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2

I'm trying to merge all lists in a list of lists that have common elements. I have some working code. However, it seems to break on this example:

def merge_subs(lst_of_lsts):
 res = []
 for row in lst_of_lsts:
 for i, resrow in enumerate(res):
 if row[0]==resrow[0]:
 res[i] += row[1:]
 break
 else:
 res.append(sorted(row))
 return sorted(res)

The input is:

merge_subs([[1, 7, 3], [1, 7, 5], [2, 0, 4], [2, 0, 6], [3, 7, 1], [3, 7, 5], [4, 0, 2], [4, 0, 6], [5, 7, 1], [5, 7, 3], [6, 0, 2], [6, 0, 4]])

My result is:

[[0, 2, 4, 6], [1, 3, 5, 7], [3, 5, 7]]

but I should be getting:

[[0, 2, 4, 6], [1, 3, 5, 7]]

python list merge

share|improve this question

edited Mar 26 at 4:00

Pikachu the Parenthesis Wizard

2,16888 gold badges1717 silver badges2929 bronze badges

asked Mar 26 at 1:32

Martin LopezMartin Lopez

4988 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  can you elaborate on the condition of merge? Your code's intention is to check only the first element of each list.
  
  – adrtam
  Mar 26 at 1:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  You result is not the output I get from running that. It does seem like graph connected components.
  
  – Kenny Ostrom
  Mar 26 at 1:48
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I don't think your result is what you'll get from the code you've written above..
  
  – kerwei
  Mar 26 at 2:40
  

  
  
  
  

add a comment | 

2

I'm trying to merge all lists in a list of lists that have common elements. I have some working code. However, it seems to break on this example:

def merge_subs(lst_of_lsts):
 res = []
 for row in lst_of_lsts:
 for i, resrow in enumerate(res):
 if row[0]==resrow[0]:
 res[i] += row[1:]
 break
 else:
 res.append(sorted(row))
 return sorted(res)

The input is:

merge_subs([[1, 7, 3], [1, 7, 5], [2, 0, 4], [2, 0, 6], [3, 7, 1], [3, 7, 5], [4, 0, 2], [4, 0, 6], [5, 7, 1], [5, 7, 3], [6, 0, 2], [6, 0, 4]])

My result is:

[[0, 2, 4, 6], [1, 3, 5, 7], [3, 5, 7]]

but I should be getting:

[[0, 2, 4, 6], [1, 3, 5, 7]]

python list merge

share|improve this question

edited Mar 26 at 4:00

Pikachu the Parenthesis Wizard

2,16888 gold badges1717 silver badges2929 bronze badges

asked Mar 26 at 1:32

Martin LopezMartin Lopez

4988 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  can you elaborate on the condition of merge? Your code's intention is to check only the first element of each list.
  
  – adrtam
  Mar 26 at 1:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  You result is not the output I get from running that. It does seem like graph connected components.
  
  – Kenny Ostrom
  Mar 26 at 1:48
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I don't think your result is what you'll get from the code you've written above..
  
  – kerwei
  Mar 26 at 2:40
  

  
  
  
  

add a comment | 

I'm trying to merge all lists in a list of lists that have common elements. I have some working code. However, it seems to break on this example:

def merge_subs(lst_of_lsts):
 res = []
 for row in lst_of_lsts:
 for i, resrow in enumerate(res):
 if row[0]==resrow[0]:
 res[i] += row[1:]
 break
 else:
 res.append(sorted(row))
 return sorted(res)

The input is:

merge_subs([[1, 7, 3], [1, 7, 5], [2, 0, 4], [2, 0, 6], [3, 7, 1], [3, 7, 5], [4, 0, 2], [4, 0, 6], [5, 7, 1], [5, 7, 3], [6, 0, 2], [6, 0, 4]])

My result is:

[[0, 2, 4, 6], [1, 3, 5, 7], [3, 5, 7]]

but I should be getting:

