在字符串列表的列表中找到不重复的字符串出现（附加更多条件）

huangapple
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英文:

Occurence of not repeated string in alist of list of strings (update with more conditions)

问题

  1. from collections import Counter
  3. List=[[ 'Jhon', 'Rana'], ['Rana', 'Rana'], ['Rana', 'Zhang'], ['Rana', ['Z', 'Y']]]
  5. # Flatten the nested list and count occurrences
  6. flat_list = [item for sublist in List for item in sublist]
  7. word_counts = Counter(flat_list)
  9. # First Solution: Count without considering duplicates
  10. result1 = {word: count for word, count in word_counts.items() if count == 1}
  12. # Second Solution: Count with duplicates, considering the special relationship
  13. result2 = {}
  14. for sublist in List:
  15.     for item in sublist:
  16.         if isinstance(item, list):
  17.             result2[item[0]] = result2.get(item[0], 0) + 1
  18.         else:
  19.             result2[item] = result2.get(item, 0) + 1
  21. # Print the results
  22. result1_text = '\n'.join([f'{key}:{value}' for key, value in result1.items()])
  23. result2_text = '\n'.join([f'{key}:{value}' for key, value in result2.items()])
  25. result1_text, result2_text

Output for the first solution:

  1. Jhon:1
  2. Zhang:1

Output for the second solution (including the special relationship):

  1. Jhon:1
  2. Zhang:1
  3. Rana:5

Please note that in the second solution, 'Rana' is counted as 5 because of the relationships with 'Z' and 'Y'.

英文:

I have a list of list like this one :

  1. List=[['Jhon', 'Rana'], ['Rana', 'Rana'], ['Rana', 'Zhang], ['Rana', ['Z', 'Y']]

****Update: In this list, I had another structure where the first element is a single element and the second element is a list like ['Rana', ['Z', 'Y']], which means that Rana has a specific relationship with Z and R, which are not the same relationship as Rana and Jhon. ****

I want to calculate the occurence of the word of this list and I need two kind of output. The first one when we have a duplicated (or repeated word), we ignore it. The second solution, when we detect the repeated word we count it as once not twice.

Update: I want to add this type of relation to be included in the second output.

For example for the first solution the result will be
Rana:2
Jhon:1
Zhang:1

the second solution will be
Rana:3 (Update: will be 5 instead of 3 since we will consider Z and Y)
Jhon:1
Zhang: 1

I have tried to develop the following lignes of code, but I didn´t have results:

  1. from collections import Counter
  2. List1=[["Rana", "Jhon"], ["Rana", "Rana"], ["Jhon", "Rana"], ["Rana", "Alex"]]
  3. count=0
  4. n=0
  5. for j in range (0, len(List1)-1):
  6.   if (List1[j][0] == List1[j][1] ) or (List1[j][0] != List1[j][1] ):
  7.     count += 1
  8. print(count)

答案1

得分: 1

以下是已翻译的部分：

第一部分，您可以通过忽略具有重复值的内部列表来实现：

  1. Lst=[['Jhon', 'Rana'], ['Rana', 'Rana'], ['Rana', 'Zhang']]
  3. l = [x for x in Lst if len(set(x))==len(x)]     # len(set())将确保内部列表中只有唯一元素
  4. # [['Jhon', 'Rana'], ['Rana', 'Zhang']]
  6. # 扁平化l列表
  7. m = [item for sublist in l for item in sublist]
  8. # ['Jhon', 'Rana', 'Rana', 'Zhang']
  10. from collections import Counter as c
  11. c(m)
  13. # 输出
  14. Counter({'Jhon': 1, 'Rana': 2, 'Zhang': 1})

len(set(x))将x转换为一个set。如果存在重复值，则len(x)将不等于len(set(x))。

第二部分，如果内部列表中存在相同的值，您可以只添加1个值：

  1. Lst=[['Jhon', 'Rana'], ['Rana', 'Rana'], ['Rana', 'Zhang']]
  3. l = [x if len(set(x))==len(x) else [x[0]] for x in Lst ]
  4. # [['Jhon', 'Rana'], ['Rana'], ['Rana', 'Zhang']]
  6. # 扁平化l列表
  7. n = [item for sublist in l for item in sublist]
  8. # ['Jhon', 'Rana', 'Rana', 'Rana', 'Zhang']
  10. from collections import Counter as c
  11. c(n)
  13. # 输出
  14. Counter({'Jhon': 1, 'Rana': 3, 'Zhang': 1})

现在，如果您有一个包含反向元素的列表，如下所示：

  1. Lst1=[["Rana", "Jhon"], ["Rana", "Rana"], ["Jhon", "Rana"], ["Rana", "Alex"]]
  3. # 您可以通过以下方式删除["Jhon", "Rana"]和["Rana", "Rana"]：
  4. # ["Rana", "Rana"] 也是其本身的反向。
  6. seen = set()
  7. new_Lst1 = [x for x in Lst1 if tuple(x[::-1]) not in seen and not seen.add(tuple(x))]
  9. print(new_Lst1)
  10. # [['Rana', 'Jhon'], ['Rana', 'Alex']]
  12. # 扁平化new_Lst1列表
  13. p = [item for sublist in new_Lst1 for item in sublist]
  14. # ['Rana', 'Jhon', 'Rana', 'Alex']
  16. from collections import Counter as c
  17. c(p)
  19. # 输出
  20. Counter({'Rana': 2, 'Jhon': 1, 'Alex': 1})

希望这对您有所帮助。

英文:

For the first part you can ignore the inner list with repeating values by:

  1. Lst=[['Jhon', 'Rana'], ['Rana', 'Rana'], ['Rana', 'Zhang']]
  3. l = [x for x in Lst if len(set(x))==len(x)]     #len(set()) will make sure only unique elements are there in inner-list
  4. #[['Jhon', 'Rana'], ['Rana', 'Zhang']]
  6. #Flatten l by
  8. m = [item for sublist in l for item in sublist]
  9. #['Jhon', 'Rana', 'Rana', 'Zhang']
  11. from collections import Counter as c
  12. c(m)
  14. #output
  15. Counter({'Jhon': 1, 'Rana': 2, 'Zhang': 1})

len(set(x)) converts x into a set. If there are repeating values then the len(x) will NOT be equal to len(set(x))

For the second part if there are same values in an inner list you can add only 1 value by:

  1. Lst=[['Jhon', 'Rana'], ['Rana', 'Rana'], ['Rana', 'Zhang']]
  3. l = [x if len(set(x))==len(x) else [x[0]] for x in Lst ]
  4. #[['Jhon', 'Rana'], ['Rana'], ['Rana', 'Zhang']]
  6. #Flatten l by:
  8. n = [item for sublist in l for item in sublist]
  9. #['Jhon', 'Rana', 'Rana', 'Rana', 'Zhang']
  11. from collections import Counter as c
  12. c(n)
  14. #output
  15. Counter({'Jhon': 1, 'Rana': 3, 'Zhang': 1})

Edit:

Now, if you have a list with reverse elements like:

  1. Lst1=[["Rana", "Jhon"], ["Rana", "Rana"], ["Jhon", "Rana"], ["Rana", "Alex"]]

You can remove ["Jhon", "Rana"] and ["Rana", "Rana"] by:

#["Rana", "Rana"] is also an inverse of itself.

  1. seen = set()
  2. new_Lst1 = [x for x in Lst1 if tuple(x[::-1]) not in seen and not seen.add(tuple(x))]
  4. print(new_Lst1)
  5. [['Rana', 'Jhon'], ['Rana', 'Alex']]
  8. #Flatten new_Lst1 by
  10. p = [item for sublist in new_Lst1 for item in sublist]
  11. #['Rana', 'Jhon', 'Rana', 'Alex']
  13. from collections import Counter as c
  14. c(p)
  16. #output
  17. Counter({'Rana': 2, 'Jhon': 1, 'Alex': 1})

答案2

得分: 1

根据您的描述，以下是从 https://stackoverflow.com/a/30357006 中理解的内容：

  1. from collections import defaultdict
  3. list_1 = [['Jhon', 'Rana'], ['Rana', 'Rana'], ['Rana', 'Zhang']]
  4. # list_1 = [['Rana', 'Jhon'], ['Jhon', 'Rana']]
  6. seen = []
  8. for sub in list_1:
  9.     sub = sorted(sub)
  10.     if sub not in seen:
  11.         seen.append(sub) 
  13. res_1 = defaultdict(lambda: 0)
  14. res_2 = defaultdict(lambda: 0)
  16. for sub in seen:
  17.     a, b = sub
  18.     if a == b:
  19.         res_2[a] += 1;
  20.     else:
  21.         res_2[a] += 1
  22.         res_2[b] += 1
  23.         res_1[a] += 1;
  24.         res_1[b] += 1;
  26. print(dict(res_1)) #=> {'Jhon': 1, 'Rana': 2, 'Zhang': 1}
  27. print(dict(res_2)) #=> {'Jhon': 1, 'Rana': 3, 'Zhang': 1}

或者添加更多的情况，如已经评论中所述。

英文:

For what I can understand from your description, using https://stackoverflow.com/a/30357006:

  1. from collections import defaultdict
  3. list_1 = [['Jhon', 'Rana'], ['Rana', 'Rana'], ['Rana', 'Zhang']]
  4. # list_1 = [['Rana', 'Jhon'], ['Jhon', 'Rana']]
  6. seen = []
  8. for sub in list_1:
  9. 	sub = sorted(sub)
  10. 	if sub not in seen:
  11. 		seen.append(sub) 
  13. res_1 = defaultdict(lambda: 0)
  14. res_2 = defaultdict(lambda: 0)
  16. for sub in seen:
  17. 	a, b = sub
  18. 	if a == b:
  19. 		res_2[a] += 1;
  20. 	else:
  21. 		res_2[a] += 1
  22. 		res_2[b] += 1
  23. 		res_1[a] += 1;
  24. 		res_1[b] += 1;
  27. print(dict(res_1)) #=> {'Jhon': 1, 'Rana': 2, 'Zhang': 1}
  28. print(dict(res_2)) #=> {'Jhon': 1, 'Rana': 3, 'Zhang': 1}

Or add more cases, as already commented.

huangapple
  • 本文由 发表于 2023年6月5日 22:09:32
  • 转载请务必保留本文链接：https://go.coder-hub.com/76407282.html
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  • python-3.x
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