用dict作为容器的python字典奇怪的行为

Question

Here is a simplified code snippet of the problem

>>> dict ({'A': 58, 'B': 130} for _ in range(1))
{'A': 'B'}

I am expecting it to return the same dictionary passed in.

if I do

>>> dict({'A': 58, 'B': 130})

I get exactly what I am looking for, that is {'A': 58, 'B': 130}

Why is this behavior different, how to fix it? I cannot alter the expression there, but I can alter the input dictionary in whatever way I like, for example, I can pass it like [ {'A': 58, 'B': 130}]

Answer 1

A dict can be initialized with another dict, or with an iterable of pairs, which is what you have given it. Note that iterating over a dict yields its keys only.

>>> d = {'A': 58, 'B': 130}
>>> list(d)
['A', 'B']
>>> dict([('A', 'B'), ('C', 'D')])
{'A': 'B', 'C': 'D'}
>>> dict([d, ('C', 'D')])
{'A': 'B', 'C': 'D'}

Python is behaving exactly as specified. Your dict happens to be a pair.

Answer 2

There's something special about the dict you're passing... ({'A': 58, 'B': 130} for _ in range(1)) represents a generator sequence of length 1. What you are passing is similar to

dict([{'A': 58, 'B': 130}])
# {'A': 'B'}

These, on the other hand, will not work:

dict([{'A':58}])
# ValueError: dictionary update sequence element #0 has length 1; 2 is required

dict([{'A':58, 'B': 130, 'C': 150}])    
ValueError: dictionary update sequence element #0 has length 3; 2 is required

The first example worked because your dictionary had exactly two entries .

The sequence is passed to the dict method, which takes the two items it needs to create a key-value pair, and creates a dictionary like this:

{'A': 'B'}

IOW, it requires an iterable of pairs, which is what your sequence with a single dict of two entries is. Anything else will throw a ValueError .

Answer 3

Because it simply will make it ['A', 'B'] , with list as well:

>>> list({'A': 58, 'B': 130})
['A', 'B']
>>> 

And since it iterates in dict since it's a list of dictionaries, it would make it list for all of them, so:

>>> dict([{'A': 58, 'B': 130}])
{'A': 'B'}
>>> dict({'A': 58, 'B': 130})
{'A': 58, 'B': 130}
>>> 

Also, without dict it would be doing something like the below for a list of dictionaries:

>>> {*{'A': 58, 'B': 130}}
{'A', 'B'}
>>> 

Just in a dictionary-like way, but you should get it now.

Answer 4

{'A': 58, 'B': 130} for _ in range(1) yields [{'A': 58, 'B': 130}] , therefore you are passing a list, not a dict, to dict() .

According to the documentation, if you pass an iterable to dict() , you effectively get:

d = {}
for k, v in iterable:
   d[k] = v

Since the dict inside the list has exactly two elements, the for k,v in iterable: loop gets the first two keys in the dict. If the dict had only one element, or more than two, you would get a ValueError .