One-liner creative solution for Remove All Before by suic - python coding challenges

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One-liner solution in Creative category for Remove All Before by suic

remove_all_before=lambda i,b:b in i and i[i.index(b):] or i

March 13, 2020

Comments:

lince on March 28, 2020, 5:44 p.m.
nice, i've nearly done the same.... but not as clean as this :D

X-Yupto10 on Feb. 20, 2021, 7:28 p.m.
Hello ! Can you explain this code ?

Mrsqd on April 11, 2021, 12:17 a.m.
I'm not the author but I'll try: i - items, b - border At first we check, if a cutting element is in the list: b in i If everything's fine ("and" after that), we return a new list with the help of an inbuilt function index() in Python: i[i.index(b):] The full slice syntax is: start:stop:step. start refers to the index of the element which is used as a start of our slice (here the start is i.index(b)). stop refers to the index of the element we should stop just before to finish our slice (we have only ":" instead of stop/step, so it means "till the end of the list"). step allows you to take each nth-element within a start:stop range. And finally if a cutting element is not in the list, then the list shoudn't be changed: or i

kamilpyszkowski on Nov. 11, 2021, 6:06 p.m.
more readable variant might be <pre class='brush: python'>remove_all_before = lambda i, b: i[i.index(b):] if b in i else i </pre> it can be read as regular spoken statement: (i - items list, b - boundary item <blockquote> return sliced items list if boundry item is in items list else return items list (which is an empty list or unsliced list(if boundary item doesn't occur) in this case) </blockquote>

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