[英]Remove first occurrence of word in string
test = 'User Key Account Department Account Start Date'
我想從字符串中刪除重復的單詞。 這個問題的解決方案運行良好......
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()))
但它只保留后續的重復項。 我想刪除字符串中的第一次出現,以便測試字符串讀取“用戶密鑰部門帳戶開始日期”。
這應該可以完成這項工作:
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
如果您只想保留每個單詞的最后一次出現,那么只需從后面開始並繼續前進。
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'
這是一種方法:
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'
您可以將源字符串轉換為列表,然后在使用unique_list
函數之前反轉列表,然后在轉換回字符串之前再次反轉列表。
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))
例子:
$ python rev.py
User Key Account Department Account Start Date
User Key Department Account Start Date
如果你喜歡它的功能:
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
將所有元素放入一個集合中。
將您的句子標記為字符串並插入到集合中。
set<std::string> s;
s.insert("aa");
s.insert("bb");
s.insert("cc");
s.insert("cc");
s.insert("dd");
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.