将列表列表分为两部分的好方法，以内部列表中的值为条件

Question

目标是从以下开始：

budget = 100
projects = [('B', 60), ('A', 35), ('F', 35), ('G', 35), ('C', 20), ('E', 20), ('D', 10)]

并实现projects=[('B', 60), ('A', 35)]和remainingBudget =5。

作为一名 JS 程序员，我通过某种方式得到了以下工作：

def findProjectsThatFitTheBudget(budget, projects):
    # find the most expensive projects that fit the given budget, and note the budget that remains after allocations
    remainingBudget = budget

    def shoulIncludeProject(name, cost):
        nonlocal remainingBudget
        if(cost <= remainingBudget):
            remainingBudget -= cost
            return True
        return False
    projects = list(
        takewhile(lambda project: shoulIncludeProject(project[0], project[1]), projects))

    # we now have the projects we are working with, and also the budget that remains unallocated

我想知道最 Pythonic 的重构方式是什么？我至少坚持以下几点：

1. 如何编写简单的 lambda 而不是外部定义
2. 如何对 lambda 的参数进行解构
3. 如何以短路方式使用and 与budget=-cost

一个漂亮的解决方案可能是：

projects = list(
    takewhile(lambda name, cost: cost<= budget and budget=-cost, projects))

以short-circuit 的方式使用and。

Answer 1

一个简单的for 循环在预算用完时短路并没有错：

viable_projects = []
for project, cost in projects:
    budget -= cost
    if budget < 0:
        break
    viable_projects.append((project, cost))

viable_projects 变为：

[('B', 60), ('A', 35)]