python sort list of strings using regex with varying number of pattern matches

Question

Just starting to work on python and having difficulties sorting string list by multiple/varying number of matches. Basically, given a list of strings, I need to split each string by a given regex (user provided) then sort by given list of keys (locations). The key can either be single integer or a list in the order in which they should be sorted. For example:

regex = r'.(FF|TT|SS)_([-.\d]+v)_([-.\d]+c)_(FF|TT|SS).'

key = [2,1,3]

Would sort the list of strings by location2, location1, location3.

I have the following that works for a fixed number of locations/keys, but can't figure out how to get it to work of varying number of 'keys':

import re

strlist = ["synopsys_SS_2v_-40c_SS.lib","synopsys_SS_1v_-40c_SS.lib","synopsys_SS_2v_-40c_TT.lib","synopsys_FF_3v_-40c_FF.lib", "synopsys_TT_4v_125c_TT.lib", "synopsys_TT_1v_-40c_TT.lib"]
regex = r'.*(FF|TT|SS)_([-\.\d]+v)_([-\.\d]+c)_(FF|TT|SS).*'
key = [2,1,3]

sfids_single = sorted(strlist, key=lambda name: ( 
  re.findall(regex,name)[0][key[0]], 
  re.findall(regex,name)[0][key[1]],
  re.findall(regex,name)[0][key[2]]))

Tried the following but it does not seem to work:

fids_single = sorted(strlist, key=lambda name: (re.findall(regex,name)[0][i] for i in key))

Also tried (w/o success):

for i in key:
  strlist.sort(key=lambda name: re.findall(regex,name)[0][key[i]])

Expected result:

['synopsys_SS_1v_-40c_SS.lib', 'synopsys_TT_1v_-40c_TT.lib', 'synopsys_SS_2v_-40c_SS.lib', 'synopsys_SS_2v_-40c_TT.lib', 'synopsys_FF_3v_-40c_FF.lib', 'synopsys_TT_4v_125c_TT.lib']

Am I on the wrong track completely? Any guidance is greatly appreciated.

Are you having trouble splitting the strings or sorting the resultant list? — wwii
– wwii, Commented Jun 28, 2017 at 18:55
I'm not having trouble splitting, just sorting by a given set of keys. This is not a homework question, it's part of code I'm writing for work. — Luca
– Luca, Commented Jun 28, 2017 at 20:38

wwii · Accepted Answer · 2017-06-28 22:41:32Z

2

Write a key function that will return the relevent portions of each string, in order of precedence, and use that function for the sort key.

one = ["synopsys_SS_2v_-40c_SS.lib","synopsys_SS_1v_-40c_SS.lib",
       "synopsys_SS_2v_-40c_TT.lib","synopsys_FF_3v_-40c_FF.lib",
       "synopsys_TT_4v_125c_TT.lib", "synopsys_TT_1v_-40c_TT.lib"]    

expected = ['synopsys_SS_1v_-40c_SS.lib', 'synopsys_TT_1v_-40c_TT.lib',
            'synopsys_SS_2v_-40c_SS.lib', 'synopsys_SS_2v_-40c_TT.lib',
            'synopsys_FF_3v_-40c_FF.lib', 'synopsys_TT_4v_125c_TT.lib']

Using your regular expression to split the string;

import operator, re
pattern = r'.*(FF|TT|SS)_([-\.\d]+v)_([-\.\d]+c)_(FF|TT|SS).*'
rx = re.compile(pattern)
seq = [2,1,3]
def key(item, seq = seq):
    seq = operator.itemgetter(*seq)
    a, b, c, d = rx.findall(item)
    return seq([a, b, c, d])


one.sort(key = key)
assert one == expected

The key function can be written without using a regular expression which may make it a bit less complicated.

def key(item, seq = seq):
    seq = operator.itemgetter(*seq)
    _, a, b, c, d = item.split('_')
    d, _ = d.split('.')
    print a, b, c, d
    return seq([a, b, c, d])

You may want to use names that are more descriptive than a, b, c, d. It relies on the strings having the same pattern.

edited Jun 28, 2017 at 22:41

answered Jun 28, 2017 at 19:03

wwii

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1 Comment

a_guest Over a year ago

@Kidneys Your first attempt with for i in key was almost right, however the expression you used returns a generator. For a sorting key you want to use a tuple instead. So by using lambda name: tuple(... for i in key) it should work.

Luca · Accepted Answer · 2017-06-29 02:25:38Z

1

Many thanks to @a_guest for providing the missing piece of the puzzle. Here's the working solution:

fids_single = sorted(strlist, key=lambda name: tuple(re.findall(regex,name)[0][i] for i in key))

answered Jun 29, 2017 at 2:25

Luca

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