test = 'User Key Account Department Account Start Date'
我想从字符串中删除重复的单词。 this question 的解决方案运行良好...
def unique_list(l):
ulist = []
[ulist.append(x) for x in l if x not in ulist]
return ulist
test = ' '.join(unique_list(test.split()))
但它只保留后续的重复项。我想删除字符串中的第一个匹配项,以便测试字符串读取“用户密钥部门帐户开始日期”。
答案 0 :(得分:3)
这应该可以:
test = 'User Key Account Department Account Start Date'
words = test.split()
# if word doesn't exist in the rest of the word list, add it
test = ' '.join([word for i, word in enumerate(words) if word not in words[i+1:]])
print(test) # User Key Department Account Start Date
答案 1 :(得分:1)
如果您只想保留每个单词的最后一次出现,那么只需从后面开始,然后继续前进。
tokens = test.split()
final = []
for word in tokens[::-1]:
if word in final:
continue
else:
final.append(word)
print(" ".join(final[::-1]))
>> 'User Key Department Account Start Date'
答案 2 :(得分:1)
这是一种方法:
l=test.split()
m=set([i for i in l if test.count(i)>1])
for i in m:
l.remove(i)
res = ' '.join(l)
>>> print(res)
'User Key Department Account Start Date'
答案 3 :(得分:1)
您可以将源字符串转换为列表,然后在使用 unique_list
函数之前反转列表,然后在转换回字符串之前再次反转列表。
def unique_list(l):
ulist = []
[ulist.append(x) for x in l if x not in ulist]
return ulist
orig="User Key Account Department Account Start Date"
orig_list=orig.split()
orig_list.reverse()
uniq_rev=unique_list(orig_list)
uniq_rev.reverse()
print(orig)
print(' '.join(uniq_rev))
示例:
$ python rev.py
User Key Account Department Account Start Date
User Key Department Account Start Date
答案 4 :(得分:0)
如果你喜欢它的功能:
from functools import reduce
from collections import Counter
import re
if __name__ == '__main__':
sentence = 'User Key Account Department Account Start Date'
result = reduce(
lambda sentence, word: re.sub(rf'{word}\s*', '', sentence, count=1),
map(
lambda item: item[0],
filter(
lambda item: item[1] > 1,
Counter(sentence.split()).items()
)
),
sentence
)
print(result)
# User Key Department Account Start Date
答案 5 :(得分:-1)
将所有元素放入一个集合中。
将您的句子标记为字符串并插入到集合中。
set<std::string> s;
s.insert("aa");
s.insert("bb");
s.insert("cc");
s.insert("cc");
s.insert("dd");