python - 毫无例外地从列表中获取第一个元素

Question

考虑这段代码

description = ""
desc = hxs.select('/html/head/meta[2]/@content').extract()
if len(desc) > 0:
    description = desc[0]       
item["description"] = description

desc 是一个字符串列表。如果列表为空，描述为空字符串，否则为列表中的第一个元素。如何让它更pythonic？

忘了说我必须使用 2.7

Answer 1

你可以写：

desc = hxs.select("/html/head/meta[2]/@content").extract()
item["description"] = desc[0] if len(desc) > 0 else ""

正如下面评论中所指出的，您还可以直接在布尔上下文中评估列表：

item["description"] = desc[0] if desc else ""

Answer 2

或者，您可以使用下一个支持默认值的事实

item["description"]  = next(iter(desc), "")

Answer 3

您可以使用异常处理（尽管它比使用条件表达式更详细）。

desc = hxs.select('/html/head/meta[2]/@content').extract()
try:
    description = desc[0]
except IndexError:
    description = ""

Answer 4

I don't think you should actually do this in your case, but for completeness…</p>

For an iterator that might be empty, you just use the optional default value for next:

desc = foo.finditer(bar)
item["description"] = next(desc, "")

Meanwhile, you can always use iter() to "peek" at the iterator you would get from a sequence (or other non-iterator iterable).

So:

desc = foo.findall(bar)
item["description"] = next(iter(desc), "")

I think this is less readable than just using the list as a sequence (whether you use chepner's EAFP answer or Frédéric Hamidi's LYBL) rather than just an iterable. After all, we have the sequence API for a reason.