Python: order of dictionary initialization

Question

I have a dictionary d that I'd like to copy shallowly and then change some of its contents. I noticed that if I write the changed property first, it'll get overwritten. If I write it last, it persists:

>>> d = {1: 1, 2: 2}
{1: 1, 2: 2}
>>> d1 = {1: 11, **d}
{1: 1, 2: 2}
>>> d2 = {**d, 1: 11}
{1: 11, 2: 2}

However, I know the order in a dictionary isn't reliable. Can I assume that in {**d, 1: 11} , d[1] definitely gets overwritten by the updated value?

Answer 1

Quoting the section on dictionary displays in the Python language spec (my italics):

If a comma-separated sequence of key/datum pairs is given, they are evaluated from left to right to define the entries of the dictionary: each key object is used as a key into the dictionary to store the corresponding datum. This means that you can specify the same key multiple times in the key/datum list, and the final dictionary's value for that key will be the last one given.

A double asterisk ** denotes dictionary unpacking. Its operand must be a mapping. Each mapping item is added to the new dictionary. Later values replace values already set by earlier key/datum pairs and earlier dictionary unpackings.

Unless I'm missing something, this seems to guarantee that later values overwrite earlier ones.

Answer 2

From Python 3.6 onwards, the standard dict type maintains insertion order by default.

In this case, later occurrences overwrite previous ones.

So, if you say d1 = {1: 11, **d} , the value of the 1 would be updated with new value, and if you say d2 = {**d, 1: 11} , the value of 1 in d would be updated with 1: 11 .

So yes! You can be sure that d[1] gets overwritten by the updated value.