Convert `map(lambda x: expression, iterable)` to `(expression for x in iterable)`

Description

map(f, iterable) has great performance when f is a built-in function, and it makes sense if your function already has a name. But if you need to introduce a lambda, it’s better (i.e., more readable and faster) to use a generator expression – no function calls needed.

Good:

for y in map(my_function_that_already_exists, it):
    ...
ys = list(map(my_function_that_already_exists, it))
ys = set(map(my_function_that_already_exists, it))
etc.

Suggestions:

# Bad:
map(lambda x: x**2, iterable)
# change to:
(x**2 for x in iterable)

# Bad:
list(map(lambda x: x.attr, iterable))
# change to:
[x.attr for x in iterable]
# OR change to:
y = list(map(operator.attrgetter("attr"), iterable))

# Bad:
z = map(lambda x: f(x), iterable)
# change to:
z = map(f, iterable)

Here are some perf numbers (deque(..., maxlen=0) is a fast way to consume an iterable):

PS C:\Users\sween> py -3.8 -m pyperf timeit -s "from collections import deque" "deque(map(lambda x: -x, range(1_000_000)), maxlen=0)"
Mean +- std dev: 85.7 ms +- 1.7 ms

PS C:\Users\sween> py -3.8 -m pyperf timeit -s "from collections import deque" "deque((-x for x in range(1_000_000)), maxlen=0)"

Mean +- std dev: 63.1 ms +- 2.9 ms

Similar things hold for filter().

[joaoe]

This could also include filter(lambda x: expression, iterable) to (x for x in iterable if expression)

[tusharsadhwani]

I can make a PR.