update defaultdict count according to the occurrence of tuples in two lists

user_8457035 0
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I have two lists and would like to create a dictionary to record the occurrence of tuples.

My current code:

tup_to_find_test = [('good', 'pea'), ('leaf', 'sweet')] 
self_per_list_test = [('leaf', 'liquid'), ('leaf', 'sweet'), ('leaf', 'sweet'),('good', 'pea'),('good', 'pea'),('good', 'pea')]
from collections import defaultdict
tup_dict_test = defaultdict(int)
for tup_to_find_test in self_per_list_test:
   tup_dict_test[tup_to_find_test]+=1

My result is:

defaultdict(int, {('leaf', 'liquid'): 1, ('leaf', 'sweet'): 1, ('good', 'pea'): 3})

My desired result is:

('leaf', 'liquid'): 0, ('leaf', 'sweet'): 2, ('good', 'pea'): 3})

I do not know why the count of ('leaf', 'liquid') is 1. Isn't the default integer of defaultdict(int) zero? Why I got 1 for the ('leaf', 'liquid') tuple?

counter tuples list dictionary python
3 Answers

This line isn't doing what you think it is:

for tup_to_find_test in self_per_list_test:
   # ...

Here you are iterating a list elementwise, in this case the elements of self_per_list_test. There's no filtering taking place. As your for loop iterates, tup_to_find_test successively represents ('leaf', 'liquid'), ('leaf', 'sweet'), etc. The fact the name is the same as a variable you've defined earlier only serves to confuse.

Instead, you can use a ternary statement to differentiate operations:

for item in self_per_list_test:
    tup_dict_test[item] += 1 if item in tup_to_find_test else 0

print(tup_dict_test)

defaultdict(int, {('leaf', 'liquid'): 0, ('leaf', 'sweet'): 2, ('good', 'pea'): 3})

collections.Counter is more idiomatic with Python. It's good practice to use set for O(1) lookup within a dictionary comprehension.

from collections import Counter

tup_to_find_set = set(tup_to_find_test)
counts = Counter(self_per_list_test)

tup_dict_test = {k: v if k in tup_to_find_set else 0 for k, v in counts.items()}

print(tup_dict_test)

{('leaf', 'liquid'): 0, ('leaf', 'sweet'): 2, ('good', 'pea'): 3}
user_92095460

Isn't the default integer of defaultdict(int) zero?

Yes.

Why I got 1 for the ('leaf', 'liquid') tuple?

You wrote:

tup_dict_test[tup_to_find_test]+=1

That is, find the current value—which creates a new one set to zero—then add one to it and store the result back. The resulting value is 1.

user_12564520

Without reinventing the wheel. You can use counter from the wonderful collections standard module library for this. 

from collections import Counter

tup_to_find_test = [('good', 'pea'), ('leaf', 'sweet')] 
self_per_list_test = [('leaf', 'liquid'), ('leaf', 'sweet'), ('leaf', 'sweet'),('good', 'pea'),('good', 'pea'),('good', 'pea')]

c = Counter(self_per_list_test)

for key in c:
    if key not in tup_to_find_test:
        c[key] = 0

print(c)

>>Counter({('good', 'pea'): 3, ('leaf', 'sweet'): 2, ('leaf', 'liquid'): 0})

Here we create a counter based on self_per_list_test and updates the counts to zero if it is not found in tup_to_find_test. Hope this is a more intuitive method in solving your problem.

user_32880920
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