Python - Filtering a list of tuples with lambda

Question

I try to update tuple values using another tuple.

grade = (('a', 10), ('b', 20), ('c', 30), ('d', 40))

factors = (('a', 1), ('b', 2))

Expected result:

result  = (('a', 11), ('b', 22), ('c', 30), ('d', 40))

What would be the right way to do this? , I tried the following code but it did not work. I would be happy to help

print(list(filter(lambda list_a: list(map(lambda x, y: x+ y, list_a[0], grade[1])) not in list(map(lambda x: x[0], factors)), grade)))

To improve your question please describe shortly what the update is expected to do. We may guess it by the result but it would help to have a clear specification. — bjhend
– bjhend, Commented Dec 9, 2020 at 18:57

Juan C · Accepted Answer · 2020-12-09 18:54:50Z

1

I'd use dictionaries, seem more suited for your task:

factors = dict(factors)
grade = dict(grade)
{k: grade[k] + factors[k] if k in factors.keys() else grade[k] for k  in grade.keys()}

Output:

{'a': 11, 'b': 22, 'c': 30, 'd': 40}

answered Dec 9, 2020 at 18:54

Juan C

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1 Comment

Popo Over a year ago

Is there any solution to do that without using dict?

Thierry Lathuille · Accepted Answer · 2020-12-09 19:00:17Z

1

You should use a dict, and you can take advantage of its get method to get 0 for non existing keys:

grade = (('a', 10), ('b', 20), ('c', 30), ('d', 40))

factors = (('a', 1), ('b', 2))
factors = dict(factors)

new_grades = [(g[0], g[1] + factors.get(g[0], 0)) for g in grade]
print(new_grades)
# [('a', 11), ('b', 22), ('c', 30), ('d', 40)]

answered Dec 9, 2020 at 19:00

Thierry Lathuille

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3 Comments

Popo Over a year ago

Is there any solution to do that without using dict?

Thierry Lathuille Over a year ago

This could be possible, but you have to get the value corresponding to a letter in factors. If you don't build a dict, you'll need a function that iterates on your tuples each time until it finds the one that contains the right letter. This is very inefficient. On the other side, once you have built the dict once, accessing the value associated with a letter is done directly without any iteration. That's much more efficient, and doesn't require you to write a function that will not be as good. So, do the right thing and enjoy!

Thierry Lathuille Over a year ago

And, as a side note, list comprehensions are usually faster than using map when you just want to build the list.

hundredrab · Accepted Answer · 2020-12-09 18:52:25Z

0

fact = dict(factors)
result = list(
    map(
        lambda x: (x[0], x[1] + fact.get(x[0], 0)), grade
    )
)

answered Dec 9, 2020 at 18:52

hundredrab

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Popo Over a year ago

Is there any solution to do that without using dict?

chepner · Accepted Answer · 2020-12-09 18:59:37Z

0

Make factors a dict:

factors = dict(factors)

then use a dict comprehension to build a new dict using the existing association list grades:

result = {g: score + factors.get(g, 0) for g, score in grade}

If you really need a tuple of tuples, use

result = tuple((g, score + factors.get(g,0)) for g, score in grade)

edited Dec 9, 2020 at 18:59

answered Dec 9, 2020 at 18:56

chepner

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1 Comment

Juan C Over a year ago

Tried your code but it brings: {'a': 1, 'b': 2, 'c': 0, 'd': 0}

superb rain · Accepted Answer · 2020-12-09 19:22:58Z

0

How about using Counters instead?

from collections import Counter

grade = Counter(dict(grade))
factors = Counter(dict(factors))

The the update is simply

grade += factors

or

grade.update(factors)

answered Dec 9, 2020 at 19:22

superb rain

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