Merge tuples with the same key

up vote
8
down vote

favorite

How to merge a tuple with the same key

list_1 = [("AAA", [123]), ("AAA", [456]), ("AAW", [147]), ("AAW", [124])]

and turn them into

list_2 = [("AAA", [123, 456]), ("AAW", [147, 124])]

edited 16 mins ago

codeforester

15.9k73555

asked 8 hours ago

Theprofessor14

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New contributor

1

Related: Python: merge lists of tuples based on its values
â€“Â Aran-Fey
8 hours ago

2

all those years and not a single exact dupe... I had a good look. Maybe it's somewhere, well...
â€“Â Jean-FranÃ§ois Fabre
7 hours ago

add a commentÂ |Â

up vote
8
down vote

favorite

How to merge a tuple with the same key

list_1 = [("AAA", [123]), ("AAA", [456]), ("AAW", [147]), ("AAW", [124])]

and turn them into

list_2 = [("AAA", [123, 456]), ("AAW", [147, 124])]

edited 16 mins ago

codeforester

15.9k73555

asked 8 hours ago

Theprofessor14

442

New contributor

1

Related: Python: merge lists of tuples based on its values
â€“Â Aran-Fey
8 hours ago

2

all those years and not a single exact dupe... I had a good look. Maybe it's somewhere, well...
â€“Â Jean-FranÃ§ois Fabre
7 hours ago

add a commentÂ |Â

up vote
8
down vote

favorite

How to merge a tuple with the same key

list_1 = [("AAA", [123]), ("AAA", [456]), ("AAW", [147]), ("AAW", [124])]

and turn them into

list_2 = [("AAA", [123, 456]), ("AAW", [147, 124])]

edited 16 mins ago

codeforester

15.9k73555

asked 8 hours ago

Theprofessor14

442

New contributor

How to merge a tuple with the same key

list_1 = [("AAA", [123]), ("AAA", [456]), ("AAW", [147]), ("AAW", [124])]

and turn them into

list_2 = [("AAA", [123, 456]), ("AAW", [147, 124])]

python tuples

edited 16 mins ago

codeforester

15.9k73555

asked 8 hours ago

Theprofessor14

442

New contributor

edited 16 mins ago

codeforester

15.9k73555

asked 8 hours ago

Theprofessor14

442

New contributor

edited 16 mins ago

codeforester

15.9k73555

edited 16 mins ago

codeforester

15.9k73555

edited 16 mins ago

codeforester

15.9k73555

asked 8 hours ago

Theprofessor14

442

New contributor

asked 8 hours ago

Theprofessor14

442

asked 8 hours ago

Theprofessor14

442

New contributor

Theprofessor14 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

1

Related: Python: merge lists of tuples based on its values
â€“Â Aran-Fey
8 hours ago

2

all those years and not a single exact dupe... I had a good look. Maybe it's somewhere, well...
â€“Â Jean-FranÃ§ois Fabre
7 hours ago

add a commentÂ |Â

1

Related: Python: merge lists of tuples based on its values
â€“Â Aran-Fey
8 hours ago

2

all those years and not a single exact dupe... I had a good look. Maybe it's somewhere, well...
â€“Â Jean-FranÃ§ois Fabre
7 hours ago

Related: Python: merge lists of tuples based on its values
â€“Â Aran-Fey
8 hours ago

all those years and not a single exact dupe... I had a good look. Maybe it's somewhere, well...
â€“Â Jean-FranÃ§ois Fabre
7 hours ago

add a commentÂ |Â

4 Answers
4

active

oldest

votes

up vote
8
down vote

The most performant approach is to use a collections.defaultdict dictionary to store data as an expanding list, then convert back to tuple/list if needed:

import collections

list_1 = [("AAA", [123]), ("AAA", [456]), ("AAW", [147]), ("AAW", [124])]

c = collections.defaultdict(list)
for a,b in list_1:
 c[a].extend(b) # add to existing list or create a new one

list_2 = list(c.items())

result:

[('AAW', [147, 124]), ('AAA', [123, 456])]

note that the converted data is probably better left as dictionary. Converting to list again loses the "key" feature of the dictionary.

