functools.reduce doesn't work properly with itertools.chain

[VasanthaGaneshKanniappan]

BPO	35043
Nosy	@rhettinger, @MojoVampire

^{Note: these values reflect the state of the issue at the time it was migrated and might not reflect the current state.}

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GitHub fields:

assignee = 'https://github.com/rhettinger'
closed_at = <Date 2018-10-23.07:02:22.996>
created_at = <Date 2018-10-22.11:49:58.788>
labels = ['type-bug', 'library']
title = "functools.reduce doesn't work properly with itertools.chain"
updated_at = <Date 2018-10-24.01:42:55.829>
user = 'https://bugs.python.org/VasanthaGaneshKanniappan'

bugs.python.org fields:

activity = <Date 2018-10-24.01:42:55.829>
actor = 'josh.r'
assignee = 'rhettinger'
closed = True
closed_date = <Date 2018-10-23.07:02:22.996>
closer = 'Vasantha Ganesh Kanniappan'
components = ['Library (Lib)']
creation = <Date 2018-10-22.11:49:58.788>
creator = 'Vasantha Ganesh Kanniappan'
dependencies = []
files = []
hgrepos = []
issue_num = 35043
keywords = []
message_count = 4.0
messages = ['328248', '328264', '328287', '328350']
nosy_count = 3.0
nosy_names = ['rhettinger', 'josh.r', 'Vasantha Ganesh Kanniappan']
pr_nums = []
priority = 'normal'
resolution = None
stage = 'resolved'
status = 'closed'
superseder = None
type = 'behavior'
url = 'https://bugs.python.org/issue35043'
versions = ['Python 3.6']

[VasanthaGaneshKanniappan]

I'm facing this issue in python 3.6.5

How to reproduce:

my_list = [[1, 2], [3, 4]]
a = reduce(itertools.chain, my_list)

The output of `a' is [1, 2, 3, 4] as expected, but now
my_list is [[1, 2, 3, 4], [3, 4]]

[MojoVampire]

Your example code doesn't behave the way you claim. my_list isn't changed, and a is a chain generator, not a list (without a further list wrapping).

In any event, there is no reason to involve reduce here. chain already handles varargs what you're trying to do without involving reduce at all:

a = list(itertools.chain(*my_list))

or if you prefer to avoid unnecessary unpacking:

a = list(itertools.chain.from_iterable(*my_list))

Either way, a will be [1, 2, 3, 4], and my_list will be unchanged, with no wasteful use of reduce.

[VasanthaGaneshKanniappan]

You are right. It works now. I'm sorry for wasting your time.

[MojoVampire]

Blech. Copy'n'paste error in last post:

a = list(itertools.chain.from_iterable(*my_list))

should be:

a = list(itertools.chain.from_iterable(my_list))

(Note removal of *, which is the whole point of from_iterable)