Python: sort specific elements in a list

Question

I have a list of elements I want to sort, but I don't want to sort all of them, only those with a particular state. For example, let's say I have a list of peole:

lst = [Mary, John, Anna, Peter, Laura, Lisa, Steve]

Some of them have a job, let's say [Anna, Lisa, Steve]. I want to sort (ascending) these by the number of hours they work and move them to the beginning of the list, while keeping the rest in the exact same order. So let's say the number of hours they work is the following:

Anna.job.hours   # 10
Lisa.job.hours   # 5
Steve.job.hours  # 8

After the partial sort the list would look like this:

[Lisa, Steve, Anna, Mary, John, Peter, Laura]

Of course I could create two new lists out of the original one, sort the one I want to sort and the put them together again to achieve what I am after:

with_job = [person for person in lst if person.job]
without_job = [person for person in lst if not person.job]

with_job.sort(key=lambda p: p.job.hours)

lst = with_job + without_job

But I wonder if there is a straigthforward, Pythonic way of doing this.

I find your way of doing Pythonic enough. You do not need the second list but other than that.. — Ma0
– Ma0, Commented Mar 17, 2017 at 9:26

Duncan · Accepted Answer · 2017-03-17 09:53:24Z

7

Why not:

lst.sort(key=lambda p: p.job.hours if p.job else 999)

edited Mar 17, 2017 at 9:53

Duncan

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answered Mar 17, 2017 at 9:27

sureshvv

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10 Comments

Elmex80s Over a year ago

Does this solution keep the order for people without job?

dabadaba Over a year ago

Wait, now that I think about it, wouldn't this solution position the people without job at the beginning of the list, and not at the end as it is intended?

Elmex80s Over a year ago

@dabadaba No, because of the minus sign -p.job.hours.

Duncan Over a year ago

@dabadaba, yes you could avoid the sentinal by using a key pair: lst.sort(key=lambda p: (not p.job, p.job.hours if p.job else 0)) would sort the ones with a job first then the others and the 0 could be any constant value.

Duncan Over a year ago

@dabadaba if makes the sort key a tuple The first element is False if p.job is set, True if p.job is not set so all the people with jobs sort first, then the ones without jobs. The second element then acts as a secondary key and sorts in order of the number of hours but only when needed to break the tie on the first element.

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