Python | Binary Group Tuple list elements
Last Updated :
27 Apr, 2023
Sometimes, while working with tuples, we can have problems of grouping them, be it based on gender or any particular binary category. This can have applications in many domains. Let’s discuss certain ways in which this can be performed.
Method #1 : Using generator + loop + zip()
The brute force method to perform this task, in this, we perform the combination grouping with the help of zip(), and iteration logic is handled by generator and loop.
Python3
def bin_group(test_list):
for tup1, tup2 in zip (test_list[ 0 :: 2 ], test_list[ 1 :: 2 ]):
yield (tup1[ 0 ], tup1[ 1 ], tup2[ 1 ])
test_list = [( 1 , 56 , 'M' ), ( 1 , 14 , 'F' ), ( 2 , 43 , 'F' ), ( 2 , 10 , 'M' )]
print ( "The original list : " + str (test_list))
res = list (bin_group(test_list))
print ( "The list after binary grouping : " + str (res))
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Output :
The original list : [(1, 56, 'M'), (1, 14, 'F'), (2, 43, 'F'), (2, 10, 'M')]
The list after binary grouping : [(1, 56, 14), (2, 43, 10)]
Method #2: Using defaultdict() + list comprehension + sorted() + items()
Combination of the above functions can be used to perform this task. In this, we convert the list into a dictionary and then perform key-based grouping and rearrange back to tuple list.
Python3
from collections import defaultdict
test_list = [( 1 , 56 , 'M' ), ( 1 , 14 , 'F' ), ( 2 , 43 , 'F' ), ( 2 , 10 , 'M' )]
print ( "The original list : " + str (test_list))
temp = defaultdict( list )
for ele in test_list:
temp[ele[ 0 ]].append(ele[ 1 ])
res = sorted ((key, ) + tuple (val) for key, val in temp.items())
print ( "The list after binary grouping : " + str (res))
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Output :
The original list : [(1, 56, 'M'), (1, 14, 'F'), (2, 43, 'F'), (2, 10, 'M')]
The list after binary grouping : [(1, 56, 14), (2, 43, 10)]
Using itertools.groupby to group tuples by their first element:
Approach:
- Import the groupby function from the itertools module.
- Define the original list of tuples.
- Sort the original list by the first element of each tuple using the sort method.
- Use a list comprehension to create the grouped list of tuples.First, use the groupby function to group the tuples in the original list by their first element.
- Then, for each group, create a new tuple with the first element being the group key and the remaining elements being the second elements of each tuple in the group.
- The * syntax is used to unpack the elements of the resulting tuple.
- The tuple and for syntax is used to create a new tuple containing the second element of each tuple in the group.
- Print the resulting grouped list.
Python3
from itertools import groupby
original_list = [( 1 , 56 , 'M' ), ( 1 , 14 , 'F' ), ( 2 , 43 , 'F' ), ( 2 , 10 , 'M' )]
original_list.sort()
grouped_list = [(k, * tuple (t[ 1 ] for t in g)) for k, g in groupby(original_list, lambda x: x[ 0 ])]
print (grouped_list)
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Output
[(1, 14, 56), (2, 10, 43)]
Time complexity: O(nlogn) (due to sorting)
Auxiliary Space: O(n)
Method 4: Using the pandas library
Python3
import pandas as pd
test_list = [( 1 , 56 , 'M' ), ( 1 , 14 , 'F' ), ( 2 , 43 , 'F' ), ( 2 , 10 , 'M' )]
df = pd.DataFrame(test_list, columns = [ 'key' , 'value' , 'gender' ])
df_binary = df.pivot(index = 'key' , columns = 'gender' ,
values = 'value' ).fillna( 0 ).astype( int )
result = [(row[ 0 ], * row[ 1 :])
for row in df_binary.reset_index().itertuples(index = False )]
print ( "The original list:" , test_list)
print ( "The list after binary grouping:" , result)
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OUTPUT :
The original list: [(1, 56, 'M'), (1, 14, 'F'), (2, 43, 'F'), (2, 10, 'M')]
The list after binary grouping: [(1, 14, 56), (2, 43, 10)]
Time complexity of this method is O(n log n) due to the sorting operation.
Auxiliary space: is O(n) due to the creation of the Pandas DataFrame.
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