FURB118 breaks program #23076
Description
Summary
what kind of breakage is ok for unsafe fixes?
This rule suggests a change from lambda to itemgetter that will fail:
def func(rows):
rows.sort(key=lambda x: x[1])
rows.sort(key=lambda x: x[x.find("=")]) # <- Why doesn't it try to convert this line?
rows.sort(key=lambda x: (x[x.find("=")], x[1])) # <-- this remove the `x` but still keep it in the itemgetter codeThis might be a unusual case that is "ok" to break. (But it is an actual case I found in code the other day).
As the second line above doesnt get incorrectly fix, I'm guessing there is some detection to avoid applying the fix when the x itself is used in the braackets.
Suggestion: detect if the lambda argument itself (x) is used inside the brackets, and skip applying the fix.
Example output from ruff 0.14.10:
$ ruff --isolated check --fix ~/tmp/bugreport_ruff_refurb118_itemgetter.py --select FURB118 --preview --show-fixes
FURB118 Use `operator.itemgetter(1)` instead of defining a lambda
--> /home/david/tmp/bugreport_ruff_refurb118_itemgetter.py:2:19
|
1 | def func(rows):
2 | rows.sort(key=lambda x: x[1])
| ^^^^^^^^^^^^^^
3 | rows.sort(key=lambda x: x[x.find("=")])
4 | rows.sort(key=lambda x: (x[x.find("=")], x[1]))
|
help: Replace with `operator.itemgetter(1)`
FURB118 Use `operator.itemgetter(x.find("="), 1)` instead of defining a lambda
--> /home/david/tmp/bugreport_ruff_refurb118_itemgetter.py:4:19
|
2 | rows.sort(key=lambda x: x[1])
3 | rows.sort(key=lambda x: x[x.find("=")])
4 | rows.sort(key=lambda x: (x[x.find("=")], x[1]))
| ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
|
help: Replace with `operator.itemgetter(x.find("="), 1)`
Found 2 errors.Version
0.14.10