Skip to content

unnecessary-iterable-allocation-for-first-element (RUF015)

Fix is always available.

What it does

Checks the following constructs, all of which can be replaced by next(iter(...)):

  • list(...)[0]
  • tuple(...)[0]
  • list(i for i in ...)[0]
  • [i for i in ...][0]
  • list(...).pop(0)

Why is this bad?

Calling e.g. list(...) will create a new list of the entire collection, which can be very expensive for large collections. If you only need the first element of the collection, you can use next(...) or next(iter(...) to lazily fetch the first element. The same is true for the other constructs.

Example

head = list(x)[0]
head = [x * x for x in range(10)][0]

Use instead:

head = next(iter(x))
head = next(x * x for x in range(10))

Fix safety

This rule's fix is marked as unsafe, as migrating from (e.g.) list(...)[0] to next(iter(...)) can change the behavior of your program in two ways:

  1. First, all above-mentioned constructs will eagerly evaluate the entire collection, while next(iter(...)) will only evaluate the first element. As such, any side effects that occur during iteration will be delayed.
  2. Second, accessing members of a collection via square bracket notation [0] of the pop() function will raise IndexError if the collection is empty, while next(iter(...)) will raise StopIteration.

References