The comments generated are not just useless, they are harmful. They made it harder to recognize and act on actually meaningful comments.
Completely disagree. Once you get really fluent in a language it is so much faster to write something from scratch. Especially if you have certain patterns you follow.
Code review is different because that has a different goal.
If you're writing enough boilerplate that you find it useful to have a little helper write boilerplate for you then you are writing way, way, way too much boilerplate.
If you have literally any boilerplate in comments then you probably have too much.
Here's the code I see:
def get_temp(sorted_temps: list[float], el: float) -> int:
"""
Search through the sorted temperatures array to find the
index of the temperature that matches the element
:param sorted_temps: sorted list of temperatures
:param el: the target element
:return: the index of the matched element
Author: Richard Hendricks
"""
i = 0
while el != sorted_temps[i] and len(sorted_temps) > i):
i += 1
return i if i < len(sorted_temps) else -1;
Some problems I see:1. the `return ...` line needs to be dedented, otherwise, if `el == sorted_temps[0]` then the return value is None
2. even if dedented, the docstring needs to report that get_temp() returns -1 if the index isn't found
3. there's an extra `)` in `> i):`, and the semi-colon is unneeded
4. the test for `len(sorted_temps) > i` must done before `el != sorted_temps[i]`,
The corrected version would be:
def get_temp(sorted_temps: list[float], el: float) -> int:
i = 0
while len(sorted_temps) > i and el != sorted_temps[i]:
i += 1
return i if i < len(sorted_temps) else -1
Furthermore, here are some alternatives:5. I would expect the following to be faster, and more easily understood:
for i, temp in enumerate(sorted_temps):
if temp == el:
return i
return -1
5. Or since the type signature says it's supposed to be a list of floats: try:
return sorted_temps.index(el)
except ValueError:
return -1
6: Or, since the values are supposed to be a sorted list, perhaps a binary search: import bisect
i = bisect.bisect_left(sorted_temps, el)
if sorted_temps[i] == el:
return i
return -1