Python - 使用 lambda 过滤元组列表

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)))

Answer 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}

Answer 2

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)]

Answer 3

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

Answer 4

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)

Answer 5

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)