Reputation: 492
I have a sorted list of two-string tuples. I also have another list of 4 character strings that are in some of those tuples. I would like to group from one of these 4 character strings to the next in the list. It is a bit hard to explain so I will demonstrate.
original_list = [('1321', 01), ('MessageXZY', 02), ('DescriptionSKS', 03), ('S7_6', 04), ('S7_3', 05), ('0A3B', 06), ('MessageZYA', 07),
('DescriptionKAM', 08), ('9K44', 09), ('MessageYAL', 10),
('DescriptionAUS', 11), ('S7_2', 12)]
I have the terms 1321, OA3B and 9K44 saved in another list. I would like to group everything between (and including) these terms into a tuple, like so:
grouped_list = [(('1321', 01), ('MessageXZY', 02), ('DescriptionSKS', 03), ('S7_6', 04), ('S7_3', 05)), (('0A3B', 06), ('MessageZYA', 07),
('DescriptionKAM', 08)), (('9K44', 09), ('MessageYAL', 10),
('DescriptionAUS', 11), ('S7_2', 12))]
If the list that contains my 4 character terms is called code and the list containing the tuples is called original_list, what code would I need to achieve this?
Edit: This is where I have gotten up to:
grouped_list = []
for tuple in original_list:
for string in tuple:
if string in code:
grouped_list = list(zip ##zip that tuple and all consecutive tuples until the next item in code
Upvotes: 2
Views: 627
Reputation: 54223
Coming from a Ruby background, I often felt the need to use something like Enumerable#slice_before
in Python. It basically splits any iterable into chunks. The slice is done before every element for which the predicate is true.
Based on the Rubinius implementation, I ported the code to Python.
def slice_before(iterable, predicate):
chunk = None
for elem in iter(iterable):
if predicate(elem):
if chunk:
yield chunk
chunk = [elem]
else:
if not chunk:
chunk = []
chunk.append(elem)
if chunk:
yield chunk
Here are a few examples:
>>> list(slice_before(range(12), lambda i: i % 3 == 0))
[[0, 1, 2], [3, 4, 5], [6, 7, 8], [9, 10, 11]]
>>> list(slice_before('LetsTestSliceBefore', str.isupper))
[['L', 'e', 't', 's'], ['T', 'e', 's', 't'], ['S', 'l', 'i', 'c', 'e'], ['B', 'e', 'f', 'o', 'r', 'e']]
You just need to initialize your data. Note that code_list
could be a set for faster lookup:
original_list = [('1321', '01'), ('MessageXZY', '02'), ('DescriptionSKS', '03'), ('S7_6', '04'), ('S7_3', '05'), ('0A3B', '06'), ('MessageZYA', '07'), ('DescriptionKAM', '08'), ('9K44', '09'), ('MessageYAL', '10'),
('DescriptionAUS', '11'), ('S7_2', '12')]
code_list = {'1321', '0A3B','9K44'}
The required code for your problem becomes a one-liner with slice_before
:
print(list(slice_before(original_list, lambda x_y: x_y[0] in code_list)))
# [[('1321', '01'), ('MessageXZY', '02'), ('DescriptionSKS', '03'), ('S7_6', '04'), ('S7_3', '05')], [('0A3B', '06'), ('MessageZYA', '07'), ('DescriptionKAM', '08')], [('9K44', '09'), ('MessageYAL', '10'), ('DescriptionAUS', '11'), ('S7_2', '12')]]
Upvotes: 3
Reputation: 2805
I am assuming that you have list of codes with which you want to split on. According to that see if this code works for you.
original_list = [('1321', '01'), ('MessageXZY', '02'), ('DescriptionSKS', '03'), ('S7_6', '04'), ('S7_3', '05'), ('0A3B', '06'), ('MessageZYA', '07'), ('DescriptionKAM', '08'), ('9K44', '09'), ('MessageYAL', '10'),
('DescriptionAUS', '11'), ('S7_2', '12')]
code_list = ['1321', '0A3B','9K44']
grouped_tuples = []
for entry in original_list:
if entry[0] in code_list:
new_tuple = []
new_tuple.append(entry)
for i in range(original_list.index(entry)+1, len(original_list)):
if(original_list[i][0] not in code_list):
new_tuple.append(original_list[i])
else:
break
grouped_tuples.append(tuple(new_tuple))
print grouped_tuples
Upvotes: 2