Python – Concatenate Tuple to Dictionary Key

Given Tuples, convert them to dictionary with key being concatenated string.

Input : test_list = [((“gfg”, “is”, “best”), 10), ((“gfg”, “for”, “cs”), 15)]
Output : {‘gfg is best’: 10, ‘gfg for cs’: 15}
Explanation : Tuple strings concatenated as strings.

Input : test_list = [((“gfg”, “is”, “best”), 10)]
Output : {‘gfg is best’: 10}
Explanation : Tuple strings concatenated as strings.

Method #1 : Using loop + join()

In this, we perform the task of concatenation for dictionary key using join() and loop is used to render the required dictionary.

Python3

# Python3 code to demonstrate working of
# Concatenate Tuple to Dictionary Key
# Using loop + join()

# initializing list
test_list = [(("gfg", "is", "best"), 10), (("gfg", "good"), 1), (("gfg", "for", "cs"), 15)]

# printing original list
print("The original list is : " + str(test_list))

res = {}
for sub in test_list:

  # joining for making key
  res[" ".join(sub[0])] = sub[1]

# printing result
print("The computed Dictionary : " + str(res))

Output

The original list is : [(('gfg', 'is', 'best'), 10), (('gfg', 'good'), 1), (('gfg', 'for', 'cs'), 15)]
The computed Dictionary : {'gfg is best': 10, 'gfg good': 1, 'gfg for cs': 15}

Method #2 : Using dictionary comprehension

This is shorthand to above method, similar functionality, just one-liner on paper for compact solution.

Python3

# Python3 code to demonstrate working of
# Concatenate Tuple to Dictionary Key
# Using dictionary comprehension

# initializing list
test_list = [(("gfg", "is", "best"), 10), (("gfg", "good"), 1), (("gfg", "for", "cs"), 15)]

# printing original list
print("The original list is : " + str(test_list))

# one liner to solve problem
res = {' '.join(key): val for key, val in test_list}

# printing result
print("The computed Dictionary : " + str(res))

Output

The original list is : [(('gfg', 'is', 'best'), 10), (('gfg', 'good'), 1), (('gfg', 'for', 'cs'), 15)]
The computed Dictionary : {'gfg is best': 10, 'gfg good': 1, 'gfg for cs': 15}