On the other hand, if you want to retain the order of the "keys" of the original list of tuples, unless you're using python 3.6/3.7, you'd have to create a list with the original "keys" (ordered, unique), then rebuild the list from the dictionary. Or use an OrderedDict but then you cannot use defaultdict (or use a recipe)

edited 7 hours ago

answered 8 hours ago

Jean-FranÃ§ois Fabre

92.1k847102

1

Note that this answer would not necessarily retain the order of the original list, if that is a consideration.
â€“Â blhsing
7 hours ago

true, unless you're using python 3.7. To retain the original order you'd have to create a list with the original "keys" (ordered, unique), then rebuild the list from the dictionary.
â€“Â Jean-FranÃ§ois Fabre
7 hours ago

add a commentÂ |Â

up vote
4
down vote

You can use a dict to keep track of the indices of each key to keep the time complexity O(n):

list_1 = [("AAA", [123]), ("AAA", [456]), ("AAW", [147]), ("AAW", [124])]
list_2 = 
i = 
for k, s in list_1:
 if k not in i:
 list_2.append((k, s))
 i[k] = len(i)
 else:
 list_2[i[k]][1].extend(s)

list_2 would become:

[('AAA', [123, 456]), ('AAW', [147, 124])]

edited 31 mins ago

codeforester

15.9k73555

answered 8 hours ago

blhsing

16.5k2631

add a commentÂ |Â

up vote
1
down vote

Similarly to other answers, you can use a dictionary to associate each key with a list of values. This is implemented in the function merge_by_keys in the code snippet below.

import pprint

list_1 = [("AAA", [123]), ("AAA", [456]), ("AAW", [147]), ("AAW", [124])]
list_2 = [("AAA", [123, 456]), ("AAW", [147, 124])]

def merge_by_key(ts):

 d = 
 for t in ts:
 key = t[0]
 values = t[1]
 if key not in d:
 d[key] = values[:]
 else:
 d[key].extend(values)

 return d.items()



result = merge_by_key(list_1)

pprint.pprint(result)

answered 7 hours ago

Odysseas

445210

Why are you manually creating list_2?
â€“Â codeforester
20 mins ago

add a commentÂ |Â

up vote
1
down vote

You can create a dictionary and loop through the list. If the item present in dictionary append the value to already existing list else assign the value to key.

dict_1 = 
for item in list_1:
 if item[0] in dict_1:
 dict_1[item[0]].append(item[1][0])
 else:
 dict_1[item[0]] = item[1]
list_2 = list(dict_1.items())

edited 23 mins ago

codeforester

15.9k73555

answered 8 hours ago

S.Harish

644

New contributor

It does the merge but doesn't retain the original order in the list. The result I got is: [('AAW', [147, 124]), ('AAA', [123, 456])] - AAA should have been the first element as in the original list.
â€“Â codeforester
22 mins ago

add a commentÂ |Â

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

up vote
8
down vote

The most performant approach is to use a collections.defaultdict dictionary to store data as an expanding list, then convert back to tuple/list if needed:

import collections

list_1 = [("AAA", [123]), ("AAA", [456]), ("AAW", [147]), ("AAW", [124])]

c = collections.defaultdict(list)
for a,b in list_1:
 c[a].extend(b) # add to existing list or create a new one

list_2 = list(c.items())

result:

[('AAW', [147, 124]), ('AAA', [123, 456])]

note that the converted data is probably better left as dictionary. Converting to list again loses the "key" feature of the dictionary.

edited 7 hours ago

answered 8 hours ago

Jean-FranÃ§ois Fabre

92.1k847102

1

Note that this answer would not necessarily retain the order of the original list, if that is a consideration.
â€“Â blhsing
7 hours ago

true, unless you're using python 3.7. To retain the original order you'd have to create a list with the original "keys" (ordered, unique), then rebuild the list from the dictionary.
â€“Â Jean-FranÃ§ois Fabre
7 hours ago

add a commentÂ |Â

up vote
8
down vote

The most performant approach is to use a collections.defaultdict dictionary to store data as an expanding list, then convert back to tuple/list if needed:

import collections

list_1 = [("AAA", [123]), ("AAA", [456]), ("AAW", [147]), ("AAW", [124])]

c = collections.defaultdict(list)
for a,b in list_1:
 c[a].extend(b) # add to existing list or create a new one

list_2 = list(c.items())

result:

