单行保存if语句的价值？

Is there a smart way to write the following code in three or four lines?

是否有一种聪明的方法可以在三行或四行中编写以下代码？

a=l["artist"]
if a:
    b=a["projects"]
    if b:
        c=b["project"]
        if c:
            print c

So I thought for something like pseudocode:

所以我想像伪代码一样：

a = l["artist"] if True:

5 个解决方案

#1

I don't necessarily think that this is better but you could do:

我不一定认为这更好但你可以这样做：

try:
    c = l["artist"]["projects"]["project"]
except (KeyError, TypeError) as e:
    print e
    pass

#2

How about:

怎么样：

try:
    print l["artist"]["projects"]["project"]
except KeyError:
    pass
except TypeError:
    pass # None["key"] raises TypeError.

This will try to print the value, but if a KeyError is raised, the except block will be run. pass means to do nothing. This is known and EAFP: it’s Easier to Ask Forgiveness than Permission.

这将尝试打印该值，但如果引发KeyError，则将运行except块。通过意味着什么也不做。这是众所周知的和EAFP：宽恕比宽恕更容易。

#3

p = l.get('artist') and l['artist'].get('projects') and l['artist']['projects'].get('project')
if p:
     print p

You can also make a more general function for this purpose:

您还可以为此目的制作更通用的功能：

def get_attr(lst, attr):
    current = lst
    for a in attr:
        if current.get(a) is not None:
            current = current.get(a)
        else:
            break
    return current

>>> l = {'artist':{'projects':{'project':1625}}}
>>> get_attr(l,['artist','projects','project'])
1625

#4

One-liner (as in the title) without exceptions:

单行（如标题中）无例外：

if "artist" in l and l["artist"] and "projects" in l["artist"] and l["artist"]["projects"] and "project" in l["artist"]["projects"]: print l["artist"]["projects"]["project"]

#5

Since you're dealing with nested dictionaries, you might find this generic one-liner useful because it will allow you to access values at any level just by passing it more keys arguments:

由于您正在处理嵌套字典，您可能会发现这种通用的单行有用，因为它允许您通过传递更多键参数来访问任何级别的值：

nested_dict_get = lambda item, *keys: reduce(lambda d, k: d.get(k), keys, item)

l = {'artist': {'projects': {'project': 'the_value'}}}

print( nested_dict_get(l, 'artist', 'projects', 'project') ) # -> the_value

Note: In Python 3, you'd need to add a from functools import reduce at the top.

注意：在Python 3中，您需要在顶部添加functools import reduce。

#1