[[0, 2, 4, 6], [1, 3, 5, 7]]

python list merge

share|improve this question

edited Mar 26 at 4:00

Pikachu the Parenthesis Wizard

2,16888 gold badges1717 silver badges2929 bronze badges

asked Mar 26 at 1:32

Martin LopezMartin Lopez

4988 bronze badges

def merge_subs(lst_of_lsts):
 res = []
 for row in lst_of_lsts:
 for i, resrow in enumerate(res):
 if row[0]==resrow[0]:
 res[i] += row[1:]
 break
 else:
 res.append(sorted(row))
 return sorted(res)

merge_subs([[1, 7, 3], [1, 7, 5], [2, 0, 4], [2, 0, 6], [3, 7, 1], [3, 7, 5], [4, 0, 2], [4, 0, 6], [5, 7, 1], [5, 7, 3], [6, 0, 2], [6, 0, 4]])

[[0, 2, 4, 6], [1, 3, 5, 7], [3, 5, 7]]

[[0, 2, 4, 6], [1, 3, 5, 7]]

share|improve this question

edited Mar 26 at 4:00

Pikachu the Parenthesis Wizard

2,16888 gold badges1717 silver badges2929 bronze badges

asked Mar 26 at 1:32

Martin LopezMartin Lopez

4988 bronze badges

share|improve this question

edited Mar 26 at 4:00

Pikachu the Parenthesis Wizard

2,16888 gold badges1717 silver badges2929 bronze badges

asked Mar 26 at 1:32

Martin LopezMartin Lopez

4988 bronze badges

edited Mar 26 at 4:00

Pikachu the Parenthesis Wizard

2,16888 gold badges1717 silver badges2929 bronze badges

edited Mar 26 at 4:00

Pikachu the Parenthesis Wizard

2,16888 gold badges1717 silver badges2929 bronze badges

asked Mar 26 at 1:32

Martin LopezMartin Lopez

4988 bronze badges

asked Mar 26 at 1:32

Martin LopezMartin Lopez

4988 bronze badges

2 Answers
2

active

oldest

votes

0

You need to use recursion:

def group(d, _start, _c = [], _seen = [], _used=[]):
 r = [i for i in d if any(c in _start for c in i) and i not in _seen and i not in _used]
 if not r:
 yield set(_c)
 for i in d:
 if i != _start and i not in _used:
 yield from group(d, i, _c=[], _seen=[], _used=_used+[i, *r])
 else:
 yield from group(d, _start, _c=_c+[i for b in r for i in b], _seen=_seen+r, _used=_used+r)

data = [[1, 7, 3], [1, 7, 5], [2, 0, 4], [2, 0, 6], [3, 7, 1], [3, 7, 5], [4, 0, 2], [4, 0, 6], [5, 7, 1], [5, 7, 3], [6, 0, 2], [6, 0, 4]]
result = list(map(list, tuple(i) for i in group(data, data[0], _seen=[data[0]]) if i))

Output:

[[0, 2, 4, 6], [1, 3, 5, 7]]

share|improve this answer

answered Mar 26 at 1:53

Ajax1234Ajax1234

45.7k44 gold badges2929 silver badges5858 bronze badges

add a comment | 

1

I agree with @Ajax1234, this problem can be solved using recursion, specifically a tail recursion:

def merge(lists, results=None):

 if results is None:
 results = []

 if not lists:
 return results

 first = lists[0]
 merged = []
 output = []

 for li in lists[1:]:
 for i in first:
 if i in li:
 merged = merged + li
 break
 else:
 output.append(li)

 merged = merged + first
 results.append(list(set(merged)))

 return merge(output, results)

And the results look like this:

>>> lists = [[1, 7, 3], [1, 7, 5], [2, 0, 4], [2, 0, 6], [3, 7, 1], [3, 7, 5], [4, 0, 2], [4, 0, 6], [5, 7, 1], [5, 7, 3], [6, 0, 2], [6, 0, 4]]
>>> merge(lists)
[[1, 3, 5, 7], [0, 2, 4, 6]]

share|improve this answer

answered Mar 26 at 6:34

consttconstt

80411 gold badge1010 silver badges1212 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thank you, very readable!
  
