How to create List of List in Python

Question

I have a below list.

['A_1', 'A_2', 'A_3', 'A_4', 'B_1','B_2', 'C_1', 'C_2', 'C_3']

I need the output in below format.

[['A_1', 'A_2', 'A_3', 'A_4'], ['B_1', 'B_2'], ['C_1', 'C_2', 'C_3']]

I am very new to Python, is there any way to achieve the above output that it would be really helpful.

• Park: Does it mean that you want to group by the first letter?
• bereal: What if it goes like ['A_1', 'B_1', 'A_2']?
• bereal: @SharimIqbal you shouldn't have edited the code like this without confirming from the OP. It could be that they meant variable names.

Answer 1

This can be achieved directly with a dictionary combined with a loop:

1. store each type of data in the form of "letter_", e.g. {"A_": ["A_1", "A_2"]}
2. Loop through each data in the list, if the type is not in the dictionary, create a list with the prefix type of this data, otherwise add it directly to the list of this type
3. Finally, directly convert the values of the dictionary into a list to achieve the requirement

category = {}
lst = ['A_1', 'A_2', 'A_3', 'A_4', 'B_1','B_2', 'C_1', 'C_2', 'C_3']
for item in lst:
 prefix = item.split("_")[0] + "_" # Example of prefix values: A_, B_,...
 if prefix not in category:
 category[prefix] = [item] 
 else:
 category[prefix].append(item)
print(list(category.values()))
# [['A_1', 'A_2', 'A_3', 'A_4'], ['B_1', 'B_2'], ['C_1', 'C_2', 'C_3']]

• Titouan L: Using in on a dictionnary search in the keys ? Or is it because theres no str value that contains the prefix ?

Answer 2

I provided a easily understandable way (but requires more lines than other advanced thecniques)

You can use dictionary, in particular, I used defaultdict here to set default value as a list ([]).

I grouped the list by first letter in a for loop, as follows:

from collections import defaultdict
l = ['A_1', 'A_2', 'A_3', 'A_4', 'B_1', 'B_2', 'C_1', 'C_2', 'C_3']
group = defaultdict(list)
for item in l:
 first_letter = item[0] # saving items by first letter
 group[first_letter].append(item)
result = list(group.values())
print(result)
# [['A_1', 'A_2', 'A_3', 'A_4'], ['B_1', 'B_2'], ['C_1', 'C_2', 'C_3']]

Answer 3

You can use itertools.groupby for that. This assumes that the input is sorted, as in your example.

from itertools import groupby
lst = ['A_1', 'A_2', 'A_3', 'A_4', 'B_1', 'B_2', 'C_1', 'C_2', 'C_3']
output = [list(g) for _, g in groupby(lst, key=lambda x: x[0])]
print(output) # [['A_1', 'A_2', 'A_3', 'A_4'], ['B_1', 'B_2'], ['C_1', 'C_2', 'C_3']]

If for some reason you don't want to use import but only use built-ins,

lst = ['A_1', 'A_2', 'A_3', 'A_4', 'B_1', 'B_2', 'C_1', 'C_2', 'C_3']
output = {}
for x in lst:
 if x[0] in output:
 output[x[0]].append(x)
 else:
 output[x[0]] = [x]
print(list(output.values()))

Note that with an input list of ['A_1', 'B_1', 'A_2'] the two approaches will result in different outputs.

Answer 4

Refer below code:

input = ['A_1', 'A_2', 'A_3', 'A_4', 'B_1', 'B_2', 'C_1', 'C_2', 'C_3']

Use below logic:

ls_a,ls_b,ls_c = [],[],[] # Declaring empty lists
for i in input:
if 'A' in i:
 ls_a.append(i)
if 'B' in i:
 ls_b.append(i)
if 'C' in i:
 ls_c.append(i)
output = [ls_a,ls_b,ls_c]
print(output)