[('AAW', [147, 124]), ('AAA', [123, 456])]

note that the converted data is probably better left as dictionary. Converting to list again loses the "key" feature of the dictionary.

edited 7 hours ago

answered 8 hours ago

Jean-FranÃ§ois Fabre

92.1k847102

1

Note that this answer would not necessarily retain the order of the original list, if that is a consideration.
â€“Â blhsing
7 hours ago

true, unless you're using python 3.7. To retain the original order you'd have to create a list with the original "keys" (ordered, unique), then rebuild the list from the dictionary.
â€“Â Jean-FranÃ§ois Fabre
7 hours ago

add a commentÂ |Â

up vote
8
down vote

The most performant approach is to use a collections.defaultdict dictionary to store data as an expanding list, then convert back to tuple/list if needed:

import collections

list_1 = [("AAA", [123]), ("AAA", [456]), ("AAW", [147]), ("AAW", [124])]

c = collections.defaultdict(list)
for a,b in list_1:
 c[a].extend(b) # add to existing list or create a new one

list_2 = list(c.items())

result:

[('AAW', [147, 124]), ('AAA', [123, 456])]

note that the converted data is probably better left as dictionary. Converting to list again loses the "key" feature of the dictionary.

edited 7 hours ago

answered 8 hours ago

Jean-FranÃ§ois Fabre

92.1k847102

The most performant approach is to use a collections.defaultdict dictionary to store data as an expanding list, then convert back to tuple/list if needed:

import collections

list_1 = [("AAA", [123]), ("AAA", [456]), ("AAW", [147]), ("AAW", [124])]

c = collections.defaultdict(list)
for a,b in list_1:
 c[a].extend(b) # add to existing list or create a new one

list_2 = list(c.items())

result:

[('AAW', [147, 124]), ('AAA', [123, 456])]

note that the converted data is probably better left as dictionary. Converting to list again loses the "key" feature of the dictionary.

edited 7 hours ago

answered 8 hours ago

Jean-FranÃ§ois Fabre

92.1k847102

edited 7 hours ago

answered 8 hours ago

Jean-FranÃ§ois Fabre

92.1k847102

answered 8 hours ago

Jean-FranÃ§ois Fabre

92.1k847102

answered 8 hours ago

Jean-FranÃ§ois Fabre

92.1k847102

1

Note that this answer would not necessarily retain the order of the original list, if that is a consideration.
â€“Â blhsing
7 hours ago

true, unless you're using python 3.7. To retain the original order you'd have to create a list with the original "keys" (ordered, unique), then rebuild the list from the dictionary.
â€“Â Jean-FranÃ§ois Fabre
7 hours ago

add a commentÂ |Â

1

Note that this answer would not necessarily retain the order of the original list, if that is a consideration.
â€“Â blhsing
7 hours ago

true, unless you're using python 3.7. To retain the original order you'd have to create a list with the original "keys" (ordered, unique), then rebuild the list from the dictionary.
â€“Â Jean-FranÃ§ois Fabre
7 hours ago

Note that this answer would not necessarily retain the order of the original list, if that is a consideration.
â€“Â blhsing
7 hours ago

true, unless you're using python 3.7. To retain the original order you'd have to create a list with the original "keys" (ordered, unique), then rebuild the list from the dictionary.
â€“Â Jean-FranÃ§ois Fabre
7 hours ago

add a commentÂ |Â

up vote
4
down vote

You can use a dict to keep track of the indices of each key to keep the time complexity O(n):

list_1 = [("AAA", [123]), ("AAA", [456]), ("AAW", [147]), ("AAW", [124])]
list_2 = 
i = 
for k, s in list_1:
 if k not in i:
 list_2.append((k, s))
 i[k] = len(i)
 else:
 list_2[i[k]][1].extend(s)

list_2 would become:

[('AAA', [123, 456]), ('AAW', [147, 124])]

edited 31 mins ago

codeforester

15.9k73555

answered 8 hours ago

blhsing

16.5k2631

add a commentÂ |Â

up vote
4
down vote

You can use a dict to keep track of the indices of each key to keep the time complexity O(n):

list_1 = [("AAA", [123]), ("AAA", [456]), ("AAW", [147]), ("AAW", [124])]
list_2 = 
i = 
for k, s in list_1:
 if k not in i:
 list_2.append((k, s))
 i[k] = len(i)
 else:
 list_2[i[k]][1].extend(s)

list_2 would become:

[('AAA', [123, 456]), ('AAW', [147, 124])]

edited 31 mins ago

codeforester

15.9k73555

answered 8 hours ago

blhsing

16.5k2631

add a commentÂ |Â

up vote
4
down vote

You can use a dict to keep track of the indices of each key to keep the time complexity O(n):

list_1 = [("AAA", [123]), ("AAA", [456]), ("AAW", [147]), ("AAW", [124])]
list_2 = 
i = 
for k, s in list_1:
 if k not in i:
 list_2.append((k, s))
 i[k] = len(i)
 else:
 list_2[i[k]][1].extend(s)

list_2 would become:

[('AAA', [123, 456]), ('AAW', [147, 124])]

edited 31 mins ago

codeforester

15.9k73555

answered 8 hours ago

blhsing

16.5k2631

You can use a dict to keep track of the indices of each key to keep the time complexity O(n):

list_1 = [("AAA", [123]), ("AAA", [456]), ("AAW", [147]), ("AAW", [124])]
list_2 = 
i = 
for k, s in list_1:
 if k not in i:
 list_2.append((k, s))
 i[k] = len(i)
 else:
 list_2[i[k]][1].extend(s)

list_2 would become:

[('AAA', [123, 456]), ('AAW', [147, 124])]

edited 31 mins ago

codeforester

15.9k73555

answered 8 hours ago

blhsing

16.5k2631

edited 31 mins ago

codeforester

15.9k73555

edited 31 mins ago

codeforester

15.9k73555

edited 31 mins ago

codeforester

15.9k73555

answered 8 hours ago

blhsing

16.5k2631

answered 8 hours ago

blhsing

16.5k2631

answered 8 hours ago

blhsing

16.5k2631

add a commentÂ |Â

up vote
1
down vote

Similarly to other answers, you can use a dictionary to associate each key with a list of values. This is implemented in the function merge_by_keys in the code snippet below.

import pprint

list_1 = [("AAA", [123]), ("AAA", [456]), ("AAW", [147]), ("AAW", [124])]
list_2 = [("AAA", [123, 456]), ("AAW", [147, 124])]

def merge_by_key(ts):

 d = 
 for t in ts:
 key = t[0]
 values = t[1]
 if key not in d:
 d[key] = values[:]
 else:
 d[key].extend(values)

 return d.items()



result = merge_by_key(list_1)

pprint.pprint(result)

answered 7 hours ago

Odysseas

445210

Why are you manually creating list_2?
â€“Â codeforester
20 mins ago

add a commentÂ |Â

up vote
1
down vote

Similarly to other answers, you can use a dictionary to associate each key with a list of values. This is implemented in the function merge_by_keys in the code snippet below.

import pprint

list_1 = [("AAA", [123]), ("AAA", [456]), ("AAW", [147]), ("AAW", [124])]
list_2 = [("AAA", [123, 456]), ("AAW", [147, 124])]

def merge_by_key(ts):

 d = 
 for t in ts:
 key = t[0]
 values = t[1]
 if key not in d:
 d[key] = values[:]
 else:
 d[key].extend(values)

 return d.items()



result = merge_by_key(list_1)

pprint.pprint(result)

answered 7 hours ago

Odysseas

445210

Why are you manually creating list_2?
â€“Â codeforester
20 mins ago

add a commentÂ |Â

up vote
1
down vote

Similarly to other answers, you can use a dictionary to associate each key with a list of values. This is implemented in the function merge_by_keys in the code snippet below.

import pprint

list_1 = [("AAA", [123]), ("AAA", [456]), ("AAW", [147]), ("AAW", [124])]
list_2 = [("AAA", [123, 456]), ("AAW", [147, 124])]

def merge_by_key(ts):

 d = 
 for t in ts:
 key = t[0]
 values = t[1]
 if key not in d:
 d[key] = values[:]
 else:
 d[key].extend(values)

 return d.items()



result = merge_by_key(list_1)

pprint.pprint(result)

answered 7 hours ago

Odysseas

445210

Similarly to other answers, you can use a dictionary to associate each key with a list of values. This is implemented in the function merge_by_keys in the code snippet below.