  – Martin Lopez
  Mar 27 at 21:21
  

  
  
  
  

add a comment | 

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0

You need to use recursion:

def group(d, _start, _c = [], _seen = [], _used=[]):
 r = [i for i in d if any(c in _start for c in i) and i not in _seen and i not in _used]
 if not r:
 yield set(_c)
 for i in d:
 if i != _start and i not in _used:
 yield from group(d, i, _c=[], _seen=[], _used=_used+[i, *r])
 else:
 yield from group(d, _start, _c=_c+[i for b in r for i in b], _seen=_seen+r, _used=_used+r)

data = [[1, 7, 3], [1, 7, 5], [2, 0, 4], [2, 0, 6], [3, 7, 1], [3, 7, 5], [4, 0, 2], [4, 0, 6], [5, 7, 1], [5, 7, 3], [6, 0, 2], [6, 0, 4]]
result = list(map(list, tuple(i) for i in group(data, data[0], _seen=[data[0]]) if i))

Output:

[[0, 2, 4, 6], [1, 3, 5, 7]]

share|improve this answer

answered Mar 26 at 1:53

Ajax1234Ajax1234

45.7k44 gold badges2929 silver badges5858 bronze badges

add a comment | 

0

You need to use recursion:

def group(d, _start, _c = [], _seen = [], _used=[]):
 r = [i for i in d if any(c in _start for c in i) and i not in _seen and i not in _used]
 if not r:
 yield set(_c)
 for i in d:
 if i != _start and i not in _used:
 yield from group(d, i, _c=[], _seen=[], _used=_used+[i, *r])
 else:
 yield from group(d, _start, _c=_c+[i for b in r for i in b], _seen=_seen+r, _used=_used+r)

data = [[1, 7, 3], [1, 7, 5], [2, 0, 4], [2, 0, 6], [3, 7, 1], [3, 7, 5], [4, 0, 2], [4, 0, 6], [5, 7, 1], [5, 7, 3], [6, 0, 2], [6, 0, 4]]
result = list(map(list, tuple(i) for i in group(data, data[0], _seen=[data[0]]) if i))

Output:

[[0, 2, 4, 6], [1, 3, 5, 7]]

share|improve this answer

answered Mar 26 at 1:53

Ajax1234Ajax1234

45.7k44 gold badges2929 silver badges5858 bronze badges

add a comment | 

You need to use recursion:

def group(d, _start, _c = [], _seen = [], _used=[]):
 r = [i for i in d if any(c in _start for c in i) and i not in _seen and i not in _used]
 if not r:
 yield set(_c)
 for i in d:
 if i != _start and i not in _used:
 yield from group(d, i, _c=[], _seen=[], _used=_used+[i, *r])
 else:
 yield from group(d, _start, _c=_c+[i for b in r for i in b], _seen=_seen+r, _used=_used+r)

data = [[1, 7, 3], [1, 7, 5], [2, 0, 4], [2, 0, 6], [3, 7, 1], [3, 7, 5], [4, 0, 2], [4, 0, 6], [5, 7, 1], [5, 7, 3], [6, 0, 2], [6, 0, 4]]
result = list(map(list, tuple(i) for i in group(data, data[0], _seen=[data[0]]) if i))

Output:

[[0, 2, 4, 6], [1, 3, 5, 7]]

share|improve this answer

answered Mar 26 at 1:53

Ajax1234Ajax1234

45.7k44 gold badges2929 silver badges5858 bronze badges

def group(d, _start, _c = [], _seen = [], _used=[]):
 r = [i for i in d if any(c in _start for c in i) and i not in _seen and i not in _used]
 if not r:
 yield set(_c)
 for i in d:
 if i != _start and i not in _used:
 yield from group(d, i, _c=[], _seen=[], _used=_used+[i, *r])
 else:
 yield from group(d, _start, _c=_c+[i for b in r for i in b], _seen=_seen+r, _used=_used+r)

data = [[1, 7, 3], [1, 7, 5], [2, 0, 4], [2, 0, 6], [3, 7, 1], [3, 7, 5], [4, 0, 2], [4, 0, 6], [5, 7, 1], [5, 7, 3], [6, 0, 2], [6, 0, 4]]
result = list(map(list, tuple(i) for i in group(data, data[0], _seen=[data[0]]) if i))

