Print matching item in list python

Question

I am just starting to learn python, I have a simple question.

l=['aa123','aa122','aa124','bb125','bb180']

#form above list i want to print a result as following:

Group 1
aa123
aa122
aa124

Group 2
bb125
bb180

and I want that if I update the list with 'cc000', it will print also a 'Group 3'

Thanks and regards, Giovanni

Answer 1

You can use a dictionary to better group your values:

import string
from collections import defaultdict
l=['aa123','aa122','aa124','bb125','bb180']
key = {a:b+1 for a, b in zip(string.ascii_lowercase, range(26))}
d = defaultdict(list)
for val in l:
   d[key[val[0]]].append(val)
for a, b in d.items():
   print("Group {}".format(a), b)

Output:

Group 1 ['aa123', 'aa122', 'aa124']
Group 2 ['bb125', 'bb180']

Or, using groupby in list comprehension:

import itertools
final_vals = {"Group {}".format(a):list(b) for a, b in itertools.groupby(sorted(l, key=lambda x:x[0]), key=lambda x:x[0])}
print(final_vals)

Output:

{'Group b': ['bb125', 'bb180'], 'Group a': ['aa123', 'aa122', 'aa124']}

Answer 2

i = ['bb334', 'aa341', 'cc555', 'aa342', 'aa337']
x = []

# Creating new ordered list
for j in range(0, 26):
    for k in i:
        if ord(k[0]) - 97 == j:
        x.append(k)

# Printing out list by groups
current_ord = ord(x[0][0])
counter = 1

print 'Group ' + str(counter) + ':'

for j in x:
    if ord(j[0]) != current_ord:
    current_ord = ord(j[0])
    counter += 1
        print '\nGroup ' + str(counter) + ':'

    print j

Kind of slow since you're doing an 26 * len(i) iterations to create the sorted list. I'm not sure if this answers your question fully... what happens when an item like 'ab111' is in the list? Will it ignore this?