Python list of list initialize

Question

I know that

[[]] * 10

will give 10 references of the same empty, And

[[] for i in range(10)]

will give 10 empty lists. But in this example:

def get_list(thing):
  return [thing for i in range(10)]
a = get_list([])
a[0].append(1)
print(a)

Why the result is again

[[1], [1], [1], [1], [1], [1], [1], [1], [1], [1]]

EDIT:

Unlike question List of lists changes reflected across sublists unexpectedly , I understand that Python doesn't do copy in [x]*10.

But why [] is special in [[] for i in range(10)] ? I think this is inconsistent. Instead of creating a empty list [] then pass to [ ___ for i in range(10)], Python take "[]" literally and execute "[]" for every i.

Possible duplicate of [List of lists changes reflected across sublists unexpectedly](https://stackoverflow.com/questions/240178/list-of-lists-changes-reflected-across-sublists-unexpectedly) — fredtantini, Jul 29 '17 at 08:44
Because in this case your list consists of 10 references to the list referenced by `thing` parameter. — Błotosmętek, Jul 29 '17 at 08:44

score 5 · Answer 1 · answered Jul 29 '17 at 08:44

5

That is because thing is the same list.

In [[] for i in range(10)] a new list is generated every time.

In [thing for i in range(10)] it's always the list named thing.

answered Jul 29 '17 at 08:44

Bharel

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score 0 · Answer 2 · answered Jul 29 '17 at 23:32

I now understand why I'm wrong. I answer my question:

I misunderstood the list comprehension in python

[ ___ for i in range(10)]

It's not like passing a augment to a function. List comprehension execute your expression literally each time in its loop.

The expression [] in [ [] for i in range(10)] is executed each time so you get a new empty list for each entry.

Same for [thing for i in range(10)]. expression thing is executed, and the result is alway the same: a reference to one empty list.

Python list of list initialize

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