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])]
Combining tuples in list of tuples in Python 1 With for loop. In the below approach we create for loops that will create a pair of elements by taking each element of... 2 Example. 3 Output. 4 With product. The itertools module has iterator named product which creates Cartesian product of parameters passed onto... 5 Example. 6 Output. More ...
Iterate over the list of tuples. Check if the first element of the tuple is present as a key in the dictionary or not. If it is present, then append the current tuple values without the first one to the previous values. If not present, then initialize the key with the current tuple elements including the first element.
A list can contain tuples as its elements. In this article we will see how we can combine each element of a tuple with another given element and produce a list tuple combination. In the below approach we create for loops that will create a pair of elements by taking each element of the tuple and looping through the element in the list.
Iterate over the list of tuples. Check if the first element of the tuple is present as a key in the dictionary or not. If it is present, then append the current tuple values without the first one to the previous values.
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)
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])]
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())
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])]
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)
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