lead', 'expedition', 'dan', 'an', 'to', 'pole', 'during', 'went', 'summer', 'the'])

评论后，我编辑：此解决方案保留输入顺序并过滤分隔符

import re
from collections import OrderedDict
line = "Dan went to the north pole to lead an expedition during summer."
list(OrderedDict.fromkeys(re.findall(r"[\w']+", line)))
# ['Dan', 'went', 'to', 'the', 'north', 'pole', 'lead', 'an', 'expedition', 'during', 'summer']

Answer 3

"to"已在"d"。因此，您的循环会跳过"to"和"lead"之间的空格，但会继续连接;一旦它到达下一个空格，就会发现"tolead"中d不在>>> import string >>> set("Dan went to the north pole to lead an expedition during summer.".translate(None, string.punctuation).lower().split()) set(['summer', 'north', 'lead', 'expedition', 'dan', 'an', 'to', 'pole', 'during', 'went', 'the'])，所以它会附加它。

更简单的解决方案;它还会删除所有形式的标点符号：

{{1}}

对字符串进行标记可以合并一些单词

3 个答案: