Multi tool use

Sort list of dict by key with different types in Python 3

I have a method that group a list of dict by a key. To do it I found here that I have to use the groupby function but before I have to sort the list. Here is my method right now:

groupby

def group_list_by_key(data, key):
data.sort(key=lambda x: x[key])
result =
for k, v in groupby(data, key=lambda x: x[key]):
result.append(list(v))
return result


This piece of code works only if every key is defined in all the dicts and the values are all of the same type. However, where I use this method I don't know if the key is defined everywhere and if they are of the same type. On Python 2.x I know that exists sorted function with cmp parameter that could do a custom sort but from Python 3.x this isn't possible anymore. Is there a way to make a custom sort? I am thinking about use the classic sort by < and sorting also by typename.

sorted

cmp

<

By now I thought about use the get function and cast to string in the sort like

data.sort(key=lambda x: str(x.get(key)))
...
for k, v in groupby(data, key=lambda x: x.get(key)):


It only overcomes in case of string, numeric and None content but not a generic object and it breaks easily if for example I execute

a = [{'b': 0, 'c': 1}, {'b': '0'}, {'b': 0, 'c': 2}, {'b': 1}, {'c': 3}]
group_list_by_key(a, 'b')


The output is

[[{'b': 0, 'c': 1}], [{'b': '0'}], [{'b': 0, 'c': 2}], [{'b': 1}], [{'c': 3}]]


instead of what I expect should be (order of lists is not a problem)

[[{'b': 0, 'c': 1}, {'b': 0, 'c': 2}], [{'b': '0'}], [{'b': 1}], [{'c': 3}]]


0

'0'

int

2 Answers
2

You can solve your problem by doing something like this

data = [{'b': 0, 'c': 1}, {'b': '0'}, {'b': 0, 'c': 2}, {'b': 1}, {'c': 3}]
key='b'

def f(x):
ret = x.get(key, -1)
return ret if type(ret) == int else -2

result = [list(v) for k, v in groupby(sorted(data, key=f), f)]

# result: [[{'b': '0'}], [{'c': 3}], [{'b': 0, 'c': 1}, {'b': 0, 'c': 2}], [{'b': 1}]]


But if you still need a custom comparison function, you can do it using functools.cmp_to_key

import functools
sorted(x, key=functools.cmp_to_key(custom_cmp_function))


0

'0'

None

f

ret

Thanks to @Sunitha and @njzk2 for pointing out the cmp_to_key function, it did totally what I wanted. So, my grouping now is:

from functools import cmp_to_key
from itertools import groupby

def group_list_by_key(data, key):
def compare_values_types(a, b):
a = a.get(key)
b = b.get(key)
if a.__class__ == b.__class__:
if a < b:
return -1
elif a > b:
return 1
else:
return 0
else:
if a.__class__.__name__ < b.__class__.__name__:
return -1
elif a.__class__.__name__ > b.__class__.__name__:
return 1
else:
return 0
data.sort(key=cmp_to_key(compare_values_types))
return [list(v) for k, v in groupby(data, key=lambda x: x.get(key))]


Calling on the sample list

a = [{'b': 0, 'c': 1}, {'b': '0'}, {'b': 0, 'c': 2}, {'b': 1}, {'c': 3}]
group_list_by_key(a, 'b')


It returns the expected list

[[{'c': 3}], [{'b': 0, 'c': 1}, {'b': 0, 'c': 2}], [{'b': 1}], [{'b': '0'}]]


What I did is to compare in the classic way the keys of the same type, otherwise I simply do a string comparison between the classes' names (using a.__class__.__name__ instead of type(a).__name__, check out to this answer).
Thanks to all!

a.__class__.__name__

type(a).__name__

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