Lambda sorting in Python 3

Question

I was working on python2 for my project.Now I want to migrate from python2 to python3. I am getting syntax error while writing the for loop code snippet given below.I'm bit confused where I went wrong.

Code :

for k, val in sorted(datasource_columns.iteritems(), key=lambda(
                            k, v): sort_order.index(k)):
                        # columns.add_column(key, sorted(val))
                        columns.add_column(k, val)

where , sort_order = ['Name', 'Data Type', 'Type', 'Role']

sort_order.index(k) sounds like a very inefficient sort key... — Jean-François Fabre
– Jean-François Fabre ♦, Commented Dec 11, 2017 at 9:15

Guillaume · Accepted Answer · 2017-12-11 09:42:34Z

1

As already said, iteritems() will be a problem, but you mention a syntax error, which comes from the lambda declaration with parenthesis:

Change:

key=lambda(k, v): sort_order.index(k)

To:

key=lambda k, v: sort_order.index(k)

But you may as well write this more efficient code:

for key in sort_order:
    val = datasource_columns.get(key)
    if val is not None: # to avoid getting None values in your columns
        columns.add_column(key, val)

In the future, please post the actual stacktrace whenever you get an exception, this makes it a lot easier to understand any issue.

edited Dec 11, 2017 at 9:42

answered Dec 11, 2017 at 9:15

Guillaume

6,1293 gold badges28 silver badges48 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

PM 2Ring Over a year ago

Your more efficient code has a slight problem: there may be keys in sort_order that aren't present in datasource_columns. Your code will create a (key, None) item for such keys. That may not be desirable.

Guillaume Over a year ago

Indeed, but that can be easily fixed with an if val is not None: statement, if required.

PM 2Ring Over a year ago

Certainly! So you should add that info to your answer. And it's probably worthwhile mentioning why the original key function is so horrible.

eljur · Accepted Answer · 2021-06-05 03:50:51Z

1

    key=lambda k, v: sort_order.index(k)

does not work. TypeError: () missing 1 required positional argument: v

The needed change for Python 3 would be:


    test_list.sort(key = lambda args : sort_order.index(args[0]))

edited Jun 5, 2021 at 3:50

answered Aug 19, 2019 at 15:31

eljur

213 bronze badges

Comments

Stop harming Monica · Accepted Answer · 2017-12-11 09:19:17Z

0

Not sure what changed exactly but lambda (k, v): sort_order.index(k) is invalid syntax in python3. You can use lambda k, v: sort_order.index(k) instead.

answered Dec 11, 2017 at 9:19

Stop harming Monica

12.7k1 gold badge40 silver badges63 bronze badges

Collectives™ on Stack Overflow

Lambda sorting in Python 3

3 Answers 3

3 Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

3 Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related