Python: Arguments to sorting function

Question

I am trying to sort a python list of strings. I know that I can use the method sorted and set the attribute key to be a function that implements the behavior I need to sort the elements of the dictionary on. My problem is that this method needs an argument.

UPDATE: I want the method to generalize to multiple arguments.

Example: I want to sort a list of strings based on their priorities in 2 dictionaries . So I need to use those priority dictionaries in sorting the list.

I want something like:

sorted(myList, key=sortingAlgorithm(priorityDictionary1, priorityDictionary2), reverse=True)

This can be done if I set the priority dictionaries as global variables and then I will not need to have it as an argument to the sorting algorithm. But, I want to be able to do it without global variables.

If this is not possible, can you please recommend the pythonic way of doing this.

Thanks in advance.

What is sortingAlgorithm() in your code ? The sorted() function has its own sorting algorithm, what it needs is instructions where each item should go, compared to the others. Key= needs to be "a function which takes a list item, and returns a number/string/sortable value". And as the answers come in, there are many approaches to that - but none of them involve passing a sorting algorithm to sort... — TessellatingHeckler
– TessellatingHeckler, Commented May 19, 2016 at 20:21

hilberts_drinking_problem · Accepted Answer · 2016-05-19 20:40:36Z

4

Try key=priorityDictionary.get

get is a method on a dictionary that takes a key and returns a value, if one is found.

Edit: The above solution applies to a case of a single dict. Here is a generalization where values of priorityDict2 are keys for priorityDict1.

And, as Padriac Cunningham points out, you can use a lambda to sort using nested dicts:

output = sorted(myList, key=lambda element: priorityDict1[priorityDict2[element]])

edited May 19, 2016 at 20:40

answered May 19, 2016 at 20:15

hilberts_drinking_problem

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8 Comments

Charles Duffy Over a year ago

Oooh, point -- I was thinking we needed a comparator, but that's not the case here.

Hossam Over a year ago

Thanks for your answer, but can you please check my update?

hilberts_drinking_problem Over a year ago

@Hossam do you want a lexicographic sort on tuples in this case?

Hossam Over a year ago

@YakymPirozhenko No, I want to use the values from both dictionaries and incorporate them in a function that returns the value that should be used for sorting.

Hossam Over a year ago

@YakymPirozhenko Thank you.

|

Ignacio Vazquez-Abrams · Accepted Answer · 2016-05-19 20:15:27Z

1

Use a closure.

def something(foo):
  def somethingsomething(bar):
    return foo(bar)
  return somethingsomething

baz = something(len)
print baz('quux')

answered May 19, 2016 at 20:15

Ignacio Vazquez-Abrams

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Comments

Vedang Mehta · Accepted Answer · 2016-05-19 20:20:59Z

1

Code -

priorities = {'xyz' : 1, 'cde' : 3, 'abc' : 4, 'pqr' : 2}

arr = ['abc', 'cde', 'pqr', 'xyz', 'cde', 'abc']

arr.sort(key = lambda s : priorities[s])

print(arr)

Output -

['xyz', 'pqr', 'cde', 'cde', 'abc', 'abc']

answered May 19, 2016 at 20:20

Vedang Mehta

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Comments

solidpixel · Accepted Answer · 2016-05-19 20:21:16Z

1

Some slightly more standalone examples:

stringList = ["A1", "B3", "C2", "D4"]
priorityDict = {"A1": 1, "B3": 3, "C2": 2, "D4": 4}

# Reference an accessor function
print sorted(stringList, key=priorityDict.get)

# Provide a lambda function (with "free" closure)
print sorted(stringList, key=lambda x: priorityDict[x])

# Provide a normal function (helper definition needs scope visibility 
# on priority dict, but doesn't have to be global - you can define 
# functions in the scope of other functions a little like lambdas)
def myKey(x):
    return priorityDict[x]

print sorted(stringList, key=myKey)

answered May 19, 2016 at 20:21

solidpixel

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Comments

Padraic Cunningham · Accepted Answer · 2016-05-19 20:37:20Z

1

If you want to pass the value from one dict as the key to the next you need a function or a lambda:

l = ["foo", "Foo", "Bar", "bar", "foobar", "Foobar"]

d1 = {"f": 1, "F": 0, "b": 1, "B": 0}
d2 = {1: 10, 0: 20}
print(sorted(l, key=lambda x: d2[d1[x[0]]]))

If a key may not exist you can still use get:

sorted(l, key=lambda x: d2.get(d1.get(x[0], {}), some_default))

Just make sure some_default makes sense and can be compared to the other values.

answered May 19, 2016 at 20:37

Padraic Cunningham

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Charles Duffy · Accepted Answer · 2016-05-19 20:15:32Z

0

def sortingAlgorithm(primaryDictionary):
  def __cmp__(a, b):
    pass # do something with a, b, and primaryDictionary here
  return __cmp__

answered May 19, 2016 at 20:15

Charles Duffy

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