One-liner for updating a value in a list of dictionaries - Python

Question

I have a list of dictionaries like so:

l = [{"integer":"1"},{"integer":"2"},{"integer":"3"},{"integer":"4"}]

I would like to do something similar to the following so that each of the numbers which are the value pair for the "integer" key are returned as integers:

l = [{"integer":"1"},{"integer":"2"},{"integer":"3"},{"integer":"4"}]
r = map(lambda x: x["integer"]=int(x["integer"]), l)
print r 
#[{"integer":1},{"integer":2},{"integer":3},{"integer":4}]

But this causes an error:

SyntaxError: lambda cannot contain assignment

Does anyone know of a clean way to do this in python? Preferably a oneliner using map or something similar?

What's wrong with your solution? Do you have other keys which correspond to values that are not integers? — pault, Sep 11 '18 at 18:31
@pault - My apologies, I should have mentioned, lambdas don't like assignment, so this code doesn't run. — Justin Gardner, Sep 11 '18 at 18:33
@Barmar one-liners make you smarter obviously haha jk but I myself think there is a certain appeal to one liners, don't you think? — vash_the_stampede, Sep 11 '18 at 18:35
If you're entering code golf competitions. For production code, less so. — Barmar, Sep 11 '18 at 18:37

Nick Dima · Answer 1 · 2018-09-11T18:37:24.640

3

Use a list comprehension comprehension

You will iterate through the dictionaries in the list and have them returned as x, then insert a new dictionary with your desired key and the integer value of the return within a new list

r = [{'integer': int(x['integer'])} for x in l]

edited Sep 11 '18 at 18:37

answered Sep 11 '18 at 18:29

Nick Dima

348
1
10

Do you know of a way to do that without having to redefine the whole dictionary? – Justin Gardner Sep 11 '18 at 18:31
@JustinGardner Still just in 1 line? – Nick Dima Sep 11 '18 at 18:36

pault · Answer 2 · 2018-09-11T19:05:47.380

You should just use a loop:

l = [{"integer":"1"},{"integer":"2"},{"integer":"3"},{"integer":"4"}]
for d in l:
    d["integer"] = int(d["integer"])
print(l)
#[{'integer': 1}, {'integer': 2}, {'integer': 3}, {'integer': 4}]

However, here is a one-liner that should work for you:

l = [{"integer":"1"},{"integer":"2"},{"integer":"3"},{"integer":"4"}]
[d.update({"integer": int(d["integer"])}) for d in l]
print(l)
#[{'integer': 1}, {'integer': 2}, {'integer': 3}, {'integer': 4}]

Be aware that dict.update() returns None, so if you assigned the output of the list comprehension to a variable it would be a list containing all Nones.

print([d.update({"integer": int(d["integer"])}) for d in l])
#[None, None, None, None]

score 1 · Answer 3 · answered Sep 11 '18 at 18:29

1

The following works in one line:

r = [{'integer':int(x['integer'])} for x in l]
print(r)
# [{'integer': 1}, {'integer': 2}, {'integer': 3}, {'integer': 4}]

This utilizes a dict comprehension inside a list comprehension.

answered Sep 11 '18 at 18:29

Joel

1,564
7
12
20

Do you know of a way to do that without having to redefine the whole dictionary? – Justin Gardner Sep 11 '18 at 18:33
You could just say `l = [{'integer':int(x['integer'])} for x in l]` followed by `print(l)`. I just followed your code where you defined a new variable `r`, but you don't necessarily need to do this. – Joel Sep 11 '18 at 18:35
@Joel His code assigns a new variable, but it also modifies the original dictionaries in place (or would, if it were legal). It seems that the in-place modification is what he's really looking for. – Barmar Sep 11 '18 at 18:40

score 1 · Answer 4 · answered Sep 11 '18 at 18:47

1

[i.update({w:int(k)}) for i in l for w,k in i.items()]

it the second loop is only looping over one key set, so take the two loops with gram of salt :)

answered Sep 11 '18 at 18:47

Jonas Wolff

2,034
1
12
17

score 0 · Answer 5 · answered Sep 11 '18 at 19:00

0

Awesome solution by one of my friends in a chat:

>>> listOfDict = [{1:1338}, {1:1338}, {1:1338}]
>>> y = 1337
>>> value = 1
>>> map(lambda x: x.update({value: y}), listOfDict)
[None, None, None]
>>> listOfDict
[{1: 1337}, {1: 1337}, {1: 1337}]

answered Sep 11 '18 at 19:00

Justin Gardner

605
2
9
17

1

This is essentially equivalent to [my answer](https://stackoverflow.com/a/52282310/5858851), except you replaced the list comp with a call to `map`. BTW, this won't work in python3. In any case, I still say that you should just use a loop. – pault Sep 11 '18 at 20:10

score 0 · Answer 6 · answered Mar 17 '22 at 09:35

You can try the following

l = [{"integer":"1"},{"integer":"2"},{"integer":"3"},{"integer":"4"}]
a = [dict(d, **{'abcd':5}) for d in l]
print(a)
[{'integer': '1', 'abcd': 4}, {'integer': '2', 'abcd': 4}, {'integer': '3', 'abcd': 4}, {'integer': '4', 'abcd': 4}]

Michael Seltenreich · Answer 7 · 2023-03-07T22:58:21.610

0

Use **kwargs to copy keyword arguments from the existing dicts, then overwrite the ones you want like so:

[{**element, 'value_to_overwrite':'new value'} for element in list_of_dicts]

edited Mar 07 '23 at 22:58

answered Mar 06 '23 at 22:51

Michael Seltenreich

3,013
2
29
55

One-liner for updating a value in a list of dictionaries - Python

7 Answers7