Question

我有以下代码。

sum_review = reduce(add,[book['rw'] for book in books])
sum_rating = reduce(add,[book['rg'] for book in books])
items = len(books)
avg_review = sum_review/items
avg_rating = sum_rating/items

我想要的是这个。

sum_review,sum_rating = reduce(add,([book['rw'],[book['rg']) for book in books])
items = len(books)
avg_review = sum_review/items
avg_rating = sum_rating/items

显然这不起作用。如果没有常规循环，我怎样才能解决这种冗余？

Answer 1

我在这里避免使用reduce。对于这么简单的使用sum：

sum_review = sum(book['rw'] for book in books)
sum_rating = sum(book['rg'] for book in books)

在我看来，这个更简单的版本不需要重构来删除冗余。只有两个项目（rw和rg）我认为最好保持原样。

Answer 2

有两种简化代码的典型方法：

自上而下：首先获取值，然后使用zip(*iterable)转置它们。它也很酷，因为它只迭代集合一次：

values = ((book["rw"], book["rg"]) for book in books)
avg_review, avg_rating = [sum(xs) / len(books) for xs in zip(*values)]

自下而上：创建一个抽象操作的函数：

get_avg = lambda xs, attr: sum(x[attr] for x in xs) / len(xs)
avg_review = get_avg(books, "rw")
avg_rating = get_avg(books, "rg")

Answer 3

您应该优先考虑清晰度而不是优化。在使用Python的3年中，我只需要分析两次发现性能瓶颈。您的原始代码清晰有效。将前两行压缩为一个会损害可读性并且几乎不会影响性能。

如果我有来修改你的代码，它会这样：

avg_review = sum(book['rw'] for book in books) / len(books)
avg_rating = sum(book['rg'] for book in books) / len(books)

（这是五行代码，一到两行，提高了清晰度。）

Answer 4

sum_review, sum_rating = reduce(lambda a,b: (a[0] + b[0], a[1]+b[1]), ((book['rw'], book['rg']) for book in books), (0,0) )
items = len(books)
avg_review = sum_review/items
avg_rating = sum_rating/items

（测试）

Answer 5

如何解决此冗余问题

通过制作一个功能，当然：

def average_value(items, key):
  values = [x[key] for x in items]
  return sum(items) / len(items)

avg_review, avg_rating = average_value(books, 'rw'), average_value(books, 'rg')

我可以根据两个值将列表理解缩减为两个列表吗？

5 个答案: