I'm working with .csv
files, so I need to sort by specific column
this answer doesn't work:
sorting with two key= arguments
thus using the idea from
How do I sort unicode strings alphabetically in Python?
we have
in python2
import icu # conda install -c conda-forge pyicu
collator = icu.Collator.createInstance(icu.Locale('el_GR.UTF-8'))
parts = [('3', 'ά', 'C'),
('6', 'γ', 'F'),
('5', 'β', 'E'),
('4', 'Ἀ', 'D'),
('2', 'Α', 'B'),
('1', 'α', 'A')]
foo = sorted(parts, key=lambda s: (s[1]), cmp=collator.compare)
for c in foo:
print c[0], c[1].decode('utf-8'), c[2]
with the correct result:
1 α A
2 Α B
4 Ἀ D
3 ά C
5 β E
6 γ F
but in python3
import icu # conda install -c conda-forge pyicu
from functools import cmp_to_key
collator = icu.Collator.createInstance(icu.Locale('el_GR.UTF-8'))
parts = [('3', 'ά', 'C'),
('6', 'γ', 'F'),
('5', 'β', 'E'),
('4', 'Ἀ', 'D'),
('2', 'Α', 'B'),
('1', 'α', 'A')]
foo = sorted(parts, key=lambda s: (s[1], collator.getSortKey))
#foo = sorted(parts, key=lambda s: (s[1], collator.compare))#the same result as collator.getSortKey
for c in foo:
print (c[0], c[1], c[2])
with wrong result:
2 Α B
1 α A
5 β E
6 γ F
4 Ἀ D
3 ά C
I think your calling sorted with the wrong key function.
From docs.python.org:
The value of the key parameter should be a function that takes a single argument and returns a key to use for sorting purposes. This technique is fast because the key function is called exactly once for each input record.
Your key lambda returns a tuple containing the character and a function.
python3 sorts tuples by the first item first, so "Α" is compared to "α" (byte order, not alphabetical), and if they are equal, collator.getSortKey is compared to collator.getSortKey.
I think you want to use the following lambda, I belief it conveys what you want to happen.
foo = sorted(parts, key=lambda s: collator.getSortKey(s[1]))
This should sort alphabetical instead of with byte order.