[o for o in a if o['k'] == v][0]
This miserable line of code requires 3 references to the dict o
. Compare the JavaScript version, with not a single direct reference to o
:
a.find(({ k }) => k == v)
It not only requires less references, it uses less characters without broaching unreadable code-golf territory.
Is there a shorter way to find a dict in a list in Python comparable to this?
Perhaps Python doesn't even have an equivalent to JavaScript's Array.prototype.find
. The first line of code seems to be more equivalent to Array.prototype.filter
.
CodePudding user response:
You can write it like this instead:
result = next(filter(lambda o: o['k'] == v, a))
It's not exactly shorter, but it is in some ways less repetitive.
Documentation for next
and filter
is on this page; you can find a tutorial on lambda
functions here.
Even if you decide to stick with the comprehension-style syntax, by the way, you should probably refactor into into a generator expression rather than a list comprehension, since there is no need to build the list in memory if you only need one item from it:
result = next(o for o in a if o['k'] == v)