[[0, 2, 4, 6], [1, 3, 5, 7]]

share|improve this answer

answered Mar 26 at 1:53

Ajax1234Ajax1234

45.7k44 gold badges2929 silver badges5858 bronze badges

answered Mar 26 at 1:53

Ajax1234Ajax1234

45.7k44 gold badges2929 silver badges5858 bronze badges

answered Mar 26 at 1:53

Ajax1234Ajax1234

45.7k44 gold badges2929 silver badges5858 bronze badges

1

I agree with @Ajax1234, this problem can be solved using recursion, specifically a tail recursion:

def merge(lists, results=None):

 if results is None:
 results = []

 if not lists:
 return results

 first = lists[0]
 merged = []
 output = []

 for li in lists[1:]:
 for i in first:
 if i in li:
 merged = merged + li
 break
 else:
 output.append(li)

 merged = merged + first
 results.append(list(set(merged)))

 return merge(output, results)

And the results look like this:

>>> lists = [[1, 7, 3], [1, 7, 5], [2, 0, 4], [2, 0, 6], [3, 7, 1], [3, 7, 5], [4, 0, 2], [4, 0, 6], [5, 7, 1], [5, 7, 3], [6, 0, 2], [6, 0, 4]]
>>> merge(lists)
[[1, 3, 5, 7], [0, 2, 4, 6]]

share|improve this answer

answered Mar 26 at 6:34

consttconstt

80411 gold badge1010 silver badges1212 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thank you, very readable!
  
  – Martin Lopez
  Mar 27 at 21:21
  

  
  
  
  

add a comment | 

1

I agree with @Ajax1234, this problem can be solved using recursion, specifically a tail recursion:

def merge(lists, results=None):

 if results is None:
 results = []

 if not lists:
 return results

 first = lists[0]
 merged = []
 output = []

 for li in lists[1:]:
 for i in first:
 if i in li:
 merged = merged + li
 break
 else:
 output.append(li)

 merged = merged + first
 results.append(list(set(merged)))

 return merge(output, results)

And the results look like this:

>>> lists = [[1, 7, 3], [1, 7, 5], [2, 0, 4], [2, 0, 6], [3, 7, 1], [3, 7, 5], [4, 0, 2], [4, 0, 6], [5, 7, 1], [5, 7, 3], [6, 0, 2], [6, 0, 4]]
>>> merge(lists)
[[1, 3, 5, 7], [0, 2, 4, 6]]

share|improve this answer

answered Mar 26 at 6:34

consttconstt

80411 gold badge1010 silver badges1212 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thank you, very readable!
  
  – Martin Lopez
  Mar 27 at 21:21
  

  
  
  
  

add a comment | 

I agree with @Ajax1234, this problem can be solved using recursion, specifically a tail recursion:

def merge(lists, results=None):

 if results is None:
 results = []

 if not lists:
 return results

 first = lists[0]
 merged = []
 output = []

 for li in lists[1:]:
 for i in first:
 if i in li:
 merged = merged + li
 break
 else:
 output.append(li)

 merged = merged + first
 results.append(list(set(merged)))

 return merge(output, results)

And the results look like this:

>>> lists = [[1, 7, 3], [1, 7, 5], [2, 0, 4], [2, 0, 6], [3, 7, 1], [3, 7, 5], [4, 0, 2], [4, 0, 6], [5, 7, 1], [5, 7, 3], [6, 0, 2], [6, 0, 4]]
>>> merge(lists)
[[1, 3, 5, 7], [0, 2, 4, 6]]

share|improve this answer

answered Mar 26 at 6:34

consttconstt

80411 gold badge1010 silver badges1212 bronze badges

def merge(lists, results=None):

 if results is None:
 results = []

 if not lists:
 return results

 first = lists[0]
 merged = []
 output = []

 for li in lists[1:]:
 for i in first:
 if i in li:
 merged = merged + li
 break
 else:
 output.append(li)

 merged = merged + first
 results.append(list(set(merged)))

 return merge(output, results)

>>> lists = [[1, 7, 3], [1, 7, 5], [2, 0, 4], [2, 0, 6], [3, 7, 1], [3, 7, 5], [4, 0, 2], [4, 0, 6], [5, 7, 1], [5, 7, 3], [6, 0, 2], [6, 0, 4]]
>>> merge(lists)
[[1, 3, 5, 7], [0, 2, 4, 6]]

share|improve this answer

answered Mar 26 at 6:34

consttconstt

80411 gold badge1010 silver badges1212 bronze badges

answered Mar 26 at 6:34

consttconstt

80411 gold badge1010 silver badges1212 bronze badges

answered Mar 26 at 6:34

consttconstt

80411 gold badge1010 silver badges1212 bronze badges