AishwaryaKulkarni AishwaryaKulkarni - 1 month ago 10
Python Question

Conditional statment regarding various regex and length of a list in python

I have following list :

['E12.2', 'E16.1', 'E15.1']
['E10.1', 'I11.2', 'I10.1_27353757']
['E16.1', 'E18.1', 'E17.3']
['E1.8', 'I12.1_111682336', 'I12.1_111682195']
['E55.1', 'E57.1', 'E56.1','E88.1']
['U22.3', 'U22.6_13735517', 'U23.1']


and I want to put a condition to filter out the lists that have a) length equal to 3 b) not containing '_' c) not containing alphabet 'U'
I am trying to implement in one line, how do I do that? I have following condition working and I know you can use regex module for matching regex in lists but can I do all the conditions in single line?

if(len(fin_list) == 3)

Answer

This is one possible way:

lists = [['E12.2', 'E16.1', 'E15.1'],
         ['E10.1', 'I11.2', 'I10.1_27353757'],
         ['E16.1', 'E18.1', 'E17.3'],
         ['E1.8', 'I12.1_111682336', 'I12.1_111682195'],
         ['E55.1', 'E57.1', 'E56.1','E88.1'],
         ['U22.3', 'U22.6_13735517', 'U23.1']]

for lst in lists:
    if len(lst) != 3 and not any('_' in item or 'U' in item for item in lst):
        print(lst)

# Output:
# ['E55.1', 'E57.1', 'E56.1', 'E88.1']

The interesting bit here is the use of any over a generator expression. To break it down, this iterates over each item in lst and applies a test to see if _ or U are in it. That list comprehension results in True/False for each item in the list. any then looks for the first True. If it finds one, it immediately returns True. If it doesn't find one, it returns False.

Comments