import pprint

list_1 = [("AAA", [123]), ("AAA", [456]), ("AAW", [147]), ("AAW", [124])]
list_2 = [("AAA", [123, 456]), ("AAW", [147, 124])]

def merge_by_key(ts):

 d = 
 for t in ts:
 key = t[0]
 values = t[1]
 if key not in d:
 d[key] = values[:]
 else:
 d[key].extend(values)

 return d.items()



result = merge_by_key(list_1)

pprint.pprint(result)

answered 7 hours ago

Odysseas

445210

answered 7 hours ago

Odysseas

445210

answered 7 hours ago

Odysseas

445210

answered 7 hours ago

Odysseas

445210

Why are you manually creating list_2?
â€“Â codeforester
20 mins ago

add a commentÂ |Â

Why are you manually creating list_2?
â€“Â codeforester
20 mins ago

Why are you manually creating list_2?
â€“Â codeforester
20 mins ago

add a commentÂ |Â

up vote
1
down vote

You can create a dictionary and loop through the list. If the item present in dictionary append the value to already existing list else assign the value to key.

dict_1 = 
for item in list_1:
 if item[0] in dict_1:
 dict_1[item[0]].append(item[1][0])
 else:
 dict_1[item[0]] = item[1]
list_2 = list(dict_1.items())

edited 23 mins ago

codeforester

15.9k73555

answered 8 hours ago

S.Harish

644

New contributor

It does the merge but doesn't retain the original order in the list. The result I got is: [('AAW', [147, 124]), ('AAA', [123, 456])] - AAA should have been the first element as in the original list.
â€“Â codeforester
22 mins ago

add a commentÂ |Â

up vote
1
down vote

You can create a dictionary and loop through the list. If the item present in dictionary append the value to already existing list else assign the value to key.

dict_1 = 
for item in list_1:
 if item[0] in dict_1:
 dict_1[item[0]].append(item[1][0])
 else:
 dict_1[item[0]] = item[1]
list_2 = list(dict_1.items())

edited 23 mins ago

codeforester

15.9k73555

answered 8 hours ago

S.Harish

644

New contributor

It does the merge but doesn't retain the original order in the list. The result I got is: [('AAW', [147, 124]), ('AAA', [123, 456])] - AAA should have been the first element as in the original list.
â€“Â codeforester
22 mins ago

add a commentÂ |Â

up vote
1
down vote

You can create a dictionary and loop through the list. If the item present in dictionary append the value to already existing list else assign the value to key.

dict_1 = 
for item in list_1:
 if item[0] in dict_1:
 dict_1[item[0]].append(item[1][0])
 else:
 dict_1[item[0]] = item[1]
list_2 = list(dict_1.items())

edited 23 mins ago

codeforester

15.9k73555

answered 8 hours ago

S.Harish

644

New contributor

You can create a dictionary and loop through the list. If the item present in dictionary append the value to already existing list else assign the value to key.

dict_1 = 
for item in list_1:
 if item[0] in dict_1:
 dict_1[item[0]].append(item[1][0])
 else:
 dict_1[item[0]] = item[1]
list_2 = list(dict_1.items())

edited 23 mins ago

codeforester

15.9k73555

answered 8 hours ago

S.Harish

644

New contributor

edited 23 mins ago

codeforester

15.9k73555

edited 23 mins ago

codeforester

15.9k73555

edited 23 mins ago

codeforester

15.9k73555

answered 8 hours ago

S.Harish

644

New contributor

answered 8 hours ago

S.Harish

644

answered 8 hours ago

S.Harish

644

New contributor

S.Harish is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

It does the merge but doesn't retain the original order in the list. The result I got is: [('AAW', [147, 124]), ('AAA', [123, 456])] - AAA should have been the first element as in the original list.
â€“Â codeforester
22 mins ago

add a commentÂ |Â

It does the merge but doesn't retain the original order in the list. The result I got is: [('AAW', [147, 124]), ('AAA', [123, 456])] - AAA should have been the first element as in the original list.
â€“Â codeforester
22 mins ago

It does the merge but doesn't retain the original order in the list. The result I got is: [('AAW', [147, 124]), ('AAA', [123, 456])] - AAA should have been the first element as in the original list.
â€“Â codeforester
22 mins ago

add a commentÂ |Â

Theprofessor14 is a new contributor. Be nice, and check out our Code of Conduct.

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