Can this conditional be shortened?

Question

Is it possible to just use .get() function from dictionary for this? Is this the best we can do to shorten this piece of code?

n_dict is a Dict type (uppercase D) and NES is just a list of str

eted_info = {}
for key in n_dict:
    if key in NES:
        eted_info[key] = n_dict[key]

I'm just curious if there is a better / more pythonic way to retrieve a value, like C# has with TryGetValue.

Most pythonic way would probably be to use a dict comprehension. — Ted Klein Bergman, Dec 14 '21 at 01:48
`eted_info` **only** contains keys that are in `NES`? ... No keys with null values? How many items in `NES`? — wwii, Dec 14 '21 at 01:55

score 5 · Accepted Answer · edited Dec 14 '21 at 18:51

5

I think a dictionary comprehension and using n_dict.items() is the cleanest way of doing this

n_dict = {'a': 1, 'b': 2, 'c': 3}

NES = ['a', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j']

eted_info = {key:value for key,value in n_dict.items() if key in NES}

print(eted_info)

RESULT

{'a': 1, 'c': 3}

edited Dec 14 '21 at 18:51

Chris

26,361
5
21
42

answered Dec 14 '21 at 01:57

Buddy Bob

5,829
1
13
44

Andrew Wei · Answer 2 · 2021-12-14T02:04:40.430

2

You can do something with dictionary comprehension like this:

eted_info = {key: n_dict[key] for key in n_dict if key in NES}

edited Dec 14 '21 at 02:04

answered Dec 14 '21 at 01:49

Andrew Wei

870
1
7
12

thanks man, that definitely looks better :) is there any performance difference? – caasswa Dec 14 '21 at 01:52
It is slightly faster, but not by much in most cases: https://stackoverflow.com/questions/52542742/why-is-this-loop-faster-than-a-dictionary-comprehension-for-creating-a-dictionar – Andrew Wei Dec 14 '21 at 01:59
1

Slight variation using `items()`: `eted_info = {k: v for k, v in n_dict.items() if k in NES}` – Samwise Dec 14 '21 at 02:05
1

Another using sets (might be more performant if `NES` and `ndict` are very large, since it avoids O(n^2) lookups): `eted_info = {k: n_dict[k] for k in set(n_dict) & set(NES)}` – Samwise Dec 14 '21 at 02:06
1

@Samwise Would you mind posting that second one as an answer? I don't find it as readable for likely use, but as you said, for very large `NES` and `ndict`, it's quite slick. – CrazyChucky Dec 14 '21 at 02:19

score 2 · Answer 3 · answered Dec 14 '21 at 02:23

2

If you want to avoid the O(n^2) operation of iterating through each item in NES for each item in n_dict, you can build a list of keys as a set intersection and iterate through that:

eted_info = {k: n_dict[k] for k in set(n_dict) & set(NES)}

answered Dec 14 '21 at 02:23

Samwise

68,105
3
30
44

wwii · Answer 4 · 2021-12-14T02:19:16.960

1

Iterate over keys in NES; use n_dict.get.get with a default of None; conditionally add to eted_info.

for key in NES:
    v = n_dict.get(key,None)
    if v: eted_info[key] = v

This will only iterate over the list once regardless of the length of n_dict.
Assumes all values in n_dict are truthy. Other placeholders could be used for the default value.

edited Dec 14 '21 at 02:19

answered Dec 14 '21 at 02:01

wwii

23,232
7
37
77

You could write that as a dict-comprehension:: eted_info = {key:n_dict[key] for key in NES if key in n_dict} Pretty much the same as @AndrewWei but the small difference saves a bit of complexity. – Anne Aunyme Dec 14 '21 at 02:26
@AnneAunyme - yes you could! Feel free to add (edit) that to my answer. – wwii Dec 14 '21 at 15:10

Can this conditional be shortened?

4 Answers4