Updating a list of lists in Python

Question

I have the following list of lists:

MyList = [[130, 10], [131, 15], [132, 1]]

Then i have some inputs. If i get, for example: Data = [130, 3], [135, 10], i need to update the list like this:

MyList = [[130, 3], [131, 15], [132, 1], [135, 10]]

So, if in MyList there is already a sublist where the first element is the same as the first element of a sublist in Data, update the same element. Instead, if there isn't one, append it.

I managed to do this but i was wondering if there was a cleaner solution, as i really don't like the actual one:

Temp = [x[0] for x in MyList]

for x in Data:
    if x[0] not in Temp:
        Sublist = []
        Sublist.append(x[0])
        Sublist.append(x[1])
        MyList.append(Sublist)
    else:
        for y in MyList:
            if x[0] == y[0]:
                x[1] = y[1]

Is there any better way to do this? I feel like this code can be improved, i also don't like editing elements while looping. Any kind of help is welcome!

I think the code you have is the most efficient it can be, sorry for the unhelpfulness — Evorage
– Evorage, Commented Jun 18, 2020 at 20:04
If this is working code that you think could be improved, see Code Review. If not, please clarify the problem with a minimal reproducible example. — jonrsharpe
– jonrsharpe, Commented Jun 18, 2020 at 20:05
A cleaner solution would be to not use a list, probably better of with a dict. — juanpa.arrivillaga
– juanpa.arrivillaga, Commented Jun 18, 2020 at 20:13

juanpa.arrivillaga · Accepted Answer · 2020-06-26 18:40:24Z

4

The cleanest solution would be to use a dict to begin with. So,

>>> data = dict([[130, 10], [131, 15], [132, 1]])
>>> data
{130: 10, 131: 15, 132: 1}
>>> for x, y in [130, 3], [135, 10]:
...     data[x] = y
...
>>> data
{130: 3, 131: 15, 132: 1, 135: 10}

If you really need a list at the end:

>>> list(data.items())
[(130, 3), (131, 15), (132, 1), (135, 10)]

Or even just:

>>> data = {130: 10, 131: 15, 132: 1}
>>> new_data = [130, 3], [135, 10]
>>> data.update(new_data)
>>> data
{130: 3, 131: 15, 132: 1, 135: 10}

edited Jun 26, 2020 at 18:40

answered Jun 18, 2020 at 20:16

juanpa.arrivillaga

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3 Comments

Jack022 Over a year ago

i think that, indeed, using a dict is the best way to go here, rather than a list. Thank you!

juanpa.arrivillaga Over a year ago

@Jack022 yes, one of the most important considerations for clean code is to choose the right data structures that allow you to write the code cleanly in the first place

Jack022 Over a year ago

One last thing: wouldn't i get the same output here using the update() method?

Toma Margishvili · Accepted Answer · 2020-06-18 20:25:34Z

0

If I got your problem right, this is what I have in mind, similar to yours.

myList = [[130, 10], [131, 15], [132, 1]]
data = [130, 3], [135, 10]

if data[0][0] == myList[0][0]:
    myList.pop(0)
    myList.insert(0, data[0])
    myList = myList + list(data[1:])
else:
    myList = myList + list(data)

print(myList)

output:

[[130, 3], [131, 15], [132, 1], [135, 10]]

answered Jun 18, 2020 at 20:25

Toma Margishvili

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Comments

Hamza · Accepted Answer · 2020-06-18 20:29:03Z

0

A cleaner solution (if you insist to use lists instead of the preferred structure dictionary) might be:

MyList = [item for item in MyList if item[0] not in [i[0] for i in Data]]
MyList.extend(Data)

edited Jun 18, 2020 at 20:29

answered Jun 18, 2020 at 20:10

Hamza

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Comments

sahasrara62 · Accepted Answer · 2020-06-18 20:38:32Z

0

As per your this question, what you can do is extend your main list with all the data, and then use dict operation to make a dictionary

MyList = [[130, 10], [131, 15], [132, 1]]


new_data = [[130, 3], [135, 10]]

MyList.extend(new_data)

final_data=  dict(MyList)
print(final_data)
# {130: 3, 131: 15, 132: 1, 135: 10}

edited Jun 18, 2020 at 20:38

answered Jun 18, 2020 at 20:17

sahasrara62

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1 Comment

juanpa.arrivillaga Over a year ago

this isn't a clean way to use a dict at all. There's no need to check if a in my_dic.keys(), and even if there were, you should use if x in my_dict. In any case, that is pointless, because you always do my_dic.update({a:b}), which is also not a clean way of doing that, that should just be my_dic[a] = b

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