Python: order of dictionary initialization

Green绿色 0
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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?

dictionary python
2 Answers

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.

Ture Pålsson0

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.

usercE8H0
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